If a fair die is rolled 4 times, then what is the probability that there are exactly 2 sixes?

Concept:

Binomial distribution:

\(b\;\left( {x;n\;,\;p} \right) = \;{\;^n}{C_x} \times {p^x} \times {q^{n - x}}\) where p is the probability of success, q is the probability of failure, n is the total no. of attempts and x is the no. of successful attempts.

Calculation:

Given: A fair die is rolled 4 times.

Let p represent the probability of getting 6 when a dice is rolled = 1 / 6

Let q represent the probability of not getting 6 when a dice is rolled

= 1 – (1 / 6) = 5 / 6

As we know that, according to binomial distribution:

\(b\;\left( {x;n\;,\;p} \right) = \;{\;^n}{C_x} \times {p^x} \times {q^{n - x}}\)

Here, n = 4, x = 2, p = 1 / 6 and q = 5 / 6.

So, the probability of getting exactly 2 sixes when a fair dice is rolled 4 times

⇒ \(b\;\left( {x;n\;,\;p} \right) = \;{\;^4}{C_2} \times {\left( {\frac{1}{6}} \right)^2} \times {\left( {\frac{5}{6}} \right)^2} = \frac{{25}}{{216}}\)