Question 9 of 10 10.0 Points In the game where A, B and C play together...

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Question 9 of 10 10.0 Points In the game where A, B and C play together with dice. A rolls first, then B, then C and again A,

math Statistics-And-Probability

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Answer 1

Answer #1

Here, probability of winning is throwing 6 hence prob of throwing 6 is 1/6.

Therefore,

prob of winning = 1/6

prob of losing is 1-(1/6)= 5/6.

Provided that A start the game i.e A throws the die first and then B and then C and again A , B, ....and the game goes on.

When A throws and if 6 comes then the game will be stop and A will win. But if A loses then B and C will throw the die (here B and C will have to loose because according to question we want A to win the game) and then A will throw . and so on.

it will be like

A+ A'B'C'A+ A'B'C'A'B'C'A+..........

Where,

A denotes A win and A' denotes A loses.

Similarly for B denotes B win and B' denotes B loses.

C denotes C win and C' denotes C loses.

This will form an infinite geometric progession series

After converting it into probabilities we have,

(1/6) + (5/6)(5/6)(5/6)(1/6) + (5/6)(5/6)(5/6)(5/6)(5/6)(5/6)(1/6)+..........

probability of losing for A,B and C is throwing other than 6 i.e 5/6.

Here first term is 1/6 and common ration is 125/216.

Applying formula for the sum of an infinite geometric progression = (first term) / (1-common ratio) = (1/6) / (1-(125/216)) =

Probability of A winning is 36/91 = 0.3956

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Answer 2

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