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Question -

A fair coin with 1 marked on one face and 6 on the other and a fair die are both tossed. Find the probability that the sum of numbers that turn up is (i) 3 (ii) 12



Answer -

Let us assume that 1, 2, 3, 4, 5 and 6 are the possible outcomes when the die is thrown.

So, the sample space S = {(1, 1), (1, 2), (1, 3), (1, 4),(1, 5), (1, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5),(6, 6)}

Then, n(S) = 12

(i) Let us assume тАШPтАЩ be the event having sum of numbers as3.

P = {(1, 2)},

Then, n (P) = 1

P(Event) = Number of outcomes favorable to event/ Totalnumber of possible outcomes

тИ┤P(P) = n(P)/n(S)

= 1/12

(ii) Let us assume тАШQтАЩ be the event having sum of number as12.

Then Q = {(6, 6)}, n (Q) = 1

P(Event) = Number of outcomes favorable to event/ Totalnumber of possible outcomes

тИ┤P(Q) = n(Q)/n(S)

= 1/12

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