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 .55
B .07
C .45
D .48
E .62
To find the probability that at least one of them wins on a given Wednesday, we can find the complement of the probability that both Leroy and Fred either lose or stalemate.
Let's start by finding the probability that both Leroy and Fred either lose or stalemate:
Probability of Leroy losing or stalemate = Probability of Leroy losing + Probability of Leroy stalemate = 0.3 + 0.5 = 0.8
Probability of Fred losing or stalemate = Probability of Fred losing + Probability of Fred stalemate = 0.25 + 0.4 = 0.65
Now, to find the probability that both Leroy and Fred either lose or stalemate, we multiply their individual probabilities together:
Probability that both Leroy and Fred either lose or stalemate = Probability of Leroy losing or stalemate * Probability of Fred losing or stalemate = 0.8 * 0.65 = 0.52
Finally, to find the probability that at least one of them wins, we take the complement of the probability that both Leroy and Fred either lose or stalemate:
Probability that at least one of them wins = 1 - Probability that both Leroy and Fred either lose or stalemate = 1 - 0.52 = 0.48
Therefore, the probability that at least one of them wins on a given Wednesday is 0.48.
The answer is D) 0.48.