A die is thrown twice and the sum of the numbers appearing is observed

1.	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?
Answer» Let E be the event that ‘number 4 appears at least once’ and F be the event that ‘the sum of the numbers appearing is 6’. Then, E = {(4,1), (4,2), (4,3), (4,4), (4,5), (4,6), (1,4), (2,4), (3,4), (5,4), (6,4)} and F = {(1,5), (2,4), (3,3), (4,2), (5,1)} We have P(E) = 11/36 and P(E) = 5/36 Also E ∩ F = {(2, 4), (4, 2)} Therefore P(E ∩ F) = 2/36 Hence, the required probability P(E/F) = (P(E ∩ F))/P(F) = (2/36)/(5/36) = 2/5

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?

Answer»

Let E be the event that ‘number 4 appears at least once’ and F be the event that ‘the sum of the numbers appearing is 6’.

Then, E = {(4,1), (4,2), (4,3), (4,4), (4,5), (4,6), (1,4), (2,4), (3,4), (5,4), (6,4)}

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

We have P(E) = 11/36 and P(E) = 5/36

Also E ∩ F = {(2, 4), (4, 2)}

Therefore P(E ∩ F) = 2/36

Hence, the required probability P(E/F) = (P(E ∩ F))/P(F) = (2/36)/(5/36) = 2/5

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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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