Question

Question: What is the probability that Abby, Barry and Sylvia win the first, second, and third prizes, respectively, in a drawing, if 200 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 Abby, Barry and Sylvia win the first, second, and third prizes, respectively, in a drawing, if 200 people enter a contest and no one can win more than one prize is$P\left(E\right)=\frac{1}{7,880,400}$.

(b)The Probability that Abby, Barry and Sylvia win the first, second, and third prizes, respectively, in a drawing, if 200 people enter a contest and winning more than one prize is allowed is$P\left(E\right)=\frac{1}{8,000,000}$.

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 that Abby, Barry and Sylvia win the first, second, and third prizes, respectively, in a drawing, if 200 people enter a contest and no one can win more than one prize.

The Probability that any individual wins among 200 contestants$=\frac{1}{200}$.

The Probability that Abby wins first prize among 200 contestants$=\frac{1}{200}$.

Now, Abby can't win any more prizes.

The Probability that Barry wins the second prize among $(200-1)=199$contestants $=\frac{1}{199}$

Now, Barry can't win any more prizes.

The Probability that Sylvia wins the third prize among $(199-1)=198$contestants $=\frac{1}{198}$

The Probability that Abby, Barry and Sylvia win the first, second, and third prizes, respectively,

$$\begin{gathered} P\left(E\right)=\frac{1}{200}\times \frac{1}{199}\times \frac{1}{198} \\ P\left(E\right)=\frac{1}{7,880,400} \end{gathered}$$

The Probability that Abby, Barry and Sylvia win the first, second, and third prizes, respectively, in a drawing, if 200 people enter a contest and no one can win more than one prize is$P\left(E\right)=\frac{1}{7,880,400}$ .

Determine the probability (b) 

As per the problem we have been asked to find the probability that Abby, Barry and Sylvia win the first, second, and third prizes, respectively, in a drawing, if 200 people enter a contest and winning more than one prize is allowed.

The Probability that any individual wins among 200 contestants$=\frac{1}{200}$.

The Probability that Abby wins first prize among 200 contestants$=\frac{1}{200}$.

Now, Abby can still win more prizes.

The Probability that Barry wins the second prize among 200 contestants$=\frac{1}{200}$

Now, Barry can still win more prizes.

The Probability that Sylvia wins the third prize among contestants$=\frac{1}{200}$

The Probability that that Abby, Barry and Sylvia win the first, second, and third prizes, respectively,

$$\begin{gathered} P\left(E\right)=\frac{1}{200}\times \frac{1}{200}\times \frac{1}{200} \\ P\left(E\right)=\frac{1}{8,000,000} \end{gathered}$$

The Probability that Abby, Barry and Sylvia win the first, second, and third prizes, respectively, in a drawing, if 200 people enter a contest and winning more than one prize is allowed is$P\left(E\right)=\frac{1}{8,000,000}$ .