Chapter 10.5, Problem 73E

To determine

To calculate Probability that at least $2$ of the $6$ cast members bring the same item.

Answer to Problem 73E

Probability that at least $2$ of the $6$ cast members bring the same item is $0.3$ .

Explanation of Solution

Given information:

Number of cast member $=6$

List of items $=10$

Calculation:

Number of possible outcomes as the number of combinations of $10$ items taken $1$ at a time, $C{}_1$ because the order is not important.

Need to apply combination,

The expression for combination for $n$ objects taken $r$ at a time is as given by,

  $C{}_r=\frac{n!}{\left(n-r\right)!.r!}$

The required number of combinations is as below,

  $\begin{array}{l}C{}_r=\frac{n!}{\left(n-r\right)!.r!}\\ C{}_r=\frac{10!}{\left(10-1\right)!.1}\\ C{}_r=\frac{10.9!}{9!.1}\\ C{}_r=10\end{array}$

To find the favorable outcomes, that is $2$ of the $6$ cast members bring the same food item.

To find the probability,

  $P$ ( at least $2$ cast members bring some food item),

  $=1-P\left(C_{lessthan2}\right)$

Where,

  $C_{lessthan2}=$ at least $2$ cast members bring some food item.

  $\begin{array}{l}=1-P\left(C_{lessthan2}\right)\\ =1-\left(\frac{C{}_0}{10}+\frac{C{}_1}{10}\right)\\ =1-0.7\\ =0.3\end{array}$

Thus, answer is $0.3$ .