multiplying probabilities homework answers

• school Campus Bookshelves
• menu_book Bookshelves
• perm_media Learning Objects
• login Login
• how_to_reg Request Instructor Account
• hub Instructor Commons

Margin Size

• Download Page (PDF)
• Download Full Book (PDF)
• Periodic Table
• Physics Constants
• Scientific Calculator
• Reference & Cite
• Tools expand_more
• Readability

selected template will load here

This action is not available.

7.10: Conditional Probability and the Multiplication Rule

• Last updated
• Save as PDF
• Page ID 129601

\( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)

\( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)

\( \newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\)

( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\)

\( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)

\( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\)

\( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)

\( \newcommand{\Span}{\mathrm{span}}\)

\( \newcommand{\id}{\mathrm{id}}\)

\( \newcommand{\kernel}{\mathrm{null}\,}\)

\( \newcommand{\range}{\mathrm{range}\,}\)

\( \newcommand{\RealPart}{\mathrm{Re}}\)

\( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)

\( \newcommand{\Argument}{\mathrm{Arg}}\)

\( \newcommand{\norm}[1]{\| #1 \|}\)

\( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\AA}{\unicode[.8,0]{x212B}}\)

\( \newcommand{\vectorA}[1]{\vec{#1}}      % arrow\)

\( \newcommand{\vectorAt}[1]{\vec{\text{#1}}}      % arrow\)

\( \newcommand{\vectorB}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)

\( \newcommand{\vectorC}[1]{\textbf{#1}} \)

\( \newcommand{\vectorD}[1]{\overrightarrow{#1}} \)

\( \newcommand{\vectorDt}[1]{\overrightarrow{\text{#1}}} \)

\( \newcommand{\vectE}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{\mathbf {#1}}}} \)

Learning Objectives

After completing this section, you should be able to:

• Calculate conditional probabilities.
• Apply the Multiplication Rule for Probability to compute probabilities.

Back in Example 7.18, we constructed the following table (Figure 7.38) to help us find the probabilities associated with rolling two standard 6-sided dice:

For example, 3 of these 36 equally likely outcomes correspond to rolling a sum of 10, so the probability of rolling a 10 is 3 36 = 1 12 3 36 = 1 12 . However, if you choose to roll the dice one at a time, the probability of rolling a 10 will change after the first die comes to rest. For example, if the first die shows a 5, then the probability of rolling a sum of 10 has jumped to 1 6 1 6 —the event will occur if the second die also shows a 5, which is 1 of 6 equally likely outcomes for the second die. If instead the first die shows a 3, then the probability of rolling a sum of 10 drops to 0—there are no outcomes for the second die that will give us a sum of 10.

Understanding how probabilities can shift as we learn new information is critical in the analysis of our second type of compound events: those built with “and.” This section will explain how to compute probabilities of those compound events.

Conditional Probabilities

When we analyze experiments with multiple stages, we often update the probabilities of the possible final outcomes or the later stages of the experiment based on the results of one or more of the initial stages. These updated probabilities are called conditional probabilities .

In other words, if O O is a possible outcome of the first stage in a multistage experiment, then the probability of an event E E conditional on O O (denoted P ( E | O ) P ( E | O ) , read “the probability of E E given O O ”) is the updated probability of E E under the assumption that O O occurred.

In the example that opened this section, we might consider rolling two dice as a multistage experiment: rolling one, then the other. If we define E E to be the event “roll a sum of 10,” O O to be the event “first die shows 5,” and Q Q to be the event “first die shows 3,” then we computed P ( E ) = 1 12 P ( E ) = 1 12 , P ( E | O ) = 1 6 P ( E | O ) = 1 6 , and P ( E | Q ) = 0 P ( E | Q ) = 0 .

Example 7.31

Computing conditional probabilities.

• April is playing a coin-flipping game with Ben. She will flip a coin 3 times. If the coin lands on heads more than tails, April wins; if it lands on tails more than heads, Ben wins. Let A A be the event “April wins,” H H be “first flip is heads,” and T T be “first flip is tails.” Compute P ( A ) P ( A ) , P ( A | H ) P ( A | H ) , and P ( A | T ) P ( A | T ) .
• You are about to draw 2 cards without replacement from a deck containing only these 10 cards: A ♡ A ♡ , A ♠ A ♠ , A ♣ A ♣ , A ♢ A ♢ , K ♠ K ♠ , K ♣ K ♣ , Q ♡ Q ♡ , Q ♠ Q ♠ , J ♡ J ♡ , J ♠ J ♠ . We’ll define the following events: F F is “both cards are the same rank,” A A is “first card is an ace,” and K K is “first card is a king.” Compute P ( F | A ) P ( F | A ) and P ( F | K ) P ( F | K ) .
• Jim’s sock drawer contains 5 black socks and 3 blue socks. To avoid waking his partner, Jim doesn’t want to turn the lights on, so he puts on 2 socks at random. Let M M be the event “Jim’s 2 socks match,” let K K be the event “the sock on Jim’s left foot is black,” and let L L be the event “the sock on Jim’s left foot is blue.” Compute P ( M ) P ( M ) , P ( M | K ) P ( M | K ) , and P ( M | L ) P ( M | L ) .

