Probability — Class 12 Maths Solution

:

Let $p$ be probability of getting a six $= \cfrac{1}{6}$

and $q$ be probability of not getting a six $= 1 - \cfrac{1}{6} = \cfrac{5}{6}$

$X$ has a binomial distribution with $n = 5,p = \cfrac{1}{6},q = \cfrac{5}{6}$

$P(\dot X = r){ = ^n}{C_r}{(q)^{n - r}}{p^r}$

$\Rightarrow P$ (Obtaining the third six in sixth throw)

$= P\left( {2{\rm{ success in the first }}5{\rm{ throws) }} \times P\left( {{\rm{ success in }}{6^{{\rm{th }}}}{\rm{ throw) }}} \right.} \right.$

$= {(^5}{C_2}{q^3}{p^2})p$
${ = ^5}{C_2}{\left( {\cfrac{5}{6}} \right)^3}{\left( {\cfrac{1}{6}} \right)^2} \times \cfrac{1}{6} = \cfrac{{625}}{{23328}}$