# Binomial Distribution Problem

• Feb 24th 2010, 04:35 PM
Saggittarius
Binomial Distribution Problem
Matt has a special coin where, when tossed, heads comes up 62% of the time. He plays a gambling game with Roy. Each time the coin is tossed, if it comes up heads, Roy pays Matt $500; else if it comes up tails, Matt pays Roy$500. The coin is tossed 100 times.

What is the probability that Matt comes out ahead more than $9,000? I assume you would use a binomial calculator where p=.62, n=100, k=18(after dividing 9000/500).. But I'm not sure if that is correct since if it comes up tails Matt pays$500. So in that case, he is not just gaining money.. he also loses money. Is this correct?
• Feb 24th 2010, 08:17 PM
Soroban
Hello, Saggittarius!

Quote:

Matt has a special coin where, when tossed, heads comes up 62% of the time.
He plays a gambling game with Roy.
Each time the coin is tossed, if it comes up heads, Roy pays Matt $500. . . If it comes up tails, Matt pays Roy$500.
The coin is tossed 100 times.

What is the probability that Matt comes out ahead more than $9,000? If Matt wins 59 games (and loses 41), he will win exactly$9,000.

To win more than $9,000, Matt must win at least 60 games. So we have: . . .$\displaystyle \begin{array}{ccc}P(\text{60 wins}) &=& {100\choose60}(0.62)^{60}(0.38)^{40} \\ \\[-3mm]
P(\text{61 wins}) &=& {100\choose61}(0.62)^{61}(0.38)^{39} \\ \\[-3mm]
P(\text{62 wins}) &=& {100\choose62}(0.62)^{62}(0.38)^{38} \\
\vdots & & \vdots \\
P(\text{99 wins}) &=& {100\choose99}(0.62)^{99}(0.38)^1 \\ \\[-3mm]
P(\text{100 wins}) &=& {100\choose100}(0.62)^{100}(0.38)^0
\end{array}$The answer is the sum of these probabilities: .$\displaystyle \sum^{100}_{n=60} {100\choose n}(0.62)^n(0.38)^{100-n} \$