Question

In a game, a person flips a fair coin twice, and based on the number of heads observed, he will be allowed to shoot so many times (equal to the number of heads observed) on a target. Assume the probability of hitting a target in one shot is 0.2.

What is the probability of not hitting the target? Answer [The answer should be a number rounded to five decimal places, don't use symbols such as %]

Answer 1

Required probability = P(not hitting the target) = 1-P(hitting the target)

= 1 - [P(hitting the target when 0 heads were observed) + P(hitting the target when 1 heads were observed) +

P(hitting the target when 2 heads were observed)]

Now,

P(hitting the target when 0 heads were observed) = 0

P(hitting the target when 1 heads were observed) = 0.2

P(hitting the target when 2 heads were observed) = P(hitting the target in at least 1 shot) = P(not hitting the target in any of the 2 shots) = (1-0.2)(1-0.2) = 0.64

Also,

P(0 heads) = P(TT) = $\frac{1}{4}$ , P(1 heads) = P(HT, TH) =
4
, P(2 heads) = P(HH) = $\frac{1}{4}$

So, required probability = $1-[(\frac{1}{4})(0)+(\frac{1}{2})(0.2)+(\frac{1}{4})(0.64)]=0.74000$