Leroy and fred play chess at a club every wednesday. the probability that leroy will lose is .3, that he will stalemate is .5, and that he will win is .2. the probability that fred will lose is .25, that he will stalemate is .4, and that he will win is .35. what is the probability that at least one of them wins on a given wednesday?
a .55b .07c .45d .48e .62

A .55

Step-by-step explanation:

Given that Leroy and Fred play chess at a club every Wednesday. The probability that Leroy will lose is .3, that he will stalemate is .5, and that he will win is .2. The probability that Fred will lose is .25, that he will stalemate is .4, and that he will win is .35.

losestalewin

Leroy0.30.50.2

Fred0.250.40.35

Probability that atleast one wins = Prob Leroy wins + Prob fred wins - prob both wins

(using addition theorem on probability)

But prob of both winning is 0

Hence required prob = 0.2 + 0.35 = 0.55

Option a is right

answer: 20

step-by-step explanation: