Please i have an issue solving this question

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victorashie

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three men shoot at a terget, the probabiity that three men can hit the target is 1/3,2/5,1/2 respectively. find the probability if only one hit the target, then it was first. (consider each makes one shot)
 
Let's see if we can clarify the problem, which is not stated quite right. I take it to be something like this:

Three men, A, B, and C, shoot at a target, taking one shot each, one at a time. Suppose that the probability that each hits the target on any one shot is 1/3, 2/5, and 1/2, respectively. If only one of the three shots hits the target, what is the probability that ...​

Here I don't know what was intended! What does "it was first" mean? Perhaps, that it was A, the first of the men, who hit the target?

After you clarify the problem, please show us what you have tried, and where you are stuck. What do you know about conditional probability?
 
victorashie said:
three men shoot at a terget, the probabiity that three men can hit the target is 1/3,2/5,1/2 respectively. find the probability if only one hit the target, then it was first. (consider each makes one shot)
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Please show us what you have tried and exactly where you are stuck.

Please follow the rules of posting in this forum, as enunciated at:

https://www.freemathhelp.com/forum/threads/read-before-posting.109846/#post-486520

Please share your work/thoughts about this assignment.
 
Does each person shoot one at a time? Does A go 1st, then B, then C.
You want to know the probability that if only one person hits the target then it is 1st person?
 
So A hits the target 1/3 of the time, B hits it 2/5 of the time, and C hits it 1/2 of the time. We can, perhaps by using a "Venn diagram", divide this into 7 "mutually exclusive" cases:
A, B, and C all hit the target
A and B hit the target and C misses.
A and C hit the target and B misses.
B and C hit the target and A misses.
A hits the target and B and C miss.
B hits the target and A and C miss.
C hits the target and A and B miss.

The question asks for the probability that IF only one person hits that person is A.

Of course, only A hitting the target is precisely
"A hits the target and B and C miss".
The probability A hits the target is given as 1/3 while the probability B misses is 1- 2/5= 3/5 and the probability C misses is 1- 1/2= 1/2. What is the probability A hits while B and C miss?

But we want the probability this happens given that exactly one of the three hits. We need to divide by the probability exactly one hits which is the sum of the probabilities
"A hits the target and B and C miss" (this is, of course, the same as calculated above)
"B hits the target and A and C miss"
"C hits the target and A and B miss".

Calculate the probability of each of those and add them. Then divide the first probability, above, by that sum.
 
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