The probability of a man hitting a target is 0.25. He shoots 7 times. What is he probability of his hitting at least twice?

Given that, the probability of a man hitting a target is 0.25.


And he shoots 7 times in total.


Let p be the probability of hitting the target.


Then,


p = 0.25




Then, q be the probability of not hitting the target.


And we know that,


p + q = 1


q = 1 – p





Let X be a random variable representing the number of times the man hits the target out of n shoots.


Then, the probability of hitting the target r times out of n times is given by,


P (X = r) = nCrprqn-r


Here, n = 7


Putting the values of n, p, and q in the above equation, we get


…(i)


We need to find the probability of hitting the target at least twice.


This can be expressed as,


Probability = P (X ≥ 2)


Or


P (X ≥ 2) = 1 – P (X < 2)


P (X ≥ 2) = 1 – [P (X = 0) + P (X = 1)]


So, put r = 0, 1 in equation (i) to get P (X = 0) and P (X = 1) respectively and then, substitute in the above formula.


We get











Thus, the probability of hitting the target at least twice is 4547/8192.


52