Consider the following scenario when answering the following questions:
Imagine a game show on television where one lucky contestant is presented with four upside-down buckets that are numbered 1, 2, 3, and 4. Under one of the buckets is a $100 bill. Under each of the other three buckets is a $10 bill. After the game ends, the contestant will receive the amount of money that is under his or her bucket.
The host of the game show asks the contestant to choose one of the four buckets. After the contestant makes a choice, the host lifts one of the remaining three buckets to reveal a $10 bill under it. At this point, three buckets remain uncovered: the bucket that the contestant originally chose and the two buckets that were not uncovered by the host.
The host subsequently asks the contestant if he or she would like to keep the original bucket or change buckets to one of the two other buckets remaining.



If the contestant changes buckets from the original bucket to one of the other buckets remaining, what is the expected value of the game?


Want to confirm my answer of $40. Thank you.

Respuesta :

Answer:

It has been delivered that out of three buckets, under two buckets, there is a 1-oz gold bar and under the outstanding one there is 5-oz gold bar.

After a choice has been finished by the contender, host has elevated one bucket and under it 1-oz gold bar was originate.

So, after this, two buckets had been left.

Out of this two buckets, one covers 5-oz gold bar and second cover 1-oz gold bar.

Selecting out of these two buckets makes 50-50 chances of either winning 5-oz gold bar or 1-oz gold bar.

Compute prospect of participant winning 1-oz gold bar -

Number of buckets under which 1-oz gold bar is placed = 1

Total number of buckets = 2

P (winning 1-oz gold bar) = Number of buckets under which 1-oz gold bar is placed/Total number of buckets

P (winning 1-oz gold bar) = 1/2

Hence, the correct answer is option (3).