SOLUTION: The probability of a shooter hitting a target is 2/3. How many minimum numbers of times must he fire so that the probability of hitting the target at least is more than 0.98?
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-> SOLUTION: The probability of a shooter hitting a target is 2/3. How many minimum numbers of times must he fire so that the probability of hitting the target at least is more than 0.98?
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Question 1022186: The probability of a shooter hitting a target is 2/3. How many minimum numbers of times must he fire so that the probability of hitting the target at least is more than 0.98? Answer by Theo(13342) (Show Source):
You can put this solution on YOUR website! if the probability of hitting the target is 2/3, then the probability of missing the target is 1/3.
the probability missing one time is 1/3.
the probability of missing two times is 1/3 * 1/3 = 1/9
the probability of missing three times is 1/3 * 1/3 * 1/3 = 1/27
etc.
if you want the probability of hitting the target to be at least .98, then the probability of missing the target has to be at most .02.
you are therefore looking for a probaiblity of missing that is less than .02.
it looks like he would have to shoot at least 4 times.
that's because 1/3 * 1/3 * 1/3 = 1/27 which is equal to .037...
multiply that by 1/3 again and you get .012... which is less than .02.
the probability of hitting the target at least one of those times has to be greater than .98.
what is the probability of hitting the target at least one out of 4 times?
that would be based on the binomial probability equation of p(x) = c(n,x) * p^x * q^(n-x).
n is the total amount of trials which is 4.
x is the number of times you want success.
n-x is the number of time you don't get success (you fail).
c(n,x) is the number of ways you can get sets of x out of n elements.
