Question

The probability of a man hitting a target is
1/4 :
i) If he fires 7 times what is the probability of his hitting the target at least twice?
ii) How many times must he fire so that the probability of his hitting the target at
least once is greater than 2/3 ?

Answer 1

i) Let X be the random variable denoting the number of times the man hits the target.

$P(X\geq 2) = 1-P(X<2)$

$P(X<2) = P(X =0)+P(X=1) = {7\choose0}\cdot(\frac{3}{4})^7 + {7\choose1}\cdot(\frac{3}{4})^6\cdot(\frac{1}{4})$

$P(X<2) = \frac{2187+5103}{16384} = 0.445$

$P(X \geq 2) = 1-0.445 = 0.555$

ii) Let us say the man shot N shots.

$P(X \geq 1) \geq\frac{2}{3}$ or equivalently $P(X = 0) \leq\frac{1}{3}$

This reduces to finding N such that $(\frac{3}{4})^N \leq \frac{1}{3}$

Now, taking log on both sides we get $N \geq \frac{\log{3}}{\log{4}-\log{3}} = 3.81$

So, he should fire at least 4 shots to have his probability of shooting at least once greater than $\frac{2}{3}$