Step 2. Now, let’s compute P ( A | H ) P ( A | H ) . We are assuming the result of the first flip is heads. That leaves us with 4 possible outcomes: HHH, HHT, HTH, and HTT. Of those, April wins 3 (HHH, HHT, HTH) and loses one (HTT). So, P ( A | H ) = 3 4 P ( A | H ) = 3 4 .

Step 3. If the result of the first flip is instead tails, the 4 possible outcomes are THH, THT, TTH, and TTT. Of those, April wins 1 (THH) and loses 3 (THT, TTH, TTT). So, P ( A | T ) = 1 4 P ( A | T ) = 1 4 .

Step 2. If the event K K happens instead, then the first card drawn is a king. That leaves 4 aces, 1 king, 2 queens, and 2 jacks in the deck. Under the assumption that the first card is a king, the event F F will occur only if the second card is also a king. Since only one of the remaining 9 cards is a king, we have P ( F | K ) = 1 9 P ( F | K ) = 1 9 .

Step 2. If the sock on Jim’s left foot is black (i.e., K K occurred), then there are 4 remaining black socks of the 7 in the drawer. So, P ( M | K ) = 4 7 P ( M | K ) = 4 7 .

Step 3. If the sock on Jim’s left foot is blue ( L L occurred), then there are 2 blue socks among the 7 remaining in the drawer. So, P ( M | L ) = 2 7 P ( M | L ) = 2 7 .

Your Turn 7.31

In Tree Diagrams, Tables, and Outcomes, we introduced the concept of dependence between stages of a multistage experiment. We stated at the time that two stages were dependent if the result of one stage affects the other stage. We explained that dependence in terms of the sample space, but sometimes that dependence can be a little more subtle; it’s more properly understood in terms of conditional probabilities. Two stages of an experiment are dependent if P ( E | F ) ≠ P ( E | F ′ ) P ( E | F ) ≠ P ( E | F ′ ) for some outcome of the second stage E E and outcome of the first stage F F .

Protecting Bombers in World War II

In his book How Not to Be Wrong , Jordan Ellenberg recounts this anecdote: During World War II, the American military wanted to add additional armor plating to bomber aircraft, in order to reduce the chances that they get shot down. So, they collected data on planes after returning from missions. The data showed that the fuselage, wings, and fuel system had many more bullet holes (per unit area) than the engine compartments, so the military brass wanted to add additional armor to the parts of the plane that were hit most often. Luckily, before they added the armor to the planes, they asked for a second opinion. Abraham Wald, a Jewish mathematician who had fled the rising Nazi regime, pointed out that it was far more important that the armor plating be added to areas where there were fewer bullet holes. Why? The planes they were studying had already completed their missions, so the military was essentially looking at conditional probabilities: the probability of suffering a bullet strike, given that the plane made it back safely. More bullet holes in an area on the plane indicated that was a region that wasn’t as important for the plane’s survival!

Compound Events Using “And” and the Multiplication Rule

For multistage experiments, the outcomes of the experiment as a whole are often stated in terms of the outcomes of the individual stages. Commonly, those statements are joined with “and.” For example, in the sock drawer example just above, one outcome might be “the left sock is black and the right sock is blue.” As with “or” compound events, these probabilities can be computed with basic arithmetic.

Multiplication Rule for Probability: If E E and F F are events associated with the first and second stages of an experiment, then P ( E and ⁢ F ) = P ( E ) × P ( F | E ) P ( E and ⁢ F ) = P ( E ) × P ( F | E ) .

In The Addition Rule for Probability, we considered probabilities of events connected with “and” in the statement of the Inclusion/Exclusion Principle. These two scenarios are different; in the statement of the Inclusion/Exclusion Principle, the events connected with “and” are both events associated with the same single-stage experiment (or the same stage of a multistage experiment). In the Multiplication Rule, we’re looking at events associated with different stages of a multistage experiment.

Example 7.32

Using the multiplication rule for probability.

You are president of a club with 10 members: 4 seniors, 3 juniors, 2 sophomores, and 1 first-year. You need to choose 2 members to represent the club on 2 college committees. The first person selected will be on the Club Awards Committee and the second will be on the New Club Orientation Committee. The same person cannot be selected for both. You decide to select these representatives at random.

