Chapter 13.2, Problem 61STP

To determine

Find the probability of at least $8$ successes and exactly $5$ failures.

Answer to Problem 61STP

The probability for both the cases are $0.05731$ & $0.1009$ .

Explanation of Solution

Given:

It is given in the question that a binomial distribution has a $40\%$ rate of success and there are $12$ trails.

Concept Used:

In this, use the concept of binomial distribution $P(x)=\frac{n!}{x!(n-x)!}{p}^x{q}^{n-x}$ where $q=1-p$

Calculation: Since, For at least $8$ successes, we take the trail after $8$ .

  $P=\frac{40}{100}=0.40$ , $q=1-0.40=0.60$

  $\begin{array}{l}$ =\frac{12!}{8!(12-8)!}{(0.40)}^8{(0.6)}^{12-8}+\frac{12!}{9!(12-9)!}{(0.40)}^9{(0.6)}^{12-9}+\frac{12!}{10!(12-10)!}{(0.40)}^{10}{(0.6)}^{12-10}+\frac{12!}{11!(12-11)!}{(0.40)}^{11}{(0.6)}^{12-11}$$ +\frac{12!}{12!(12-12)!}{(0.40)}^{12}{(0.6)}^{12-12}$$ =\frac{12!}{8!(4)!}{(0.40)}^8{(0.6)}^4+\frac{12!}{9!(3)!}{(0.40)}^9{(0.6)}^3+\frac{12!}{10!(2)!}{(0.40)}^{10}{(0.6)}^2+\frac{12!}{11!(1)!}{(0.40)}^{11}{(0.6)}^1$$ +\frac{12!}{12!(0)!}{(0.40)}^{12}{(0.6)}^0$=0.05731\end{array}$

For, exactly $5$ failures,

  $\begin{array}{l}$ =\frac{12!}{5!(12-5)!}{(0.40)}^{12-5}{(0.6)}^5$$ =\frac{12!}{5!(12-5)!}{(0.40)}^7{(0.6)}^5$=0.1009\end{array}$

Conclusion:

The probability for both are $0.05731$ & $0.1009$ .