Chapter 2.6, Problem 51E

A local fraternity is conducting a raffle where 50 tickets are to be sold—one per customer. There are three prizes to be awarded. If the four organizers of the raffle each buy one ticket, what is the probability that the four organizers win

1. a all of the prizes?
2. b exactly two of the prizes?
3. c exactly one of the prizes?
4. d none of the prizes?

To determine

Compute the probability that the four organizers win all of the prizes.

Answer to Problem 51E

The probability that the four organizers win all of the prizes is 0.0002.

Explanation of Solution

Probability:

For every event (say A) in sample space S, a number is called the probability of event A P(A), if it follows the following axioms.

$\begin{array}{l}Axiom1:P\left(A\right)\ge 0\\ Axiom2:P\left(S\right)=1\\ Axiom3:\text{For a pairwise mutually exclusive events in }S,\left(\text{say }{A}_1,A_2,\dots \right),\\ P\left(A_1\cup A_2\cup A_3\cup \dots \right)={\displaystyle \sum _{i=1}^{\infty }P\left(A_i\right)}\end{array}$

Note that there are 3 prizes in 50 tickets. That is the number of possibilities to select 3 winners are $C_3^{50}$.

The number of possibilities that four organizers win all of the prizes is $C_3^4$.

The probability that the four organizers win all of the prizes is calculated as follows:

$\begin{array}{c}P\left(alloftheprizes\right)=\frac{C_3^4}{C_3^{50}}\\ =\frac{\frac{4!}{3!\left(4-3\right)!}}{\frac{50!}{3!\left(50-3\right)!}}\\ =\frac{4}{19,600}\\ =0.0002\end{array}$

Therefore, the probability that the four organizers win all of the prizes is 0.0002.

To determine

Compute the probability that the four organizers win exactly two of the prizes.

Answer to Problem 51E

The probability that the four organizers win exactly two of the prizes is 0.0141.

Explanation of Solution

The number of possibilities that four organizers win exactly two of the prizes is $C_2^4\times C_{3-2}^{50-4}$.

The probability that the four organizers win exactly two of the prizes is calculated as follows:

$\begin{array}{c}P\left(exactlytwooftheprizes\right)=\frac{C_2^4{C}_{3-2}^{50-4}}{C_3^{50}}\\ =\frac{\frac{4!}{2!\left(4-2\right)!}\times \frac{46!}{1!\left(46-1\right)!}}{\frac{50!}{3!\left(50-3\right)!}}\\ =\frac{276}{19,600}\\ =0.0141\end{array}$

Therefore, the probability that the four organizers win exactly two of the prizes is 0.0141.

To determine

Compute the probability that the four organizers win exactly one of the prizes.

Answer to Problem 51E

The probability that the four organizers win exactly one of the prizes is 0.2112.

Explanation of Solution

The number of possibilities that four organizers win exactly one of the prizes is $C_1^4\times C_{3-1}^{50-4}$.

The probability that the four organizers win exactly one of the prizes is calculated as follows:

$\begin{array}{c}P\left(exactlyoneoftheprizes\right)=\frac{C_1^4\times C_{3-1}^{50-4}}{C_3^{50}}\\ =\frac{\frac{4!}{1!\left(4-1\right)!}\times \frac{46!}{2!\left(46-2\right)!}}{\frac{50!}{3!\left(50-3\right)!}}\\ =\frac{4,140}{19,600}\\ =0.2112\end{array}$

Therefore, the probability that the four organizers win exactly one of the prizes is 0.2112.

To determine

Compute the probability that the four organizers win none of the prizes.

Answer to Problem 51E

The probability that the four organizers win none of the prizes is 0.7745.

Explanation of Solution

The number of possibilities that four organizers win none of the prizes is $C_0^4\times C_{3-0}^{50-4}$.

The probability that the four organizers win none of the prizes is calculated as follows:

$\begin{array}{c}P\left(noneoftheprizes\right)=\frac{C_0^4\times C_{3-0}^{50-4}}{C_3^{50}}\\ =\frac{\frac{4!}{0!\left(4-0\right)!}\times \frac{46!}{3!\left(46-3\right)!}}{\frac{50!}{3!\left(50-3\right)!}}\\ =\frac{15,180}{19,600}\\ =0.7745\end{array}$

Therefore, the probability that the four organizers win none of the prizes is 0.7745.