Question

A student has probability \(\dfrac{2}{3}\) of getting distinction in a test. Out of 5 tests, the probability that he gets distinction in at least 3 tests is

• A. \(\dfrac{112}{243}\)
• B. \(\dfrac{17}{81}\)
• C. \(\dfrac{131}{243}\)
• D. \(\dfrac{64}{81}\)

Hint:

Use binomial distribution \(P(X \geq r) = 1 - \sum_{k=0}^{r-1} P(X=k)\) for "at least" problems.

Answer 1

The Correct Option is C

This is binomial: \(n = 5, p = \dfrac{2}{3}\), need \(P(X \geq 3)\)
\[ = 1 - P(X \leq 2) = 1 - \left[P(0) + P(1) + P(2)\right] \Rightarrow 1 - \left[\binom{5}{0}(\dfrac{1}{3})^5 + \binom{5}{1}(\dfrac{2}{3})(\dfrac{1}{3})^4 + \binom{5}{2}(\dfrac{2}{3})^2(\dfrac{1}{3})^3\right] = \dfrac{131}{243} \]