Question

The probability of a boy winning a game is \( \frac{2}{3} \). Let \( n \) denote 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 wins the game. Hence \( n \), mean, variance, and standard deviation of the random variable \( X \) are respectively:
(A) 3
(B) \( \frac{2}{3} \)  
(C) 2  
(D) 0.81  
Choose the correct answer from the options given below: 

• A. (A), (C), (B), (D)
• B. (A), (B), (C), (D)
• C. (B), (A), (D), (C)
• D. (C), (A), (D), (B)

Answer 1

The Correct Option is A

The objective is to determine the minimum number of games \( n \) a boy must play to achieve a win probability exceeding 90%. The probability of winning a single game is \( p = \frac{2}{3} \), and the probability of losing is \( 1-p = \frac{1}{3} \). Consequently, the probability of losing all \( n \) games is \( \left(\frac{1}{3}\right)^n \). The probability of winning at least one game is given by the complement of losing all games:

\(1-\left(\frac{1}{3}\right)^n > 0.9\)

Solving the inequality for \( n \):

• \(1-\left(\frac{1}{3}\right)^n > 0.9\)
• \(\left(\frac{1}{3}\right)^n < 0.1\)

Applying logarithms to both sides to isolate \( n \):

\(\log\left(\left(\frac{1}{3}\right)^n\right) < \log(0.1)\)

\(n \log\left(\frac{1}{3}\right) < \log(0.1)\)

Solving for \( n \):

\(n > \frac{\log(0.1)}{\log\left(\frac{1}{3}\right)}\)

Using base-10 logarithms for calculation:

• \(\log(0.1) \approx -1\)
• \(\log\left(\frac{1}{3}\right) \approx -0.4771\)

The inequality becomes:

\(n > \frac{-1}{-0.4771} \approx 2.09\)

The smallest integer value for \( n \) satisfying this condition is 3.

For a binomial distribution \( X \sim Bin(n,p) \) with \( n=3 \) and \( p=\frac{2}{3} \):

• Mean \( = np = 3 \times \frac{2}{3} = 2\)
• Variance \( = np(1-p) = 3 \times \frac{2}{3} \times \frac{1}{3} = \frac{2}{3}\)
• Standard deviation \( = \sqrt{\text{Variance}} = \sqrt{\frac{2}{3}} \approx 0.81\)

The computed values are: \( n = 3 \), mean = 2, variance = \(\frac{2}{3}\), and standard deviation \(\approx 0.81\).

The correct answer is indicated by (A), (C), (B), (D).