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

a) For the dice as we are given here that there is one face with 1, two faces with 2 and three faces with number 3. Therefore the probability of getting 3 coins here are computed as:

P(c1) = 1/6
P(c2) = 2/6 = 1/3
P(c3) = 3/6 = 0.5

Also the conditional probabilities of getting heads for each of the 3 coins here is given to be:
P(H | c1) = 0.2,
P(H | c2) = 1/3
P(H | c3) = 1/2 = 0.5

Using law of total probability, we have here:
P(H) = P(H | c1)P(c1) + P(H | c2)P(c2) + P(H | c3)P(c3)

which is the sum product of the probabilities of getting a particular coin and the condtional probability of getting a head given that the perticular coin is selected.

= 0.2*(1/6) + (1/3)*(1/3) + 0.5*0.5 = 0.3944

Therefore 0.3944 is the required probability here.

b) Given that a heads came up, the probability that the die landed on a 2 is computed here using Bayes theorem here as:

P(c2|H) P(H|c2)P(c2) (1/3) P(H) 0.3944 = 0.2817

Therefore 0.2817 is the required probability here.

c) Probability of getting a head first time on third attempt is computed here as:

= Probability of getting tails on first two tosses * Probability of getting a heads on the third toss

= [1 - P(H)]²*P(H)

= (1 - 0.3944)²*0.3944 = 0.1446

Therefore 0.1446 is the required probability here.

d) The expected winnings for the game here is computed as:

= 0*(1 - P(H) ) + 1*P(c1 | H) + 2*P(c2 | H) + 3*P(c3 | H)

= [0.2*(1/6)] + 2*(1/9) + 3*0.25

= 1.0056

Therefore 1.0056 is the expected winning amount here.