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4 years ago

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One concern of a gambler is that she will go broke before achieving her first win. Suppose that she plays a game in which the pr

obability of winning is .1 (and is unknown to her). It costs her $10 to play and she receives $80 for a win. If she commences with $30, what is the probability that she wins exactly once before she loses her initial capital?

1 answer:

Ray Of Light [21]4 years ago

4 0

Answer:

The probability that she wins exactly once before she loses her initial capital is 0.243.

Step-by-step explanation:

The gambler commences with $30, i.e. she played 3 games.

Let <em>X</em> = number of games won by the gambler.

The probability of winning a game is, <em>p</em> = 0.10.

The random variable <em>X</em> follows a Binomial distribution, with probability mass function:

$P(X=x)={n\choose x}p^{x}(1-p)^{n-x};\ x=0, 1, 2,...$

Compute the probability of exactly one winning as follows:

$P(X=1)={3\choose 1}(0.10)^{1}(1-0.10)^{3-1}\\=3\times0.10\times0.81\\=0.243$

Thus, the probability that she wins exactly once before she loses her initial capital is 0.243.

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