Question

Find chances of getting a sum of 9 atleast twice in ten throws with two dices.

Answer 1

$n\left(s\right)=$ number of element in throw of two dice $=6\times 6=36$

$n=10$

$P=$ success=getting sum $9$

$sum9$ occurs in cases

on $(4,5),(5,4),(3,6),(6,3)$

$\therefore p=\frac{4}{36}=\frac{1}{9}$

$q=1-p=1-\frac{1}{9}=\frac{8}{9}$

$x$=no of success

$\therefore$ required probability

$=P(x\ge z)$

$=1-p(x=0)-p(x=1)$

$=1{-}^{10}{C}_0\left(\frac{1}{9}{)}^0\right(\frac{8}{9}{)}^{10}{-}^{10}{C}_1\left(\frac{1}{9}{)}^1\right(\frac{8}{9}{)}^9$

$=1-1\times 1\times \frac{8^{10}}{9^{10}}-10\times \frac{1}{9}\times \frac{8^9}{9^9}$

$=1-(\frac{8}{9}{)}^{10}-10\times \frac{8^9}{9^{10}}$