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Problem 92: An archer hits a bull’s-eye with a probability of 0.09, and the results of different attempts can be taken to be independent of each other.

Problem 92: An archer hits a bull’s-eye with a probability of 0.09, and the results of different attempts can be taken to be independent of each other. If the archer shoots nine arrows, calculate the probability that: a. Exactly two arrows score bull’s-eyes. b. At least two arrows score bull’s-eyes. What is the expected number of bull’s-eyes scored?

Solution:


Given that $n=9,$ and $p=0.09$
a. Exactly two arrows score bull’s-eyes

$P(x= 2) = b(2;9,0.09)$
$b(x;n,p)=\left(\begin{array}{c}n\\ x\end{array}\right){p}^{x}(1-p)^{n-x}$
$P(x= 2) = b(2;9,0.09) = \left(\begin{array}{c}9\\ 2\end{array}\right){0.09}^{2}(1-0.09)^{9-2} = 0.01568$
b. At least two arrows score bull’s-eyes.
$P(x \ge 2) = 1- P(x<2) = 1 - [b(0;9,0.09)+b(1;9,0.09)]$
$b(x;n,p)=\left(\begin{array}{c}n\\ x\end{array}\right){p}^{x}(1-p)^{n-x}$
$b(0;9,0.09) = \left(\begin{array}{c}9\\ 0\end{array}\right){0.09}^{0}(1-0.09)^{9-0} = 0.42793$
$b(1;9,0.09) = \left(\begin{array}{c}9\\ 1\end{array}\right){0.09}^{1}(1-0.09)^{9-1} = 0.380905$
$P(x \ge 2) = 1 - [b(0;9,0.09)+b(1;9,0.09)] $
$= 1-(0.42793+0.380905) = 1- 0.8088 = 0.1912$
Expected number of bull's eyes scored.
$E(x) = np = 9*0.09 = 0.81$

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