• What is the probability that a senior is chosen for both positions?
• What is the probability that a junior is chosen first and a sophomore is chosen second?
• What is the probability that a sophomore is chosen first and a senior is chosen second?
• We need the probability that a senior is chosen first and a senior is chosen second. These are two stages of a multistage experiment, so we’ll apply the Multiplication Rule for Probability: P ( senior chosen first and senior chosen second ) = P ( senior chosen first ) × P ( senior chosen second | senior chosen first ) . P ( senior chosen first and senior chosen second ) = P ( senior chosen first ) × P ( senior chosen second | senior chosen first ) . Since there are 4 seniors among the 10 members, P ( senior chosen first ) = 4 10 = 2 5 P ( senior chosen first ) = 4 10 = 2 5 . Next, assuming a senior is chosen first, there are 3 seniors among the 9 remaining members. So, P ( senior chosen second | senior chosen first ) = 3 9 = 1 3 P ( senior chosen second | senior chosen first ) = 3 9 = 1 3 . Putting this all together, we get P ( senior chosen first and senior chosen second ) = 2 5 × 1 3 = 2 15 P ( senior chosen first and senior chosen second ) = 2 5 × 1 3 = 2 15 .
• There are 3 juniors among the 10 members, so P ( junior chosen first ) = 3 10 P ( junior chosen first ) = 3 10 . Assuming a junior is chosen first, there are 2 sophomores among the remaining 9 members, so P ( sophomore chosen second | junior chosen first ) = 2 9 P ( sophomore chosen second | junior chosen first ) = 2 9 . Thus, using the Multiplication Rule for Probability, we have P ( junior chosen first and sophomore chosen second ) = 3 10 × 2 9 = 1 15 P ( junior chosen first and sophomore chosen second ) = 3 10 × 2 9 = 1 15 .
• The probability that a sophomore is chosen first is 2 10 = 1 5 2 10 = 1 5 , and the probability that a senior is chosen second given that a sophomore was chosen first is 4 9 4 9 . Thus, using the Multiplication Rule for Probability, we have: P ( sophomore chosen first and senior chosen second ) = 1 5 × 4 9 = 4 45 P ( sophomore chosen first and senior chosen second ) = 1 5 × 4 9 = 4 45 .

Your Turn 7.32

Work it out, the birthday problem.

One of the most famous problems in probability theory is the Birthday Problem, which has to do with shared birthdays in a large group. To make the analysis easier, we’ll ignore leap days, and assume that the probability of being born on any given date is 1 365 1 365 . Now, if you have 366 people in a room, we’re guaranteed to have at least one pair of people who share a single birthday. Imagine filling the room by first admitting someone born on January 1, then someone born on January 2, and so on… The 365th person admitted would be born on December 31. If you add one more person to the room, that person’s birthday would have to match someone else’s.

Let’s look at the other end of the spectrum. If you choose two people at random, what is the probability that they share a birthday? As with many probability questions, this is best addressed by find out the probability that they do not share a birthday. The first person’s birthday can be anything (probability 1), and the second person’s birthday can be anything other than the first person’s birthday (probability 364 365 364 365 ). The probability that they have different birthdays is 1 × 364 365 = 364 365 1 × 364 365 = 364 365 . So, the probability that they share a birthday is 1 − 364 365 = 1 365 1 − 364 365 = 1 365 .

What if we have three people? The probability that they all have different birthdays can be obtained by extending our previous calculation: The probability that two people have different birthdays is 364 365 364 365 , so if we add a third to the mix, the probability that they have a different birthday from the other two is 363 365 363 365 . So, the probability that all three have different birthdays is 364 365 × 363 365 ≈ 0.9918 364 365 × 363 365 ≈ 0.9918 , and thus the probability that there’s a shared birthday in the group is 1 − 0.9918 ≈ 0.0082 1 − 0.9918 ≈ 0.0082 .

The big question is this: How many people do we need in the room to have the probability of a shared birthday greater than 1 2 1 2 ? Make a guess, then with a partner keep adding hypothetical people to the group and computing probabilities until you get there!

It is often useful to combine the rules we’ve seen so far with the techniques we used for finding sample spaces. In particular, trees can be helpful when we want to identify the probabilities of every possible outcome in a multistage experiment. The next example will illustrate this.

Example 7.33

Using tree diagrams to help find probabilities.

The board game Clue uses a deck of 21 cards: 6 suspects, 6 weapons, and 9 rooms. Suppose you are about to draw 2 cards from this deck. There are 6 possible outcomes for the draw: 2 suspects, 2 weapons, 2 rooms, 1 suspect and 1 weapon, 1 suspect and 1 room, or 1 weapon and 1 room. What are the probabilities for each of these outcomes?

Step 1: Let’s start by building a tree diagram that illustrates both stages of this experiment. Let’s use S, W, and R to indicate drawing a suspect, weapon, and room, respectively (Figure 7.39).

Step 2: We want to start computing probabilities, starting with the first stage. The probability that the first card is a suspect is 6 21 = 2 7 6 21 = 2 7 . The probability that the first card is a weapon is the same: 2 7 2 7 . Finally, the probability that the first card is a room is 9 21 = 3 7 9 21 = 3 7 .

Step 3: Let’s incorporate those probabilities into our tree: label the edges going into each of the nodes representing the first-stage outcomes with the corresponding probabilities (Figure 7.40).

Note that the sum of the probabilities coming out of the initial node is 1; this should always be the case for the probabilities coming out of any node!

