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.What is the probability that the tossed coin lands Heads?
2. What is the probability that the die landed on a 2 given the tossed coin lands Heads?
3. 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. 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?
1.
Probability that the tossed coin lands Heads
die lands 1 * C_1 heads + die lands 2 * C_2 heads + die lands 3* C_3 heads
= (1/6) * (1/5) + (2/6)*(1/3) + (3/6)*(1/2) = 1/30 + 1/9 + 1/4 = 2/15 + 17/36 = 71/180
Answer = 71/180
2.
According to bayes theorom
P(A|B) = P(B|A) / P(A)*P(B)
We need probability that the die landed on 2, given tossed coin
lands Head
Probability that the tossed coin is heads, given dice landed on 2
is 2/6 * 1/3 = 1/9
Probability of die landing on 2 = 2/6 = 1/3
Probability of the coin landing on heads is 71/180
So, answer is (1/9) * (1/3) * 180/71 = 20/213
3.
Probability of winning, i.e probability of getting heads on 3rd attemp is probability of getting tails on the first 2 attemps and heads on 3rd attemp.
= 109/180 * 109/180 * 71/180 = 1295029/5832000
4.
Expected winnings is
1 * die lands 1 * C_1 heads + 2 * die lands 2 * C_2 heads + 3 * die lands 3* C_3 heads
= 1/30 + 2 / 9 + 3 / 4 = (6 + 40 + 135)/180 = 181/180
Answer is
181/180
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