Question

Please answer all parts to this 4 part question

Suppose you have a six sided die. One face is printed with the number 1. Two faces are printed with the number 2. Three faces are printed with the number 3. You also have 3 coins: C_1, C_2, and C_3. C_1 will land Heads with probability 1/5. C_2 will land Heads with probability 1/3. C_3 will land Heads with probability 1/2. You roll the die. If the die lands with a 1 face up, flip coin C_1 If the die lands with a 2 face up, flip coin C_2 If the die lands with a 3 face up, flip coin C_3 1.(2.5 points) What is the probability that the tossed coin lands Heads? 2.(2.5 points) What is the probability that the die landed on a 2 given the tossed coin lands Heads? 3.(2.5 points) Every time you play the game, you win when you get Heads. What is the probability of winning the game for the first time on the 3rd attempt? 4.(2.5 points) Every time you play the game, you win when you get Heads. The amount you win is the number shown on the toss of the die. What are your expected winnings for playing the game?

Answer 1

ANSWER:

1)

Given

P(C1) = 1/6
P(C2) = 2/6
P(C3) = 3/6

Also, we are given that:
P(heads | C1) = 1/5
P(heads | C2) = 1/3
P(heads | C3) = 1/2

Using law of total probability, we get here:
P(heads) = P(heads | C1)P(C1) + P(heads | C2)P(C2) + P(heads | C3)P(C3)

P(heads) = (1/5)*(1/6) + (1/3)*(2/6) + (1/2)*(3/6) = 0.3944

Therefore 0.3944 is the required probability here.

2)

Given the toss is heads, probability that the die landed on a 2 is computed as:

P(coin_2|heads) = P(2 on die and heads) / P(heads)

(2/6) + (1/3) 03944

= 0.2817

Therefore 0.2817 is the required probability here.

3)

P(winning first on 3^rd attempt) = (1 - 0.2817)³ * 0.2817

= 0.1044

Therefore 0.1044 is the required probability here.

4)

The expected winnings here is computed as:

=  1(1/5)*(1/6) + 2(1/3)*(2/6) + 3(1/2)*(3/6)

= 0.0333 + 0.2222 + 0.75

= 1.005

Therefore expected winnings = 1.005

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