Chapter 11, Problem 17CT

Solution

To calculate: The probability that exactly two students buy rugs and at least two students buy rings.

The probability that exactly two students buy rings is 0.3087 and the probability that at least two students buy rings is 0.4718.

Given:

The probability that a student buys class ring $=0.3$

Total number of students selected at random $n=5$

Formula used:

The probability for x success among n independent trials where probability of each success is p is calculated using binomial probability distribution as shown below

  $\begin{array}{l}$ P(X=x)=\left(\begin{array}{c}n\\ x\end{array}\right)p^x{(1-p)}^{n-x}$\left(\begin{array}{c}n\\ x\end{array}\right)=\frac{n!}{x!\left(n-x\right)!}\end{array}$

Calculation:

The probability that exactly two students buy a class ring is calculated as shown below

  $\begin{array}{l}$ P(X=x)=\left(\begin{array}{c}n\\ x\end{array}\right)p^x{(1-p)}^{n-x}$$ P(X=2)=\left(\begin{array}{c}5\\ 2\end{array}\right){0.3}^2{(1-0.3)}^{5-2}$$ P(X=2)=\frac{5!}{2!\left(5-2\right)!}\times {0.3}^2\times {0.7}^3$P(X=2)=0.3087\end{array}$

The probability that exactly two students buy a class ring is 0.3087.

The total sum of all possible outcomes is always 1, thus subtracting the probabilities that exactly 0 rings bought and 1 ring bought from 1, the probability that at least 2 rings are bought is calculated as shown below

  $\begin{array}{l}P(X\ge 2)=1-P(X=0)-P(X=1)\\ P(X\ge 2)=1-\left(\left(\begin{array}{c}5\\ 0\end{array}\right){0.3}^0{(1-0.3)}^{5-0}\right)-\left(\left(\begin{array}{c}5\\ 1\end{array}\right){0.3}^1{(1-0.3)}^{5-1}\right)\\ P(X\ge 2)=1-\left(\frac{5!}{0!\left(5-0\right)!}\times {0.3}^0\times {0.7}^5\right)-\left(\frac{5!}{1!\left(5-1\right)!}\times {0.3}^1\times {0.7}^4\right)\\ P(X\ge 2)=1-0.16807-0.36015\\ P(X\ge 2)=0.47178\end{array}$

The probability that at least two students buy a class ring is 0.4718 (rounded to 4 decimal places).