Question

Question: What is the probability that that Bo, Colleen, Jeff, and Rohini win the first, second, third, and fourth prizes, respectively, in drawing if 50 people enter a contest and

(a) no one can win more than one prize.

(b) winning more than one prize is allowed.

Short answer

Answer

(a) The Probability that Bo, colleen, Jeff, and Rohini win the first, second, third, and fourth prizes, respectively, in a drawing, if 50 people enter a contest and no one can win more than one prize is$P\left(E\right)=\frac{1}{5,527,200}$ .

(b) The Probability that Bo, Colleen, Jeff, and Rohini win the first, second, third, and fourth prizes, respectively, in a drawing, if 50 people enter a contest and winning more than one prize is allowed is $P\left(E\right)=\frac{1}{62,500,000}$.

Step-by-step solution

 Given  

(a) No one can win more than one prize.

(b) Winning more than one prize is allowed.

The Concept of Probability

If $\mathit{S}$represents the sample space and$\mathit{E}$represents the event. Then the probability of occurrence of favourable event is given by the formula as below:

$\mathit{P}\mathbf{\left(}\mathit{E}\mathbf{\right)}\mathbf{=}\frac{\mathbf{n}\mathbf{\left(}\mathbf{E}\mathbf{\right)}}{\mathbf{n}\mathbf{\left(}\mathbf{S}\mathbf{\right)}}$

Determine the probability (a)

As per the problem we have been asked to find the probability Bo, Colleen, Jeff, and Rohini win the first, second, third, and fourth prizes, respectively, in a drawing, if 50 people enter a contest and no one can win more than one prize.

The Probability that any individual wins among 50 contestants$=\frac{1}{50}$.

The Probability that Bo wins first among 50 contestants$=\frac{1}{50}$.

Now, Bo can't win any more prizes.

The Probability that Colleen wins the second prize among$(50-1)=49$

contestants$=\frac{1}{49}$

Now, Colleen can't win any more prizes.

The Probability that Jeff wins the third prize among$(49-1)=48$

contestants$=\frac{1}{48}$

Now, Jeff can't win any more prizes.

The Probability that Rohini wins the fourth prize among $(48-1)=47$contestants $=\frac{1}{47}$

The Probability that Bo, Colleen, Jeff, and Rohini win the first, second, third, and fourth prizes, respectively,

$$\begin{gathered} P\left(E\right)=\frac{1}{50}\times \frac{1}{49}\times \frac{1}{48}\times \frac{1}{47} \\ P\left(E\right)=\frac{1}{5,527,200} \end{gathered}$$

The Probability that Bo, Colleen, Jeff, and Rohini win the first, second, third, and fourth prizes, respectively, in a drawing, if 50 people enter a contest and no one can win more than one prize is$P\left(E\right)=\frac{1}{5,527,200}$ .

Determine the probability (b)

As per the problem we have been asked to find the probability that Bo, Colleen, Jeff, and Rohini win the first, second, third, and fourth prizes, respectively, in a drawing, if 50 people enter a contest and winning more than one prize is allowed.

The Probability that any individual wins among 50 contestants$=\frac{1}{50}$.

The Probability that Bo wins the first prize among 50 contestants$=\frac{1}{50}$.

Now, Bo can still win more prizes.

The Probability that Colleen wins the second prize among 50 contestants$=\frac{1}{50}$

Now, Colleen can still win more prizes.

The Probability that Jeff wins the third prize among 50 contestants$=\frac{1}{50}$

Now, Jeff can still win more prizes.

The Probability that Rohini wins the fourth prize among 50 contestants$=\frac{1}{50}$

The Probability that that Bo, Colleen, Jeff, and Rohini win the first, second, third, and fourth prizes, respectively,

$$\begin{gathered} P\left(E\right)=\frac{1}{50}\times \frac{1}{50}\times \frac{1}{50}\times \frac{1}{50} \\ P\left(E\right)=\frac{1}{6,250,000} \end{gathered}$$

The Probability that Bo, Colleen, Jeff, and Rohini win the first, second, third, and fourth prizes, respectively, in a drawing, if 50 people enter a contest and winning more than one prize is allowed is$P\left(E\right)=\frac{1}{6,250,000}$ .