# Example 36

Last updated at March 11, 2017 by Teachoo

Last updated at March 11, 2017 by Teachoo

Transcript

` Example 36 A and B throw a die alternatively till one of them gets a ‘6’ and wins the game. Find their respective probabilities of winning, if A starts first. Winning the game is getting a 6 on the die P(getting 6) = 16 P(not getting six) = 1 – P(getting six) = 1 – 16 = 56 1st throw by A: A gets a six P(A wins) = 16 2nd throw by B: A does not get 6, B gets six So, P(B wins) = 56 × 16 3rd throw by A: A does not get 6, B does not get 6, A gets six P(A wins) = 56 × 56 × 16 4th throw by B: A does not get 6, B does not get 6, A does not get 6, B gets six P(B wins) = 56 × 56 × 56 × 16 5th throw by A: A does not get 6, B does not get 6, A does not get 6, B does not get 6, A gets six P(A wins) = 56 × 56 × 56 × 56 × 16 and so on So, probability that A wins is P(A wins) = 16 + 56 × 56 × 16 + 56 × 56 × 56 × 56 × 16 + ……. = 16 + 562 16 + 564 16 + ............ = 161 − 562 = 161 − 2536 = 16 36 − 2536 = 16 1136 = 611 ∴ P(A wins) = 𝟔𝟏𝟏 and P(B wins) = 1 – P(A wins) = 1 – 611 = 𝟓𝟏𝟏

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About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He provides courses for Mathematics from Class 9 to 12. You can ask questions here.