Step 4: Let’s look at the case where the first card is a suspect. There are 3 edges emanating from that node (leading to the outcomes SS, SW, and SR). We’ll label those edges with the appropriate conditional probabilities, under the assumption that the first card is a suspect. First, there are 5 remaining suspect cards among the 20 left in the deck, so P ( second is suspect | first is suspect ) = 5 20 = 1 4 P ( second is suspect | first is suspect ) = 5 20 = 1 4 . Using similar reasoning, we can compute P ( second is weapon | first is suspect ) = 6 20 = 3 10 P ( second is weapon | first is suspect ) = 6 20 = 3 10 and P ( second is room | first is suspect ) = 9 20 P ( second is room | first is suspect ) = 9 20 .

Step 5: Checking our work, we see that the sum of these 3 probabilities is again equal to 1. Let’s add those to our tree (Figure 7.41).

Step 6: Let’s continue filling in the conditional probabilities at the other nodes, always checking to make sure the sum of the probabilities coming out of any node is equal to 1 (Figure 7.42).

Step 7: We can compute the probability of landing on any final node by multiplying the probabilities along the path we would take to get there. For example, the probability of drawing a suspect first and a weapon second (i.e., ending up on the node labeled “SW”) is 2 7 × 3 10 = 3 35 Figure 7.43.

Step 8: Let’s fill in the rest of the probabilities (Figure 7.44).

Step 9: A helpful feature of tree diagrams is that the final outcomes are always mutually exclusive, so the Addition Rule can be directly applied. For example, the probability of drawing one suspect and one room (in any order) would be P ( S R ) + P ( R S ) = 9 70 + 9 70 = 9 35 P ( S R ) + P ( R S ) = 9 70 + 9 70 = 9 35 . We can find the probabilities of the other outcomes in a similar fashion, as shown in the following table:

Checking once again, the sum of these 6 probabilities is 1, as expected.

Your Turn 7.33

The monty hall problem.

On the original version of the game show Let’s Make a Deal , originally hosted by Monty Hall and now hosted by Wayne Brady, one contestant was chosen to play a game for the grand prize of the day (often a car). Here’s how it worked: On the stage were three areas concealed by numbered curtains. The car was hidden behind one of the curtains; the other two curtains hid worthless prizes (called “Zonks” on the show). The contestant would guess which curtain concealed the car. To build tension, Monty would then reveal what was behind one of the other curtains, which was always one of the Zonks (Since Monty knew where the car was hidden, he always had at least one Zonk curtain that hadn’t been chosen that he could reveal). Monty then turned to the contestant and asked: “Do you want to stick with your original choice, or do you want to switch your choice to the other curtain?” What should the contestant do? Does it matter?

With a partner or in a small group, simulate this game. You can do that with a small candy (the prize) hidden under one of three cups, or with three playing cards (just decide ahead of time which card represents the “Grand Prize”). One person plays the host, who knows where the prize is hidden. Another person plays the contestant and tries to guess where the prize is hidden. After the guess is made, the host should reveal a losing option that wasn’t chosen by the contestant. The contestant then has the option to stick with the original choice or switch to the other, unrevealed option. Play about 20 rounds, taking turns in each role and making sure that both contestant strategies (stick or switch) are used equally often. After each round, make a note of whether the contestant chose “stick” or “switch” and whether the contestant won or lost. Find the empirical probability of winning under each strategy. Then, see if you can use tree diagrams to verify your findings.

Check Your Understanding

Section 7.9 exercises.

• /**/P\left({\text{tile shows A}}\right)/**/
• /**/P\left(\text{tile shows A}\,|\,\text{tile shows a vowel}\right)/**/
• /**/P({\text{tile shows a vowel}})/**/
• /**/P\left({\text{tile shows a vowel}}\,|\,{\text{tile shows a letter that comes after M alphabetically}}\right)/**/

In the following exercises deal with the game “Punch a Bunch,” which appears on the TV game show The Price Is Right . In this game, contestants have a chance to punch through up to 4 paper circles on a board; behind each circle is a card with a dollar amount printed on it. There are 50 of these circles; the dollar amounts are given in this table:

Contestants are shown their selected dollar amounts one at a time, in the order selected. After each is revealed, the contestant is given the option of taking that amount of money or throwing it away in favor of the next amount. (You can watch the game being played in the video Playing “Punch a Bunch.” ) Jeremy is playing “Punch a Bunch” and gets 2 punches.

Math Worksheets Land

Math Worksheets For All Ages

• Math Topics
• Grade Levels

Multiplication Rule of Probability Worksheets

What is the Multiplication Rule of Probability? If you remember, we discussed probability and its rules quite in-depth. With the concept of probability taking the center position in statistics, learning about its rules are as important as learning about the concept. If the probability is something you find difficult and fear to deal with, we tell you that if you learn about its rules, you will get a better grasp at understanding probability. The multiplication rule states that: P(A and B) = P(A) * P(B|A) or P(B) * P(A|B). In the above rule, if A and B are two independent events, the formula can be shrunk to; P(A and B) = P(A) * P(B). Independent events refer to the events whose outcome is not affected by the occurrence or happening of another event. For instance, if two coins are flipped together, the second flip might have a chance of 0.50% of landing heads, regardless of the outcome in the first flip. What's the probability of having tails in the first flip and heads in the second when you flip the coins twice? A large selection of lessons and worksheets that show students how to use and apply the use of the Multiplication Rule of Probability.

