Question

In a game, a man wins a rupee for a six & loses a rupee for any other number when a fair die is thrown. The man decided to throw a die thrice but to quit as and when he gets a six. Find the expected value of the amount he wins /loses.

#Self attempted and explained answer needed. ​

Answer 1

Hello Mate,

Points to remember in this question:

• A man wins a rupee If number is 6.

• A man loses a rupee if number is not a 6.

• The dice is thrown thrice.

• The man will quit when he gets a 6.

SOLUTION:

We Know the man will quit when he gets a six in the dice so we have four cases here:

• 1) He gets a six in the first throw

In the first throw if he gets a six, then he will be awarded ₹1 and will quit.

• 2) He gets a six in second throw

If he gets six in second throw, it means that in first throw he had lost ₹1 and gained it again in the second throw and he will leave. So he won't be awarded with any money.

• 3) He gets a six on the third throw

This means he didn't get six in first or second throw so he will be in loss of ₹2.

Then If he gets a six, then one more rupee is gained, so he will be in a loss of ₹1 which is nothing but (- 1).

• 4) Didn't get six at all throws

If he didn't get six at all throws, it's obvious that he lost ₹3, which is nothing but (-3).

Now let's check probability of getting six in each cases,

Probability of getting 6 = 1/6

Probability of getting other number = 5/6

P(1) = 1/6

P(2) = (5/6) x (1/6)

= 5/36

P(3) = (5/6) x (5/6) x (1/6)

= 25/216

P(4) = (5/6) x (5/6) x (5/6)

= 125/216

So the maximum value the man would lose or gain is:

(Probability of being awarded) x (value of the award)

As there are total of four ways he can get the award or loss,

Add up the formula four times we get,

1(1/6) + 0(5/36) + [-1(25/216)] +

[-3(125/216)]

By solving we get,

-91/54

-91/54= -1.68

Hope it helps:)

Answer 2

Possible cases :

Case I :

He gets a six in I throw, then probability = 1/6.

Amount received = 1.

Case II :

He does not get six in first throw gets six in 2nd throw.

Probability = (5/6) * (1/6) = 5/36

Amount received = -1 + 1 = 0

Case III :

He does not get six in 1st, 2nd throws

Probability = (5/6) * (5/6) * (1/6) = 25/216

Amount received = -1 - 1 + 1 = -1

Case IV :

He does not get six in 1st, 2nd throws, 3rd throws

Probability = (5/6) * (5/6) * (5/6) = 125/216

Amount received = -1 - 1 - 1 = -3

Expected value he wins,

= (1/6) * 1 + (5/6 * 5/6) * 0 + (25/216) * (-1) + (125/216) * (-3)

=> -91/54

Therefore, Expected value he loses is 91/54.

#BeBrainly