Question

Roll a dice n times and let X be the number of times you roll 6. Assume that n is large.

(a) Compute the expectation E[X].

(b) Write down an approximation, in terms on n and , of the probability that X diers from its

expectation by less than 10%.

(c) How large should n be so that the probability in (b) is larger than 0.99?

Answer 1

(a) Let X$_i$ be the number on the face of die

Let X be the number you roll 6-

Therefore X have the random value as-

       $X=\sum_{i=1}^{n} X_i.$

By Linearity of Expectation-

  $E[X]=\sum_{i=1}^{n} E[X_i].$

Also $E[X_i]=\sum_{j=1}^6jP[X_i=j]=\sum_{j=1}^6j(\dfrac{1}{6})=\dfrac{1}{6}\times \dfrac{6(7)}{2}=\dfrac{7}{2}$

Also, $\sum_{j=1}^{n}j=\dfrac{n(n+1)}{2}$

Therefore, $E[X]=\dfrac{7}{2}\times\dfrac{n(n+1)}{2}= \dfrac{7n(n+1)}{4}$

(b) The probability that X differs from its expectation by less than 10% is given by-

$=\sum_{i=1}^{n} X.E(X)=\sum_{i=1}^{n}X.\dfrac{7n(n+1)}{4}$

(c) Probability in b is larger than 0.99-

$\dfrac{7n(n+1)}{4}>0.99\\[9pt] \Rightarrow 7n^2+7n-3.76>0$

on solving we get, $n=-0.38,3.76$

So The value of n must be 4.