Aligned Standard: HSS-CP.B.8

• Bag of Candy Step-by-step Lesson - What is the chance of getting a green candy out of a mixed bag.
• Guided Lesson - Marbles, pulling cards from a deck, and colored papers.
• Guided Lesson Explanation - It is also helpful to draw images along with the values.
• Practice Worksheet - What's the chances?
• Matching Worksheet - Sorry, so of these problems are a bit long winded.
• Answer Keys - These are for all the unlocked materials above.

Homework Sheets

We see several different form of probability floating around these problems.

• Homework 1 - Kitty has a bag of toys. In the bag there are 6 blue colored toys, 4 white colored toys and 9 purple colored toys. She takes one toy and records its color. She then puts it back in the bag. She then draws another color toy. What is the probability of taking out a blue colored toy followed by the white colored toy?
• Homework 2 - Cheri has a box with 8 blue balls and 4 red balls. Two balls are drawn without replacing them in the box. What is the probability that both of the balls are blue?
• Homework 3 - David has a basket of caps. In the basket there are 6 red caps, 3 yellow caps and 7 green caps. He takes one cap and records its color and puts it back in the basket. He then draws another cap. What is the probability of taking out a yellow caps followed by the red cap?

Practice Worksheets

You will find a mix of 3, 4, and 5 variables in these problems.

• Practice 1 - Mia has a box of pencils. In the box there are 12 red pencils, 14 yellow pencils and 10 green pencils. She takes one pencil records its color and puts it back in the box. She then draws another pencil. What is the probability of taking out a red pencil followed by a green pencil?
• Practice 2 - Julia has a bag with 10 pink hankies and 8 white hankies. Two hankies are drawn without replacement from the bag. What is the probability that both of the hankies are white?
• Practice 3 - Liam has a box of caps. In the box there are 6 yellow caps, 8 green caps and 10 blue caps. He randomly takes one cap out of the box. He records its color and puts it back in the box. He then draws another cap. What is the probability of taking out a blue cap followed by a yellow cap?

Math Skill Quizzes

Make sure you pay attention to if the item is replaced after it is chosen.

• Quiz 1 - Jacob has a bag of hair bands. In the bag there are 10 red hair bands, 9 green hair bands and 5 orange hair hands. He takes one hair band out, records its color hair band and puts it back in the bag. He then draws another hair band. What is the probability of taking out a red hair band followed by the green hair band?
• Quiz 2 - Lima has a box with 10 orange pens and 8 blue pens. Two pens are drawn without replacement from the box. What is the probability that both of the pens are orange?
• Quiz 3 - Alex has a box of mobile phones. In the box there are 5 green mobiles, 10 red mobiles and 8 pink mobiles. He takes one mobile, records its color and puts it back in the box. He then draws another mobile phone. What is the probability of taking out a red mobile followed by the pink mobile?

Get Access to Answers, Tests, and Worksheets

Become a paid member and get:

• Answer keys to everything
• Unlimited access - All Grades
• 64,000 printable Common Core worksheets, quizzes, and tests
• Used by 1000s of teachers!

Worksheets By Email:

Get Our Free Email Now!

We send out a monthly email of all our new free worksheets. Just tell us your email above. We hate spam! We will never sell or rent your email.

Thanks and Don't Forget To Tell Your Friends!

I would appreciate everyone letting me know if you find any errors. I'm getting a little older these days and my eyes are going. Please contact me, to let me know. I'll fix it ASAP.

• Privacy Policy
• Other Education Resource

© MathWorksheetsLand.com, All Rights Reserved

• Skip to secondary menu
• Skip to main content
• Skip to primary sidebar

Statistics By Jim

Making statistics intuitive

Multiplication Rule for Calculating Probabilities

By Jim Frost 7 Comments

The multiplication rule in probability allows you to calculate the joint probability of multiple events occurring together using known probabilities of those events individually. There are two forms of this rule, the specific and general multiplication rules.

In this post, learn about when and how to use both the specific and general multiplication rules. Additionally, I’ll use and explain the standard notation for probabilities throughout, helping you learn how to interpret it. We’ll work through several example problems so you can see them in action. There’s even a bonus problem at the end!

Before we get to the rules themselves, you need to know the definitions for independent and dependent events:

• Independent events : The occurrence of one event does not affect the probability of the other event. For example, when flipping a coin, getting “heads” does not change the likelihood of getting “heads” on the next coin flip.
• Dependent events : The occurrence of one event does affect the probability of the other event. For example, if you draw a King from a deck of cards and do not replace it, it causes the probability of drawing another King to decrease.

The notation for the joint probability of A and B occurring is the following: P(A ∩ B).

When events are independent, you can use the specific multiplication rule. When you have dependent events, you must use the general multiplication rule. Learn more in-depth about Independent Events .

