Testbook Test Series: Probability - Probability

Question

The probability of a shooter hitting the target is $\frac{1}{3}$ and three shots at the bull's eye are needed to win the game. What could be the least number of shots for the shooter to give him more than half-chance of winning the game$?$

$A)5$            $B)6$              $C)7$             $D)8$

Comments

@Shobhit Joshi Yes, you are right, it will be just P(X=3), that is the probability of winning, but still this will hold true i.e., 

P(winning) = 1 - P(losing)

And he will lose if he shoots either 0/1/2 times and win only if he shoots exactly 3 times,

P(X=3) = 1 - (P(X=0) + P(X=1) + P(X=2))

Both the approaches then are correct.

@Balaji Jegan Please edit your answer for just P(X=3) at the left hand side.

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@Magma i solved same like Balaji ,I m getting 7 least value.

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@Prateek Raghuvanshi  because the probability that you evaluated is "in n tries the number of hits is >=3", that's not what the problem is asking for and that's not what the winning case is.

Loosing case is also when out of n tries n-2 didn't hit the bull's eye did you included that in your loosing cases.