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Home  >>  JEEMAIN and NEET  >>  Mathematics  >>  Class12  >>  Probability

Lets say you enter a Chess tournament. The probability of winning a game against half the players is 0.3, the probability of winning against one quarter of them is 0.4 and the probability of winning against the remaining one quarter is 0.5. Let's say you play against a randomly chosen opponent. What is your probability of winning?

(A) 0.4

(B) 0.375

(C) 0.33

(D) 0.3

1 Answer

Let $A_i$ be the event of playing an opponent of type $i$.
$A_1$ = playing the half of the opponent pool where the probability of winning P (winning|$A_1$) = 0.3
$A_2$ = playing the half of the opponent pool type 2, where the probability of winning P (winning|$A_2$) = 0.4
$A_3$ = playing the half of the opponent pool type 3, where the probability of winning P (winning|$A_3$) = 0.5
By total probability theorem, we get, P(winning) = P($A_1$)P(winning|$A_1$) + P($A_2$)P(winning|$A_2$) + P($A_3$)P(winning|$A_3$)
P (winning) = 50% $\times$ 0.3 + 25% $\times$ 0.4 + 25% $\times$ 0.5
P (winning) = 0.375.
answered Feb 12, 2014 by balaji.thirumalai
 

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