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 problem involves finding the least number of times \( n \) the boy must play such that he wins at least once with a probability greater than 90%. The probability of winning one game is given by \( p = \frac{2}{3} \). Consequently, the probability of losing one game is \( 1-p = \frac{1}{3} \). Thus, the probability of losing all \( n \) games is \( \left(\frac{1}{3}\right)^n \). We want the probability of winning at least one game, which can be described as: 
\(1-\left(\frac{1}{3}\right)^n > 0.9\)
Solving this inequality:

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

To find \( n \), take the logarithm of both sides:
\(\log\left(\left(\frac{1}{3}\right)^n\right) < \log(0.1)\)
\(n \log\left(\frac{1}{3}\right) < \log(0.1)\)
Now solve for \( n \):
\(n > \frac{\log(0.1)}{\log\left(\frac{1}{3}\right)}\)
Calculating the values using base-10 logarithms:

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

Then:
\(n > \frac{-1}{-0.4771} \approx 2.09\)
Thus, the smallest integer \( n \) is 3.
For a binomial distribution \( X \sim Bin(n,p) \):

• 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 final results are:
\( n = 3 \), mean = 2, variance = \(\frac{2}{3}\), standard deviation \(\approx 0.81\).

Hence, the correct answer is (A), (C), (B), (D).

Answer 2

Finding $n$ such that $P(\text{winning at least once}) > 0.9$. The probability of losing a single game is:
$P(loss) = 1 - \frac{2}{3} = \frac{1}{3}$
The probability of losing all $n$ games is:
$P(losing all  n \text{ games}) = \left(\frac{1}{3}\right)^n$
The probability of winning at least once is:
$P(winning at least once) = 1 - P(losing all  n  games) = 1 - \left(\frac{1}{3}\right)^n$
We require:
$1 - \left(\frac{1}{3}\right)^n > 0.9$
$\left(\frac{1}{3}\right)^n < 0.1$
Taking the natural logarithm on both sides:
$n \ln\left(\frac{1}{3}\right) < \ln(0.1)$
Since $\ln\left(\frac{1}{3}\right) < 0$, dividing by it reverses the inequality:
$n > \frac{\ln(0.1)}{\ln\left(\frac{1}{3}\right)}$
Using approximate values:
$\ln(0.1) \approx -2.3026$, $\ln\left(\frac{1}{3}\right) \approx -1.0986$
$n > \frac{-2.3026}{-1.0986} \approx 2.1$
Thus, the least integer $n$ is:
$n = 3$
Mean, variance, and standard deviation of $X$. The number of wins $X$ follows a Binomial distribution with parameters $n = 3$ and $p = \frac{2}{3}$. The mean, variance, and standard deviation are given by:
- Mean:
Mean $= n \cdot p = 3 \cdot \frac{2}{3} = 2$
- Variance:
Variance $= n \cdot p \cdot (1 - p) = 3 \cdot \frac{2}{3} \cdot \frac{1}{3} = \frac{2}{3} \approx 0.67$
- Standard deviation:
Standard Deviation $= \sqrt{\text{Variance}} = \sqrt{\frac{2}{3}} \approx 0.816$
Matching options. From the above calculations:
$n = 3$ corresponds to (A).
Mean $= 2$ corresponds to (C).
Variance $= \frac{2}{3}$ corresponds to (B).
Standard deviation $= 0.816$ corresponds to (D).
Thus, the correct sequence is:
(A), (C), (B), (D)