Question 1022186
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.


the following chart shows the calculations.


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