Explanation Calculation: Define the random variable Y as gambler wins exactly once before gambler loses initial capital. Let event W represents that gambler wins the game with probability P ( W ) = 0.10 and event L represents that gambler loses the game with probability P ( L ) = 0.90 It costs $10 for playing the game and $80 can be won if gambler wins the games. The gambler commences with amount $30. The first possibility that Y can take is WLLLLLLLLLL , because for playing the first game gambler spends $10, then gambler has $20 ( $ 30 − $ 10 ) . If gambler wins the first game then gambler gains profit of $80 and has a total amount of $100. With this amount gambler can play 10 games which are lost. A total of 11 games are played. The second possibility that Y can take is LWLLLLLLLLL , because for playing the first game gambler spends $10, then gambler has $20 ( $ 30 − $ 10 ) . If gambler loses the first game then goes for second game, then gambler has $10 ( $ 20 − $ 10 ) . If gambler wins the second game gambler gains profit of $80 and has a total amount of $90. With this amount gambler can play 9 games which are lost. A total of 11 games are played. The third possibility that Y can take is LLWLLLLLLLL , because for playing the first game gambler spends $10, then gambler has $20 ( $ 30 − $ 10 ) . If gambler loses the first game then goes for second game, then gambler has $10 ( $ 20 − $ 10 )

Mathematical Statistics with Applications

7th Edition

ISBN: 9780495110811

Author: Dennis Wackerly, William Mendenhall, Richard L. Scheaffer

Publisher: Cengage Learning

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Chapter 3, Problem 194SE

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Find the probability that gambler wins exactly once before gambler loses initial capital if gambler commences with amount $30.

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Chapter 3, Problem 193SE

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Find the probability that gambler wins exactly once before gambler loses initial capital if gambler commences with amount $30.

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Find the probability that gambler wins exactly once before gambler loses initial capital if gambler commences with amount $30.

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Chapter 3 Solutions