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. If the person obtained two heads, what is the probability of hitting the target only once? [The answer should be a number rounded to five decimal places, don't use symbols such as %]

Answer 1

As the person got 2 heads, he would have 2 chances. There are two cases:

First, when he hits the target in the first shot and misses it in the last shot

Second, when he misses tha target in the first shot and hits it in the last shot

The probability of the missing the target is 1-0.2=0.8

So the required probability is

$=(0.2*0.8)+(0.8*0.2)$

== 0.16-016-032

So the required probability is 0.32000

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