Related posts : Probability Fundamentals and Joint Probability: Definition, Formula & Examples

Specific Multiplication Rule

Use the specific multiplication rule to calculate the joint probability of independent events. To use this rule, multiply the probabilities for the independent events. With independent events, the occurrence of event A does not affect the likelihood of event B. This rule is not valid for dependent events.

Using probability notation, the specific multiplication rule is the following:

P(A ∩ B) = P(A) * P(B)

Or, the joint probability of A and B occurring equals the probability of A occurring multiplied by the probability of B occurring.

Examples of the Specific Multiplication Rule

For example, to calculate the probability of obtaining “heads” during two consecutive coin flips, multiply the probability of heads on the first coin flip (0.5) by the probability of heads on the second coin flip (0.5).

0.5 X 0.5 = 0.25

The joint probability of two consecutive heads is 0.25.

You have ten pairs of pants and three are tan. Consequently, the probability of drawing a tan pair (event TP) is 0.3.

These are independent events because selecting a pair of pants doesn’t affect the likelihood of drawing a blue shirt and vice versa.

Using the specific multiplication rule for these independent events:

P(TP ∩ BS)= P(TP) * P(BS)

0.3 X 0.25 = 0.075

Or, the joint probability of randomly selecting a pair of tan pants and a blue shirt equals 0.075, which is the probability of tan pants multiplied by the probability of a blue shirt.

The likelihood of getting your preferred combination is low! You might want to drink some coffee to increase your chances!

General Multiplication Rule

Use the general multiplication rule to calculate joint probabilities for either independent or dependent events. When you have dependent events, you must use the general multiplication rule because it allows you to factor in how the occurrence of event A affects the likelihood of event B.

Using standard notation, the general multiplication rule is the following:

P(A ∩ B) = P(A) * P(B|A)

Or, the joint probability of A and B occurring equals the probability of A occurring multiplied by the conditional probability of B occurring given that A occurred.

The difference between the general and specific rules is, unsurprisingly, that you can use the general rule more generally. It works for both independent and dependent events, whereas the specific rule is valid only for independent events.

Why can you use the general form for both independent and dependent events? In the notation, focus on P(B|A), which is the conditional probability that event B occurs given that event A occurred.

In the context of independent events, P(B|A) = P(B) because event A occurring does not impact event B’s probability. That’s the very definition of independent events. Consequently, this rule becomes equivalent to the specific multiplicative rule for independent events.

However, for dependent events, P(B|A) ≠ P(B). That’s just another way of saying that event A occurring affects the probability of event B (i.e., they’re dependent events). The general multiplicative rule allows you to factor in the other event, as you will see in the next two examples!

Related post : Using Contingency Tables to Calculate Probabilities

Examples of the General Multiplication Rule

The classic example for dependent events is drawing cards from a deck of cards without replacement. As you draw cards, it affects the probability of the next card you can draw.

In notation form:

P(H1 ∩ H2) = P(H1) * P(H2|H1)

Or, the joint probability of drawing two consecutive hearts equals the probability of the first heart multiplied by the probability of the second heart given that the first card was a heart.

0.25 * 0.235 = 0.059

Dependent Events: Tan Pants and Blue Shirts Example

Let’s go back to the pants and shirt example. Imagine that we’re packing for a short trip and randomly select two pairs of pants and two pairs of shirts to include in our suitcase. We’re hoping for two pairs of tan pants and two blue shirts.

We’ll start by treating this as two sets of dependent events, one for pants and the other for shirts.

We begin with 10 pairs of pants, three of which are tan. Consequently, the probability of the first pair of pants being tan (event T1) is 0.30. The probability of the second pair being tan (T2) is 2/9 = 0.22. Hence:

P(T1 ∩ T2) = P(T1) * P(T2|T1)

0.30 * 0.22 = 0.066

The joint probability of drawing two pairs of tan pants is 0.066, which equals the probability of the first pair of tan pants multiplied by the conditional probability of the second pair of tan pants given that the first pair was tan.

And, for the shirts, we start with four blue shirts out of 16 total shirts. Using the same approach, we get the following:

P(B1 ∩ B2) = P(B1) * P(B2|B1)

0.25 * 0.20 = 0.05

We have the two joint probabilities of 0.066 for two tan pants and 0.05 for two blue shirts.

Related post : Using Permutations to Calculate Probabilities and Using Combinations to Calculate Probabilities

Bonus Example Problem!

To solve that problem, we’ll define two tan pants as event 2TP and two blue shirts as event 2BS. From our previous calculations for dependent events using the general multiplication rule, we know the following:

P(2TP) = 0.066

P(2BS) = 0.05

How do you calculate the joint probability P(2TP ∩ 2BS)?

Think back to the example of independent events where we drew one pair of pants and one shirt. Selecting pants does not affect the probabilities for shirts and vice versa. Consequently, we can treat events 2TP and 2BS as independent events even though we had dependent events when calculating probabilities for multiple pants and multiple shirts. In other words, selecting multiple pants affects the likelihood of the next pair of pants, but it does not affect shirts.

