A die is thrown twice and the sum of the numbers appearing is observed to be 6.
find the conditional probability that the number 4 has appeared at least once?  Choose one:  A) 2/5  B) 3/5  C) 3/7  D) 3/11  E) None
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 (F) = 5/36   Also EnF = {(2,4), (4,2)}   Therefore P(EnF) = 2/36   the required probability   P(E|F) = P(EnF)/P(F) = (2/36)/(5/36) = 2/5.   2/5)
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A die is thrown twice and the sum of the numbers appearing is observed to be 6.
find the conditional probability that the number 4 has appeared at least once?