A die is thrown 3 times. Let Event A be "4 on the 3rd throw" and Event B be "6 on the 1st and 5 on the 2nd throw". What is the probability of A given - Mathematics MCQ Mathematics Mock Test - 8 2026 | ExamTest.live

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A die is thrown 3 times. Let Event A be "4 on the 3rd throw" and Event B be "6 on the 1st and 5 on the 2nd throw". What is the probability of A given that B has already occurred?

Solution & Step-by-step Explanation

We need to find P (A ∣ B) = \frac{P ( A \cap B )}{P ( B )} .Total outcomes = 6 \times 6 \times 6 = 216 .Event B: 6 on 1st, 5 on 2nd. The outcomes are (6, 5, 1), (6, 5, 2), (6, 5, 3), (6, 5, 4), (6, 5, 5), (6, 5, 6) .So, n (B) = 6 and P (B) = 6/216 .Event A \cap B : 6 on 1st, 5 on 2nd, AND 4 on 3rd.There is only one such outcome: (6, 5, 4) .So, n (A \cap B) = 1 and P (A \cap B) = 1/216 .Calculating P (A ∣ B) : P (A ∣ B) = \frac{1/216}{6/216} = \frac{1}{6}

Try it yourself before checking the explanation above.

A die is thrown 3 times. Let Event A be "4 on the 3rd throw" and Event B be "6 on the 1st and 5 on the 2nd throw". What is the probability of A given that B has already occurred?

1/108

1/36

1/6

1/216

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