Hence, we can use the specific multiplication rule for independent events for this part of the solution:

P(2TP ∩ 2BS) = P(2TP) * P(2BS)

0.066 * 0.05 = 0.0033

The probability of drawing two pairs of tan pants and two blue shirts is only 0.0033 or 0.33%! That’s not too likely to occur by chance. If we genuinely want that combination, we should consider a non-random approach to packing!

Calculating joint probabilities using the multiplication rule is simple. Determine whether your events are independent or dependent, and then use the correct form of the rule!

Share this:

Reader Interactions

March 12, 2021 at 2:03 pm

Thanks Jim, wonderful article.

March 11, 2021 at 10:43 am

This is a wonderful article. Until now I had a hard time understanding the basics of probability. This is truly enlightening. You have a great way of explaining things. I had read your Introduction to Statistics book and in the process of reading Linear Regression Both these books are gems.

Thanks for everything!

March 16, 2021 at 12:32 am

Thanks so much for your kind words! I’m so glad this blog post and my books have been helpful! 🙂

March 7, 2021 at 10:05 pm

A great post breaking down dependent vs independent events and joint probabilities in a straight-forward way. These concepts are critical to interpreting and modelling business outcomes, but are often not applied correctly because of a lack of awareness and understanding.

Great examples too, especially the final example that combines both dependent and independent events together to calculate the joint probability.

BTW I also enjoyed reading your hypothesis testing book, some really intuitive and well written explanations and a well thought out structure.

Cheers, Joe

March 7, 2021 at 10:23 pm

Thanks so much for your kind words. You made my day!

Also, I’m so glad that my hypothesis testing book was helpful! 🙂

March 7, 2021 at 9:43 pm

Another well explained post – Thankyou. Is that in one of you books? – I have Intro to stats and Hypothesis testing

March 7, 2021 at 9:51 pm

Thanks! It’s not currently in one of my book. However, I will be writing a probability book down the road!

Thanks for supporting my books too! 🙂

Comments and Questions Cancel reply

Multiplication Rule for Probabilities of Independent Events

Examples on using the multiplication rule to find the probability of two or more independent events occurring are presented along with detailed solutions.

Independent Events

In probabilities, two events are independent if the occurrence of one does not affect the probability of occurrence of the other. Example 1 The following events A and B independent.

•   A = "roll a die and get a \( 1 \)" , B = "flip a coin and get a tail".
•   A = "draw a card from a deck and get a King", replace it back into the deck, B = "draw another card and get a Queen"
•   A = "roll a die and get a \( 4 \)" , B = "roll the same die (or another one) and get a "6"
•   A = "flip a coin and get a head" , B = roll the same coin (or another one) and get a tail" A jar has 3 blue balls, 2 white balls and 5 red balls
•   A = Pick a ball at random from the jar and get a red ball, replace it back into the jar, B = Pick a ball at random from the jar and get a white ball
•   C = "draw a card from a deck and get a King", D = "draw a second card from the same deck and get a Queen". A jar contains 3 blue balls, 2 white balls and 3 red balls
•   C = "Pick a ball at random from the jar and get a red ball", D = "Pick a second ball at random from the same jar and get a white ball".

Examples with Detailed Solutions

More questions with solutions, solutions to above exercises, more references and links.

• Terms of Service

• Math Article
• Multiplication Rule Probability

Multiplication Rule of Probability

The multiplication rule of probability explains the condition between two events. For two events A and B associated with a sample space S  set A∩B  denotes the events in which both events A  and event B  have occurred. Hence, (A∩B)  denotes the simultaneous occurrence of events A  and B . Event A∩B can be written as AB . The probability of event AB  is obtained by using the properties of conditional probability .

What is the Multiplication Rule of Probability?

According to the multiplication rule of probability, the probability of occurrence of both the events A and B is equal to the product of the probability of B occurring and the conditional probability that event A occurring given that event B occurs.

If A and B are dependent events, then the probability of both events occurring simultaneously is given by:

If A and B are two independent events in an experiment, then the probability of both events occurring simultaneously is given by:

We know that the conditional probability of event A  given that B  has occurred is denoted by P(A|B)  and is given by:

Where,  P(B)≠0

P(A∩B) = P(B)×P(A|B) ……………………………………..(1)

\(\begin{array}{l}P(B|A)~ = ~\frac{P(B∩A)}{P(A)}\end{array} \)

Where, P(A) ≠ 0.

P(B∩A) = P(A)×P(B|A)

Since, P(A∩B) = P(B∩A)

P(A∩B) = P(A)×P(B|A) ………………………………………(2)

From (1) and (2), we get:

P(A∩B) = P(B)×P(A|B) = P(A)×P(B|A) where,

P(A) ≠ 0,P(B) ≠ 0.

The above result is known as the multiplication rule of probability.

For independent events A  and B, P(B|A) = P(B). The equation (2) can be modified into,

P(A∩B) = P(B) × P(A)

Multiplication Theorem of Probability

We have already learned the multiplication rules we follow in probability, such as;

P(A∩B) = P(A)×P(B|A) ; if P(A) ≠ 0

P(A∩B) = P(B)×P(A|B) ; if P(B) ≠ 0

Let us learn here the multiplication theorems for independent events A and B.

