Question

If getting 5 or 6 in a throw of an unbiased die is a success and the random variable X denotes the number of successes in six throws of the die, find P (X ≥ 4).

 

Answer 1

Let X denote the number of successes, i.e. of getting 5 or 6 in a throw of die in 6 throws.
Then, X follows a binomial distribution with n =6;

\[p = \text{ of getting 5 or 6 } = \frac{1}{6} + \frac{1}{6} = \frac{1}{3}; q = 1 - p = \frac{2}{3}; \]
\[P(X = r) = ^{6}{}{C}_r \left( \frac{1}{3} \right)^r \left( \frac{2}{3} \right)^{6 - r} \]
\[P(X \geq 4) = P(X = 4) + P(X = 5) + P(X = 6)\]
\[ = ^{6}{}{C}_4 \left( \frac{1}{3} \right)^4 \left( \frac{2}{3} \right)^{6 - 4} +^{6}{}{C}_5 \left( \frac{1}{3} \right)^5 \left( \frac{2}{3} \right)^{6 - 5} + ^{6}{}{C}_6 \left( \frac{1}{3} \right)^6 \left( \frac{2}{3} \right)^{6 - 6} \]
\[ = \frac{1}{3^6}(60 + 12 + 1)\]
\[ = \frac{73}{729}\]