Mathematics - Probability Question with Solution | TestHub

There are only two outcomes, success which stands for getting either 3 or 5, and failure, stands for getting any other outcome.

Probability for success$=p=\frac{2}{6}$

Probability for failure$=q=\frac{4}{6}$

Number of trials$=n=27$

Number of dice$=4$

Using Binomial Probability,

P(at least 2 shows 3 or 5) $={}^4\mathrm{C}_2·{\left(\frac{2}{6}\right)}^2{\left(\frac{4}{6}\right)}^2+{}^4\mathrm{C}_3{\left(\frac{2}{6}\right)}^3\left(\frac{4}{6}\right)+{}^4\mathrm{C}_4{\left(\frac{2}{6}\right)}^4$

$=\frac{384+128+16}{6^4}=\frac{11}{27}$

Expectation is numerically same as mean.

$\therefore$ expectation$=np$

$=27·\frac{11}{27}=11$