Conditional Probability and the Multiplication Rule

Learning Objectives

After completing this section, you should be able to:

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

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 \(O\) is a possible outcome of the first stage in a multistage experiment, then the probability of an event \(E\) conditional on \(O\) (denoted \(P(E|O)\), read “the probability of \(E\) given \(O\)”) is the updated probability of \(E\) under the assumption that \(O\) 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 \(E\) to be the event “roll a sum of 10,” \(O\) to be the event “first die shows 5,” and \(Q\) to be the event “first die shows 3,” then we computed \(P(E)=\frac{1}{12}\), \(P(E|O)=\frac{1}{6}\), and \(P(E|Q)=0\).

Condensed — the full section is in OpenStax Contemporary Mathematics.

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.

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.

Condensed — the full section is in OpenStax Contemporary Mathematics.

Key Concepts

• Conditional probabilities are computed under the assumption that the condition has already occurred.
• The Multiplication Rule for Probability is used to find the probability that two events occur in sequence.

Formulas

• If \(E\) and \(F\) are events associated with the first and second stages of an experiment, then \(P(E\text{and}F)=P(E)\times P(F|E)\).

Practice (3)

Try each one on paper first. Reveal the answer to check; verified ones can be opened in the solver for every step.

1. 1. 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 \(A\) be the event “April wins,” \(H\) be “first flip is heads,” and \(T\) be “first flip is tails.” Compute \(P(A)\), \(P(A|H)\), and \(P(A|T)\).
   2. You are about to draw 2 cards without replacement from a deck containing only these 10 cards: \(\text{A}♡\), \(\text{A}♠\), \(\text{A}♣\), \(\text{A}♢\), \(\text{K}♠\), \(\text{K}♣\), \(\text{Q}♡\), \(\text{Q}♠\), \(\text{J}♡\), \(\text{J}♠\). We’ll define the following events: \(F\) is “both cards are the same rank,” \(A\) is “first card is an ace,” and \(K\) is “first card is a king.” Compute \(P(F|A)\) and \(P(F|K)\).
   3. 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 \(M\) be the event “Jim’s 2 socks match,” let \(K\) be the event “the sock on Jim’s left foot is black,” and let \(L\) be the event “the sock on Jim’s left foot is blue.” Compute \(P(M)\), \(P(M|K)\), and \(P(M|L)\).
   Otkrij odgovor

   1. Step 1. The sample space is {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}. The event \(A\) consists of the first 4 of those outcomes: HHH, HHT, HTH, and THH. Thus, \(P(A)=\frac{4}{8}=\frac{1}{2}\).

      Step 2. Now, let’s compute \(P(A|H)\). 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, \(P(A|H)=\frac{3}{4}\).

      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, \(P(A|T)=\frac{1}{4}\).

   2. Step 1. If the event \(A\) happens, then 1 of the 4 aces is drawn first; the remaining cards in the deck are 3 aces, 2 kings, 2 queens, and 2 jacks. In order for the event \(F\) to occur, the second card drawn has to be an ace. Since there are 3 aces among the remaining 9 cards, \(P(F|A)=\frac{3}{9}=\frac{1}{3}\).

      Step 2. If the event \(K\) 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 \(F\) will occur only if the second card is also a king. Since only one of the remaining 9 cards is a king, we have \(P(F|K)=\frac{1}{9}\).

   3. Step 1. We can view the event \(M\) as a compound event using “or”: both socks are blue or both socks are black. Let’s compute the probability that both socks are blue using combinations. We’re choosing 2 socks from a group of 8; 3 of the 8 are blue. So, \(P(\text{both socks blue})=\frac{{}_{3}{C}_{2}}{{}_{8}{C}_{2}}=\frac{3}{28}\). Similarly, \(P(\text{both socks black})=\frac{{}_{5}{C}_{2}}{{}_{8}{C}_{2}}=\frac{10}{28}\). Therefore, since these events are mutually exclusive, we can use the Addition Rule: \(P(M)=P(\text{both socks blue})+P(\text{both socks black})=\frac{3}{28}+\frac{10}{28}=\frac{13}{28}\).

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

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

2. 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.

