Question

Find the probability of throwing atmost 2 sixes in 6 throws of a single die.

Answer 1

The repeated tossing of a die are Bernoulli trials. Let X represent the number of times of getting 6 in 6 throws of the die.

p=P (throwing 6 with die) = $\frac{1}{6}$ and q=1-p=1- $\frac{1}{6}=\frac{5}{6}$

Clearly, X has the probability distribution with n=6, $p=\frac{1}{6}$ and $q=\frac{5}{6}$.

$\therefore P(X=r)={}^n{C}_r{p}^r{q}^{n-r}={}^6{C}_r{\left(\frac{1}{6}\right)}^r{\left(\frac{5}{6}\right)}^{6-r}$

$\therefore$ P (throwing atmost 2 sixes) = P $(X\le 2)=P\left(0\right)+P\left(1\right)+P\left(2\right)$

=${}^6{C}_0{p}^0{q}^6+{}^6{C}_1{p}^1{q}^5+{}^6{C}_2{p}^2{q}^4=q^6+6p{q}^5+15p^2{q}^4$

=$q^4(q^2+6pq+15p^2)={\left(\frac{5}{6}\right)}^4[{\left(\frac{5}{6}\right)}^2+6\times \frac{1}{6}\times \frac{5}{6}+15\times {\left(\frac{1}{6}\right)}^2]$

=${\left(\frac{5}{6}\right)}^4\left[\frac{25+30+15}{36}\right]=\frac{70}{36}{\left(\frac{5}{6}\right)}^4=\frac{35}{18}.{\left(\frac{5}{6}\right)}^4$