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Q.7. 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?

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Solution

Dear Student,

Let S be the event that the sum of the numbers appearing in 2 throws of the dice is 6. Then, S = {(1,5), (2,4), (3,3), (4,2), (5,1)} = 5 outcomes.Let F be the event that 4 appears at least once on the dice. Then, F = {(1,4), (4,1), (2,4), (4,2), (3,4), (4,3), (4,4), (5,4), (4,5), (6,4), (4,6)} = 11outcomes.The outcomes possible are {(1,1), (1,2), ... (1,6), (2,1), (2,2), ... (2,6), (3,1), (3,2), ... (3,6), (4,1), (4,2), ... (4,6), (5,1), (5,2), ... (5,6), (6,1), (6,2), ... (6,6)}. Thus, there are 36 possible outcomes.Now, SF = {(2,4), (4,2)}Conditional probability that 4 has appeared at least once given that the sum of the numbers in two throws of the dice is 6 = P(F/S) = P(SF)P(S) = 236536 = 25 = 0.4

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