If A and B are two independent events for a random experiment, then the probability of simultaneous occurrence of two independent events will be equal to the product of their probabilities. Hence,

P(A∩B) = P(A).P(B)

Now, from multiplication rule we know;

P(A∩B) = P(A)×P(B|A)

Since A and B are independent, therefore;

P(B|A) = P(B)

Therefore, again we get;

Hence, proved.

Solved Example of Multiplication Rule of Probability

Illustration 1: An urn contains 20 red and 10 blue balls. Two balls are drawn from a bag one after the other without replacement. What is the probability that both the balls are drawn are red?

Solution: Let A and B denote the events that the first and the second balls are drawn are red balls. We have to find P(A∩B) or P(AB).

P(A) = P(red balls in first draw) = 20/30

Now, only 19 red balls and 10 blue balls are left in the bag. The probability of drawing a red ball in the second draw too is an example of conditional probability where the drawing of the second ball depends on the drawing of the first ball.

Hence Conditional probability of B  on A  will be,

P(B|A) = 19/29

By multiplication rule of probability,

P(A∩B) = P(A) × P(B|A)

\(\begin{array}{l}P(A∩B)~ =~ \frac{20}{30} ~× ~\frac{19}{29} ~=~ \frac{38}{87}\end{array} \)

Addition Rule of Probability

The addition rule states the probability of two events is the sum of the probabilities of two events that will happen minus the probability of both the events that will happen.

Mathematically, the addition rule of probability is expressed as:

Related Articles

• Probability
• Probability worksheets
• Total Probability Theorem
• Chance And Probability
• Types of Events in Probability
• Independent Events

For a detailed discussion on the probability, download BYJU’S-the learning app.

Put your understanding of this concept to test by answering a few MCQs. Click ‘Start Quiz’ to begin!

Select the correct answer and click on the “Finish” button Check your score and answers at the end of the quiz

Visit BYJU’S for all Maths related queries and study materials

Your result is as below

Request OTP on Voice Call

Leave a Comment Cancel reply

Your Mobile number and Email id will not be published. Required fields are marked *

Post My Comment

Register with BYJU'S & Download Free PDFs

Register with byju's & watch live videos.

Probability Practice Questions

Click here for questions, click here for answers, gcse revision cards.

5-a-day Workbooks

Primary Study Cards

Privacy Policy

Terms and Conditions

Corbettmaths © 2012 – 2024

• For educators
• English (US)
• English (India)
• English (UK)
• Greek Alphabet

Your solution’s ready to go!

Our expert help has broken down your problem into an easy-to-learn solution you can count on.

Question: Calculate the theoretical probability of winning when the contestant stays with Door 1 by multiplying the probabilities along each branch of this tree diagram. Express your answers as reduced fractions: P( Door 1, Door 2, Door 1)= P( Door 1 , Door 3, Door 1)= P( Door 2 , Door 3, Door 1)= P( Door 3 , Door 2 Door 1)= What is the probability of winning when the

The answers I put in were incorrect.

Given that,

Now calculate the theoretical probability of winning when the contestant stays with Door ...

Not the question you’re looking for?

Post any question and get expert help quickly.

Mr. Valder's Courses

• Contact Information
• Functions - Unit 1
• Polynomials - Unit 2
• Logarithms - Unit 3
• Trigonometry - Unit 4
• Trigonometry - Unit 5
• Vectors - Unit 6
• Parametrics - Unit 7
• Sequences - Unit 8
• Final Review
• AFM Expectations
• Graphing Functions
• Polynomials
• Right Triangle Trig
• Laws of Trig
• Probability
• Final Exam Review
• Cumulative Review
• Additional Topics
• Problem Set # 1
• Problem Set # 2
• Problem Set # 3
• Problem Set # 4
• Problem Set # 5
• Problem Set # 6

Please ensure that your password is at least 8 characters and contains each of the following:

• a special character: @$#!%*?&

• Solve equations and inequalities
• Simplify expressions
• Factor polynomials
• Graph equations and inequalities
• Advanced solvers
• All solvers
• Arithmetics
• Determinant
• Percentages
• Scientific Notation
• Inequalities

What can QuickMath do?

QuickMath will automatically answer the most common problems in algebra, equations and calculus faced by high-school and college students.

• The algebra section allows you to expand, factor or simplify virtually any expression you choose. It also has commands for splitting fractions into partial fractions, combining several fractions into one and cancelling common factors within a fraction.
• The equations section lets you solve an equation or system of equations. You can usually find the exact answer or, if necessary, a numerical answer to almost any accuracy you require.
• The inequalities section lets you solve an inequality or a system of inequalities for a single variable. You can also plot inequalities in two variables.
• The calculus section will carry out differentiation as well as definite and indefinite integration.
• The matrices section contains commands for the arithmetic manipulation of matrices.
• The graphs section contains commands for plotting equations and inequalities.
• The numbers section has a percentages command for explaining the most common types of percentage problems and a section for dealing with scientific notation.

Math Topics

More solvers.

• Add Fractions
• Simplify Fractions