Answer to Question #189571 in Statistics and Probability for Mariam

Question

Toss a fair coin twice. You win Ghc 1 if at least one of the two tosses comes out heads.

(a) Assume that you play this game 300 times. What is, approximately, the probability that you win

at least Ghc 250?

(b) Approximately how many times do you need to play so that you win at least Ghc 250 with

probability at least 0.99?

Answer 1

Solution:

P(winning Ghc 1) = P(getting at least 1 heads in two tosses)$=\frac34$

Now, $n=300,p=\frac34,q=\frac14$

$\mu=np=225,\sigma=\sqrt{npq}=\sqrt{56.25}=7.5$

$X\sim N(\mu,\sigma)$

(a)$P(X\ge250)=1-P(X<250)=1-P(z<\frac{250-225}{7.5})$

$=1-P(z<3.3)=1-0.99957=0.00043$

(b) $P(X\ge250)\ge0.99$

$\Rightarrow 1-P(X<250)\ge0.99 \\ \Rightarrow 0.01\ge P(X<250) \\ \Rightarrow P(z<\frac{250-n(3/4)}{\sqrt{n(3/4)(1/4)}})\le 0.01$

$\Rightarrow \frac{250-n(3/4)}{\sqrt{n(3/4)(1/4)}}>0.50399 \\ \Rightarrow \frac{250-n(3/4)}{0.50399}>\sqrt{(3n/16)} \\ \Rightarrow (\frac{250-n(3/4)}{0.50399})^2>3n/16$

On solving, we get,

$n>338.68$

Or $n\ge339$