fair die is rolled twice. Let 𝐴 be the event that an even number is obtained on the first rolland let 𝐵 be the event that a 3 is obtained on the second roll.(i) Find 𝑃(𝐴 ∩ 𝐵).(ii) Hence, or otherwise, find 𝑃(𝐴 ∪ 𝐵)
Question
fair die is rolled twice. Let 𝐴 be the event that an even number is obtained on the first rolland let 𝐵 be the event that a 3 is obtained on the second roll.(i) Find 𝑃(𝐴 ∩ 𝐵).(ii) Hence, or otherwise, find 𝑃(𝐴 ∪ 𝐵)
Solution
(i) The event A and B are independent events. The probability of A, P(A), is the probability of getting an even number (2, 4, or 6) on a fair six-sided die, which is 1/2. The probability of B, P(B), is the probability of getting a 3 on a fair six-sided die, which is 1/6. The probability of both A and B occurring, P(A ∩ B), is the product of their individual probabilities, so P(A ∩ B) = P(A) * P(B) = 1/2 * 1/6 = 1/12.
(ii) The probability of either A or B or both occurring, P(A ∪ B), is the sum of their individual probabilities minus the probability of both occurring (since we're double counting that scenario), so P(A ∪ B) = P(A) + P(B) - P(A ∩ B) = 1/2 + 1/6 - 1/12 = 4/6 = 2/3.
Similar Questions
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