a.
Probability that the player wins on the first roll = Probability
that the player throws 7 or 11
We can get 7 as (1,6), (2, 5), (3, 4), (4, 3) , (5, 2) , (6, 1) out of 36 combinations.
So, probability to throw 7 = 6/36
We can get 11 as (5,6), (6, 5) out of 36 combinations.
So, probability to throw 11 = 2/36
Probability that the player wins on the first roll = Probability
that the player throws 7 or 11
= 6/36 + 2/36 = 8/36 = 2/9
b.
Probability that the player wins on the first roll = Probability that the player throws 2 or 3
We can get 2 as (1, 1) out of 36 combinations.
So, probability to throw 2 = 1/36
We can get 3 as (2,1), (1, 2) out of 36 combinations.
So, probability to throw 3 = 2/36
Probability that the player wins on the first roll = Probability
that the player throws 7 or 11
= 1/36 + 2/36 = 3/36 = 1/12
c.
If the player throws 4, the player wins on the next roll if he gets
4 on next roll.
We can get 4 as (1, 3), (2, 2) , (3, 1) out of 36 combinations.
So, probability to throw 4 = 3/36 = 1/12
If the player throws 4, probability that the player wins on the
next roll = 1/12
d.
If the player throws 4, the player loses on the next roll if he gets 7 on next roll.
If the player throws 4, probability that the player wins on the next roll = 6/36 = 1/6
e.
If the player throws 4, the game ends if player wins or loses on
the next roll. That is he gets 4 or 7 on next roll.
If the player throws 4, probability that the game ends on the next
roll = 1/12 + 1/6 = 3/12 = 1/4
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