Four fair dice are thrown independently 27 times. Then the expected...

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$= p = \dfrac{2}{6}$

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

Number of trials$= n = 27$

Number of dice$= 4$

Using Binomial Probability,

P(at least 2 shows 3 or 5) $= C2 4 \cdot {\left(\dfrac{2}{6}\right)}^{2}{\left(\dfrac{4}{6}\right)}^{2} + C3 4 {\left(\dfrac{2}{6}\right)}^{3}\left(\dfrac{4}{6}\right) + C4 4 {\left(\dfrac{2}{6}\right)}^{4}$

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

Expectation is numerically same as mean.

$∴$ expectation$= np$

$= 27 \cdot \dfrac{11}{27} = 11$