Answer to Question #127379 in Statistics and Probability for Emmason

Question

There are three boxes, one of which contains a prize. A contestant is given two chances, such that if he chooses the wrong box in the first round, that box is removed from the selection and he then chooses between the two remaining boxes.
1. What is the probability that the contestant wins?
2. Does the contestant’s probability of winning increases on the second round?

Answer 1

Solution:

1.Let A be the event that the contestant wins.

Then P(A)=m/n.

" A contestant is given two chances". So, m=2.

"There are three boxes". So, n=3.

P(A)=2/3.

2.Let B₁ be the event that the contestant wins in the first round.

Then P(B₁)=m/n.

"There are three boxes, one of which contains a prize". So m=1, n=3.

P(B₁)=1/3.

Let B₂ be the event that the contestant wins in the second round.

Then P(B₂)=P(A | $\overline{B_1})$ P($\overline{B_1}$).

$P(\overline{B_1})=1-P(B_1)=1-1/3=2/3$.

P(A|$\overline{B_1})$ =m/n - is the probability of winning in the second round after the first unsuccessful attempt.

After the first unsuccessful attempt there are two boxes left, one of which contains a prize.

So, n=2, m=1

P(A|$\overline{B_1})$ =1/2>P(B₁)=1/3.

P(B₂)=1/2$\cdot$ 2/3=1/3.

Answer:

So P(B₁)=P(B₂), and the probability of the event that the contestant wins in the second round is equal to the probability of the event that the contestant wins in the first round.

But the probability of winning in the second round after the first unsuccessful attempt is greater than the probability of the event that the contestant wins in the first round.