A target shooter - Binomial distribution

[Guest]

A targetshooter finds, that on average, the target is hit eight times out of every 10 and a bulls'eye is scored, on average, once every four rounds. Five rounds are fired. What is the probability that:

a) the target is hit each time?
b) at least two bull's eyes are scored?

 

[soroban]

Hello, americo74!

A targetshooter finds, that on average, the target is hit 8 times out of every 10
and a bulls-eye is scored, on average, once every 4 times.
Five rounds are fired. .What is the probability that:

a) the target is hit each time?
b) at least two bulls-eyes are scored?

We have: \(\displaystyle P(\text{target})\,=\,0.8,\;\;P(\text{bulls'eye})\,=\,0.25\)

\(\displaystyle (a)\;P(\text{5 targets})\:=\0.8)^5\:=\:\)0.32768


The opposite of "at least 2 bulls-eyes" is "0 bulls-eyes or 1 bulls-eye".
. . \(\displaystyle P(\text{0 bulls-eyes})\:=\0.75)^5\:\approx\:0.2373\)
. . \(\displaystyle P(\text{1 bulls-eye}) \:=\:\begin{pmatrix}5 \\1\end{pmatrix}(0.25)^1(0.75)^4\:\approx\:0.3955\)
Hence: \(\displaystyle \,P(\text{0 or 1 bulls-eye{)\:=\:0.2373\,+\,0.3955 \:=\:0.6328\)

Therefore: \(\displaystyle \,P(\text{{at least 2 bulls-eyes})\:=\:1\,-\,0.6328\:=\:\)0.3672