SOLUTION: The probability that A hits a target is 1/4 and the probability that B hits the target is 1/8.
(i) if each fires twice, what is the probability that the target will be hit at leas
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-> SOLUTION: The probability that A hits a target is 1/4 and the probability that B hits the target is 1/8.
(i) if each fires twice, what is the probability that the target will be hit at leas
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Question 1157343: The probability that A hits a target is 1/4 and the probability that B hits the target is 1/8.
(i) if each fires twice, what is the probability that the target will be hit at least once?
(ii) if each fire once and the target is hit only once, what is the probability that A hits the target?
(iii) If A can fire only twice, how many times must B fire so that there is at least a 90% probability that the target will be hit? Answer by VFBundy(438) (Show Source):
You can put this solution on YOUR website! (i) if each fires twice, what is the probability that the target will be hit at least once?
Find the probability the target will not be hit at all, then subtract this from 1. Probability the target will not be hit at all: = = The probability the target will be hit at least once: = 583/1024
----------------------------------------------------------- (ii) if each fire once and the target is hit only once, what is the probability that A hits the target? Probability A hits the target and B misses: = Probability B hits the target and A misses: =
Probability the "hit" is made by A: = = 7/10
----------------------------------------------------------- (iii) If A can fire only twice, how many times must B fire so that there is at least a 90% probability that the target will be hit?
We want to find how many times B must fire to have at most a 10% probability the target will be missed. ≤ ≤ ≤ ≤ ≤
n = log(base 7/8) 8/45
n = 12.93
You want to round this up, so B will have to shoot 13 times to ensure the chance of the target being hit is 90%.
