8. A game of chance consists of drawing one card out of ten at random. One card

Question

8. A game of chance consists of drawing one card out of ten at random. One card contains the number 70, in which a person, if they chose that card, wins $70.   A second card contains the number 5, if they chose that card the person wins $5. If any other card is chosen, the person losses $10. The game is then played by shuffling cards, facing them down, and having the person choose a card while face down. Once the card is chosen, the outcome is determined, and the card is put back and the deck reshuffled so one can play again.

Let the random variable X equal the amount of money won or lost.

X

$70

$5

-$10

P(X = x)

0.1

0.1

0.8

a. What is the probability that a person loses three games in a row?

b. On average what does the person expect to win or lose in the long run per game?

c. Calculate s, which is SD(X) to the nearest cent.

d. Consider the situation in which you play this game four times, and the total amount won or lost is summed up.   This creates a new distribution, the sum of the four independent outcomes. Let the random variable T = X1 + X2 + X3 + X4, where each subscript denotes the particular play order.   Find E(T).

e. Consider the situation in which you play this game four times, and the total amount won or lost is summed up.    This creates a new distribution, the sum of the four independent outcomes. Let the random variable T = X1 + X2 + X3 + X4, where each subscript denotes the particular play order. Find SD(T).

X

$70

$5

-$10

P(X = x)

0.1

0.1

0.8

Explanation / Answer

a)as probability of losing =0.8

probability that a person loses three games in a row =(0.8)3 =0.512

b)On average does the person expected to win=E(X)=0.1*70+0.1*5-0.8*10= $ -0.5

c)here E(X2) ==0.1*702+0.1*52+0.8*102=572.5

therefore SD(X) =(E(X2)-(E(X))2)1/2 =$23.92

d)E(T) =E(X1)+E(X2)+E(X3)+E(X4) =-0.5+-0.5+-0.5+-0.5 =$-2.0

e) SD(T)=(Var(X1)+Var(X2)+Var(X3)+Var(X4))1/2 =23.92*(4)1/2 =47.84

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