The probability of a boy winning a game is \(\displaystyle\frac{2}{3}\). Let n denotes that the least number of times he must play the game so that the probability of winning the game at least once is more than 90% and X denotes that number of times he win the game. Hence n, mean, variance and standard deviation of random variable X are respectively:
(A) 3
(B) \(\displaystyle\frac{2}{3}\)
(C) 2
(D) 0.81
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (A), (C), (B), (D)
The probability of losing one game is \(1 - \frac{2}{3} = \frac{1}{3}\). The boy wins at least once unless he loses every game:
\( \displaystyle P(\text{at least one win in } n \text{ games}) = 1 - \left(\frac{1}{3}\right)^n \)
This must be more than 0.9:
\( \displaystyle \left(\frac{1}{3}\right)^n \lt 0.1 \)
For n = 2, \(\frac{1}{9} \approx 0.111\), which is not less than 0.1. For n = 3, \(\frac{1}{27} \approx 0.037 \lt 0.1\). So the least value is n = 3.
X, the number of wins in 3 games, is a binomial variable with n = 3, \(p = \frac{2}{3}\), \(q = \frac{1}{3}\):
\( \displaystyle \text{Mean} = np = 3\times\frac{2}{3} = 2 \)
\( \displaystyle \text{Variance} = npq = 3\times\frac{2}{3}\times\frac{1}{3} = \frac{2}{3} \)
\( \displaystyle \text{Standard deviation} = \sqrt{\frac{2}{3}} \approx 0.8165 \)
The value 0.81 in (D) is this standard deviation to two decimal places (cut off, not rounded).
So n = 3, mean = 2, variance = 2/3 and standard deviation ≈ 0.81, which are (A), (C), (B), (D) in that order.