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Probability
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Conditional Probability
Practice: Conditional Probability
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Independent Events
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Glossary

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ProbabilityConditional Probability

Reading time: ~35 min

Does knowing that one event has happened change the probability of another? Let's find out with an experiment whose outcomes are equally likely: tossing three fair coins. Its sample space has eight outcomes,

S = \{HHH, HHT, HTH, THH, HTT, THT, TTH, TTT\}

and each of them has probability \frac{1}{8}. Let E be the event "at least two heads". Then E = \{HHH, HHT, HTH, THH\}, so P(E) = \frac{4}{8} = \frac{1}{2}.

Now suppose someone tells you one fact about the toss, an event F, before you guess. Tap a given event F. Everything outside F fades away; the outcomes of F turn green, and the ones that are in E as well turn purple. Complete the three missions.

Shrinking the Sample Space

Let's do the first case on paper. F is "the first coin shows a tail", so F = \{THH, THT, TTH, TTT\}.

Once we know that F has happened, every outcome starting with H is impossible. The sample space has shrunk from S to F, which has outcomes, all still equally likely.

Inside F, the event E happens only for THH, the one outcome that lies in both E and F. That outcome makes up the event E \cap F. So the probability of E, given that F has occurred, is .

We write this as P(E|F), read "the probability of E given F". Look at where the numbers came from:

P(E|F) = \dfrac{\text{number of outcomes in } E \cap F}{\text{number of outcomes in } F} = \dfrac{n(E \cap F)}{n(F)}

Divide the top and the bottom by n(S) = 8, and the counts turn into probabilities:

P(E|F) = \dfrac{n(E \cap F)/n(S)}{n(F)/n(S)} = \dfrac{P(E \cap F)}{P(F)} = \dfrac{1/8}{4/8} = \dfrac{1}{4}

Definition. If E and F are two events associated with the same sample space of a random experiment, the conditional probability of E given that F has occurred is

P(E|F) = \dfrac{P(E \cap F)}{P(F)}, \quad \text{provided } P(F) \neq 0.

Two facts follow at once. E \cap F is a part of F, so its probability is never more than P(F), and therefore 0 \le P(E|F) \le 1. And the condition P(F) \neq 0 matters: nobody can tell you that an impossible event has happened.

Tip

In a word problem, look for the words "given that", "if it is known that" or "it is found that". The event that follows them is F — the information you already have.

Properties of conditional probability

Conditional probability behaves just like ordinary probability; it simply lives on the smaller sample space F. Let E and F be events of a sample space S, with P(F) \neq 0.

Property 1. P(S|F) = P(F|F) = 1.

Given F, the whole sample space has certainly happened, and so has F itself. Indeed P(S|F) = \frac{P(S \cap F)}{P(F)} = \frac{P(F)}{P(F)} = 1, and in the same way P(F|F) = \frac{P(F \cap F)}{P(F)} = .

Property 2. If A and B are any two events of S, then

P((A \cup B)|F) = P(A|F) + P(B|F) - P((A \cap B)|F)

Why: by the distributive law, (A \cup B) \cap F = (A \cap F) \cup (B \cap F). The addition rule from Class 11 then gives

P((A \cup B)|F) = \dfrac{P(A \cap F) + P(B \cap F) - P(A \cap B \cap F)}{P(F)}

and splitting the fraction into three gives the property. When A and B are mutually exclusive, the last term is , and the rule becomes simply P((A \cup B)|F) = P(A|F) + P(B|F).

Property 3. P(E'|F) = 1 - P(E|F).

Why: E and E′ are mutually exclusive and E \cup E' = S. Properties 1 and 2 give 1 = P(S|F) = P(E|F) + P(E'|F).

Try it on the three coins. Given that the first coin shows a tail, P(E|F) = \frac{1}{4}, so the chance of fewer than two heads is P(E'|F) = .

Worked Examples

Example 1 : If P(A) = \frac{7}{13}, P(B) = \frac{9}{13} and P(A \cap B) = \frac{4}{13}, evaluate P(A|B).

Solution : By the definition,

P(A|B) = \dfrac{P(A \cap B)}{P(B)} = \dfrac{4/13}{9/13} =

The 13s cancel. Whenever every probability has the same denominator, P(A|B) is simply a ratio of counts.

Example 2 : A family has two children. What is the probability that both the children are boys given that at least one of them is a boy?

Gut feeling first. Many people reason: "one child is a boy, and the other is a boy or a girl, so it is a half." What do you think?

1/2
1/3
1/4

Solution : Write b for a boy and g for a girl, the elder child first. The sample space is S = \{(b, b), (b, g), (g, b), (g, g)\}, with four equally likely outcomes.

Let E be "both children are boys" and F "at least one child is a boy". Then E = \{(b, b)\}, F = \{(b, b), (b, g), (g, b)\} and E \cap F = \{(b, b)\}.

