Question

A pair of dice is thrown seven times.
Getting a total of numbers on the two dice to be seven is considered as a success . Find the probability of atleast six successes.

Answer 1

Let $p=$ probability of getting a total of $7=\frac{6}{36}=\frac{1}{6}$

$q=1-p=1-\frac{1}{6}=\frac{5}{6}$

Here $n=7$

$P(X=r){=}^n{C}_r{q}^{n-r}{p}^r$

$P$(atleast $6$ successes)$=P\left(6\right)+P\left(7\right)$

${=}^7{C}_6{\left(\frac{5}{6}\right)}^{7-6}{\left(\frac{1}{6}\right)}^6{+}^7{C}_7{\left(\frac{5}{6}\right)}^{7-7}{\left(\frac{1}{6}\right)}^7$

$=7\times \frac{5}{6}\times {\left(\frac{1}{6}\right)}^6+1\times 1\times {\left(\frac{1}{6}\right)}^7$

$=(35+1){\left(\frac{1}{6}\right)}^7$

$=36\times {\left(\frac{1}{6}\right)}^6$

$=6^2\times {\left(\frac{1}{6}\right)}^6$

$={\left(\frac{1}{6}\right)}^5$