Conditional Probability and Independent Events

Statistics — Third Secondary

Conditional Probability and Independent Events

Worked examples

Example 1→A fair six-sided die is rolled once. Let A be the event “an even number appears” and B be the event “a number greater than 3 appears”. Find P(A | B), and decide whether the events are independent.

The sample space is S = {1, 2, 3, 4, 5, 6}. The event B = {4, 5, 6}, so P(B) = 3 ÷ 6 = 1 ÷ 2.
A ∩ B = {4, 6}, so P(A ∩ B) = 2 ÷ 6 = 1 ÷ 3.
P(A | B) = (1 ÷ 3) ÷ (1 ÷ 2) = 2 ÷ 3. So among the numbers greater than 3, two thirds are even.
P(A) = 1 ÷ 2, and P(A) × P(B) = 1 ÷ 4 ≠ 1 ÷ 3, so the events are not independent.
Example 2 A school has 100 students: 60 boys and 40 girls. Of the boys, 36 like chess, and of the girls, 16 like chess. A student is chosen at random. Find the probability that the student is a girl given that the student likes chess. Then test whether liking chess is independent of being a boy or a girl.

Number of students who like chess = 36 + 16 = 52.
P(girl | likes chess) = 16 ÷ 52 = 4 ÷ 13 ≈ 0.31.
P(likes chess | boy) = 36 ÷ 60 = 0.6, while P(likes chess) = 52 ÷ 100 = 0.52.
Since 0.6 ≠ 0.52, the events are not independent.
Example 3 One card is drawn from a standard pack of 52 cards. Let A be the event “the card is a king” and B be the event “the card is a heart”. Show whether the events are independent.

P(A) = 4 ÷ 52 = 1 ÷ 13.
P(B) = 13 ÷ 52 = 1 ÷ 4.
P(A ∩ B) = only the king of hearts, so P(A ∩ B) = 1 ÷ 52.
P(A) × P(B) = 1 ÷ 13 × 1 ÷ 4 = 1 ÷ 52, which equals P(A ∩ B). Therefore the events are independent.
Also, P(A | B) = (1 ÷ 52) ÷ (1 ÷ 4) = 1 ÷ 13 = P(A).
Example 4 A bag contains 5 red balls and 3 blue balls. Two balls are drawn one after the other without replacement. Find the probability that both balls are red. Then find the probability that the second ball is red.

Let R1 be the event “the first ball is red” and R2 be the event “the second ball is red”. There are 8 balls in total at the start.
P(R1) = 5 ÷ 8.
After a red ball is removed, 4 red balls remain out of 7 balls, so P(R2 | R1) = 4 ÷ 7.
P(R1 ∩ R2) = 5 ÷ 8 × 4 ÷ 7 = 20 ÷ 56 = 5 ÷ 14 ≈ 0.36.
To find P(R2), we split into cases. If the first ball is blue, then P(R2 | first blue) = 5 ÷ 7. If the first ball is red, then P(R2 | R1) = 4 ÷ 7.
P(R2) = (5 ÷ 8 × 4 ÷ 7) + (3 ÷ 8 × 5 ÷ 7) = 20 ÷ 56 + 15 ÷ 56 = 35 ÷ 56 = 5 ÷ 8.
This means the probability that the second ball is red equals the probability that the first ball is red, even though the balls are drawn without replacement.

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