   1. What is the probability that a senior is chosen for both positions?
   2. What is the probability that a junior is chosen first and a sophomore is chosen second?
   3. What is the probability that a sophomore is chosen first and a senior is chosen second?
   Otkrij odgovor

   1. 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: \(P(\text{ senior chosen first and senior chosen second})=P(\text{senior chosen first})\times P(\text{senior chosen second}|\text{senior chosen first})\text{.}\) Since there are 4 seniors among the 10 members, \(P(\text{senior chosen first})=\frac{4}{10}=\frac{2}{5}\). Next, assuming a senior is chosen first, there are 3 seniors among the 9 remaining members. So, \(P(\text{senior chosen second}|\text{senior chosen first})=\frac{3}{9}=\frac{1}{3}\). Putting this all together, we get \(P(\text{senior chosen first and senior chosen second})=\frac{2}{5}\times \frac{1}{3}=\frac{2}{15}\).
   2. There are 3 juniors among the 10 members, so \(P(\text{junior chosen first})=\frac{3}{10}\). Assuming a junior is chosen first, there are 2 sophomores among the remaining 9 members, so \(P(\text{sophomore chosen second}|\text{junior chosen first})=\frac{2}{9}\). Thus, using the Multiplication Rule for Probability, we have \(P(\text{junior chosen first and sophomore chosen second})=\frac{3}{10}\times \frac{2}{9}=\frac{1}{15}\).
   3. The probability that a sophomore is chosen first is \(\frac{2}{10}=\frac{1}{5}\), and the probability that a senior is chosen second given that a sophomore was chosen first is \(\frac{4}{9}\). Thus, using the Multiplication Rule for Probability, we have: \(P(\text{sophomore chosen first and senior chosen second})=\frac{1}{5}\times \frac{4}{9}=\frac{4}{45}\).

3. 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?

   Otkrij odgovor

   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 ().

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

   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 ().

   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 \(P(\text{second is suspect}|\text{first is suspect})=\frac{5}{20}=\frac{1}{4}\). Using similar reasoning, we can compute \(P(\text{second is weapon}|\text{first is suspect})=\frac{6}{20}=\frac{3}{10}\) and \(P(\text{second is room}|\text{first is suspect})=\frac{9}{20}\).

   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 ().

   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 ().

   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 \(\frac{2}{7}\times \frac{3}{10}=\frac{3}{35}\), as illustrated in .

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

   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 \(P(SR)+P(RS)=\frac{9}{70}+\frac{9}{70}=\frac{9}{35}\). We can find the probabilities of the other outcomes in a similar fashion, as shown in the following table:

   Outcome	Probability
   2 suspects	\(\frac{1}{14}\)
   2 weapons	\(\frac{1}{14}\)
   2 rooms	\(\frac{6}{35}\)
   1 suspect and 1 weapon	\(\frac{3}{35}+\frac{3}{35}=\frac{6}{35}\)
   1 suspect and 1 room	\(\frac{9}{70}+\frac{9}{70}=\frac{9}{35}\)
   1 weapon and 1 room	\(\frac{9}{70}+\frac{9}{70}=\frac{9}{35}\)

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

Symbols used here

P(A),\ P(A \mid B)

probability, conditional probability
Chance of A; chance of A given that B happened.

\neq

not equal
The two sides are different.

\pm

plus or minus
Both signs at once: x = 3 ± 2 means 5 and 1.

\approx

approximately equal
Equal to the precision shown, not exactly.

n!

factorial
n × (n−1) × … × 1; the number of orderings of n things. 0! = 1.

\binom{n}{k}

binomial coefficient, "n choose k"
Number of k-element subsets of n things: n!/(k!(n−k)!).

\sum_{k=1}^{n} a_k

summation
Add a_k for k = 1 up to n.

A \cup B,\ A \cap B,\ A \setminus B

union, intersection, difference
In either; in both; in A but not B.

\bar{x},\ \mu

sample mean, population mean
Average of the data; average of the whole population.

\sigma,\ s,\ \sigma^2

standard deviation, sample s.d., variance
Typical distance from the mean; its square.

E[X],\ \operatorname{Var}(X)

expected value, variance
Probability-weighted average of X; its spread.

N(\mu, \sigma^2),\ z

normal distribution, z-score
The bell curve with mean μ and variance σ²; (x − μ)/σ.

How to: Conditional Probability and the Multiplication Rule

1. Calculate conditional probabilities.
2. Apply the Multiplication Rule for Probability to compute probabilities.
3. 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
4. You are about to draw 2 cards without replacement from a deck containing only these 10 cards:
5. 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
7. What is the probability that a senior is chosen for both positions?
8. What is the probability that a junior is chosen first and a sophomore is chosen second?

Questions people ask

Mean or median — which should I use?

Median when the data have outliers or a long tail (incomes, house prices); mean when the data are roughly symmetric and you want every value to count. Report both if they disagree — the gap is itself information.

What does a p-value actually say?

The probability of seeing data at least this extreme if the null hypothesis were true. It is not the probability that the null hypothesis is true.

Why divide by n − 1 for the sample variance?

The sample mean sits closer to the sample than the true mean does, so squared deviations from it are slightly too small on average; dividing by n − 1 instead of n corrects the bias.

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Parts of this page are adapted from OpenStax Contemporary Mathematics (CC BY-NC-SA 4.0). Condensed and re-explained here; errors are ours.

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