So P(F) = \frac{3}{4} and P(E \cap F) = \frac{1}{4}, which gives P(E|F) = \dfrac{1/4}{3/4} =

The clue "at least one boy" leaves three outcomes, and only one of them is two boys. Had the clue been "the elder child is a boy", F would have been \{(b, b), (b, g)\} and the answer a half. The exact wording of the information matters.

Example 3 : Ten cards numbered 1 to 10 are placed in a box, mixed up thoroughly and then one card is drawn randomly. If it is known that the number on the drawn card is more than 3, what is the probability that it is an even number?

Solution : Let A be "the number is even" and B "the number is more than 3". Then

S = \{1, 2, 3, \ldots, 10\}, A = \{2, 4, 6, 8, 10\} and B = \{4, 5, 6, 7, 8, 9, 10\}.

The even numbers that are more than 3 are , so A \cap B has 4 elements.

So P(B) = \frac{7}{10} and P(A \cap B) = \frac{4}{10}, and

P(A|B) = \dfrac{P(A \cap B)}{P(B)} = \dfrac{4/10}{7/10} =

Example 4 : In a school, there are 1000 students, out of which 430 are girls. It is known that out of 430, 10% of the girls study in class XII. What is the probability that a student chosen randomly studies in Class XII given that the chosen student is a girl?

Solution : Let E be "the student studies in Class XII" and F "the student is a girl". We want P(E|F).

The number of girls in Class XII is 10% of 430, which is .

So P(F) = \frac{430}{1000} = 0.43 and P(E \cap F) = \frac{43}{1000} = 0.043. Then

P(E|F) = \dfrac{P(E \cap F)}{P(F)} = \dfrac{0.043}{0.43} =

Of course: we were told that 10% of the girls are in Class XII. Once we know the student is a girl, the 570 boys no longer matter.

Example 5 : A die is thrown three times. Events A and B are defined as below: A : 4 on the third throw, B : 6 on the first and 5 on the second throw. Find the probability of A given that B has already occurred.

Solution : The sample space has 6 \times 6 \times 6 = equally likely outcomes.

B fixes the first two throws as 6 and 5, and the third throw can be anything: B = \{(6, 5, 1), (6, 5, 2), \ldots, (6, 5, 6)\}, so P(B) = \frac{6}{216}. The only outcome in both A and B is (6, 5, 4), so P(A \cap B) = \frac{1}{216}.

Therefore P(A|B) = \dfrac{1/216}{6/216} =

That is the same as P(A) = \frac{36}{216} = \frac{1}{6}: what happened on the first two throws tells us nothing about the third. Keep this example in mind for the section on independent events.

Example 6 : A die is thrown twice and the sum of the numbers appearing is observed to be 6. What is the conditional probability that the number 4 has appeared at least once?

Use the lab first. E — a 4 at least once — is outlined in blue. Tap the given events and complete the missions.

Solution : Let E be "4 appears at least once" and F "the sum of the numbers is 6". E has 11 outcomes: the six with a 4 on the first die, (4, 1) to (4, 6), and five more with a 4 on the second die only, (1, 4), (2, 4), (3, 4), (5, 4) and (6, 4). Also

F = \{(1, 5), (2, 4), (3, 3), (4, 2), (5, 1)\}

so P(F) = \frac{5}{36}. Since E \cap F = \{(2, 4), (4, 2)\}, we get P(E \cap F) = .

Hence P(E|F) = \dfrac{2/36}{5/36} =

When Outcomes Are Not Equally Likely

Example 7 : Consider the experiment of tossing a coin. If the coin shows head, toss it again but if it shows tail, then throw a die. Find the conditional probability of the event that 'the die shows a number greater than 4' given that 'there is at least one tail'.

Solution : The outcomes of this experiment are

S = \{(H, H), (H, T), (T, 1), (T, 2), (T, 3), (T, 4), (T, 5), (T, 6)\}

but they are not equally likely. The first toss gives H or T with probability \frac{1}{2} each. After a head, the second toss splits that \frac{1}{2} into two halves, so P(H, H) = P(H, T) = . After a tail, the die splits it into six equal parts, so each of (T, 1), …, (T, 6) has probability .

Let F be "at least one tail" and E "the die shows a number greater than 4". Then F = \{(H, T), (T, 1), (T, 2), \ldots, (T, 6)\} and E = \{(T, 5), (T, 6)\}, so E \cap F = \{(T, 5), (T, 6)\}.

Add up the probabilities of their outcomes: P(F) = \frac{1}{4} + 6 \times \frac{1}{12} = and P(E \cap F) = \frac{1}{12} + \frac{1}{12} = \frac{1}{6}.

Therefore P(E|F) = \dfrac{1/6}{3/4} =

Note

Counting outcomes would give the wrong answer here: 2 out of the 7 outcomes in F, that is \frac{2}{7}. When the outcomes are not equally likely, use the formula P(E|F) = \frac{P(E \cap F)}{P(F)} with the actual probability of each outcome.

Speed Round: Given That…

One fair die. Shrink the sample space to F in your head, then count the outcomes of E inside it.

Get five right to move on.

Sina