Question

A pair of dice is thrown $6$ times. If getting a total of $9$ is considered as a success, what is the probability of getting at least 5 successes ?

Answer 1

$n=6$ $n\left(s\right)=36$

as $sum=9$ is required

which can be obtained in following cases.

$3+6=9$ $4+5=9$

$6+3=9$ $5+4=9$

So only four cases $(3,6),(6,3),(4,5)$ and $(5,4)$ will result into sucess

$p$ =probability of success.

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

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

$x=$ no of success

$=P(x\ge 5)=P(x=5)+P(x=6)$

${=}^6{C}_5\left(\frac{1}{9}{)}^5\right(\frac{8}{9}\left){+}^6{C}_6\right(\frac{1}{9}{)}^6(\frac{8}{9}{)}^0$

$=\frac{6\times 8}{9^6}+1\times \frac{1}{9^6}$

$=\frac{49}{9^6}$