Lesson 6 Conditional Probability

Motivating Example

You know that your coworker has two children. Absent any other information, the probability that both are boys is \(P(\text{both boys}) = \frac{1}{4}\).

One day, she mentions to you, “I need to stop by St. Joseph’s after work for a PTA meeting.” St. Joseph’s is a local all-boys school. So now you know that at least one of her children is a boy. What is the probability now that both her children are boys?

Most people assume that the answer is \(\frac{1}{2}\), since the other child is equally likely to be a boy or a girl, and the gender of one child does not affect the gender of another. The actual answer may surprise you….


To quantify how probabilities change in light of new information, we calculate the conditional probability.

\[ P(\text{both boys}\ |\ \text{at least one boy}) \]

The \(|\) symbol is read “given” and the event after the \(|\) symbol represents information that we know.

In general, to calculate a conditional probability, we use the following formula.

Definition 6.1 (Conditional Probability) \[ P(B | A) = \frac{P(A \textbf{ and } B)}{P(A)}. \]

The probability \(P(A \textbf{ and } B)\) is called the joint probability of the two events \(A\) and \(B\).

So the conditional probability above is

\[\begin{align*} P(\text{both boys}\ |\ \text{at least one boy}) &= \frac{P(\text{both boys} \textbf{ and } \text{at least one boy})}{P(\text{at least one boy})} \\ &= \frac{P(\text{both boys})}{P(\text{at least one boy})} \\ &= \frac{1/4}{3/4} \\ &= \frac{1}{3}. \end{align*}\]

In the above example, the joint probability \(P(\text{both boys} \textbf{ and } \text{at least one boy})\) is easy to calculate because the two events are redundant. If we know that both are boys, then we automatically know that at least one is a boy.

The information that at least one of her children attends St. Joseph’s (and, thus, is a boy) increases the probability that she has two boys from \(1/4\) to \(1/3\).

If this result was counterintuitive to you, the video below gives some intuition.

We can rearrange the conditional probability formula to get a formula that is useful for calculating the joint probability when the conditional probability is known.
Theorem 6.1 (Multiplication Rule) \[ P(A \textbf{ and } B) = P(A) \cdot P(B | A). \]
Example 6.1 Two cards are dealt off the top of a shuffled deck of cards. What is the probability that both are queens?

Solution. In this case, we want to calculate the joint probability \[ P(\text{1st card is Q} \textbf{ and } \text{2nd card is Q}), \] and the conditional probability \[ P(\text{2nd card is Q} | \text{1st card is Q}) \] is simple. If it is known that the 1st card was a queen, then there are only 51 cards remaining in the deck, of which 3 are queens, so \[ P(\text{2nd card is Q} | \text{1st card is Q}) = \frac{3}{51}. \]

By the Multiplication Rule (6.1), \[\begin{align*} P(\text{1st card is Q} \textbf{ and } \text{2nd card is Q}) &= P(\text{1st card is Q}) \cdot P(\text{2nd card is Q} | \text{1st card is Q}) \\ &= \frac{4}{52} \cdot \frac{3}{51} \end{align*}\]

Let’s compare this with a solution based on counting the outcomes directly. There are \(52 \cdot 51\) equally likely (ordered) outcomes, of which \(4 \cdot 3\) are both queens. Therefore, the probability is \[ P(\text{1st card is queen} \textbf{ and } \text{2nd card is queen}) = \frac{4 \cdot 3}{52 \cdot 51}. \] We get the same answer; the only thing that changes is the order of operations:

  • In counting, we first multiply to get the number of outcomes, then divide to get probabilities.
  • In the multiplication rule, we first divide to get probabilities, then multiply the probabilities.


  1. You and your friend Amy are each dealt two cards: hers face up and yours face down. In which of the following scenarios are you more likely to have a pair:
    • when she has a pair of queens
    • when she has a queen and a 5?
  2. Dr. No has captured James Bond and forces him to play a game of Russian roulette. (Note: Russian roulette is very different from the casino game roulette!) Dr. No shows him an revolver with 6 chambers, all initially empty. He places 2 bullets into adjacent chambers. He makes Bond spin the cylinder, place the muzzle against his head, and pull the trigger. He survives! Luckily for Bond, the cylinder stopped on one of the empty chambers.

    Now Dr. No gives Bond two options: he can re-spin the cylinder before firing again or he can fire with the gun in its current state. (Keep in mind that the cylinder rotates to the next chamber each time the gun is fired.) What option should Bond choose to maximize his chance of surviving?

    1. Clearly write out the conditional probability of interest using \(P(B|A)\) notation.
    2. Find the probability. (Hint: You should not need to do any calculations. You should be able to find the probability just by thinking carefully about the information you have. Make sure you explain your reasoning carefully.)

Additional Exercises