Question:

The probability of a boy winning a game is 23 \frac{2}{3} . Let n 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 X denotes that number of times he wins the game. Hence n n , mean, variance, and standard deviation of the random variable X X are respectively:
(A) 3
(B) 23 \frac{2}{3}  
(C) 2  
(D) 0.81  
Choose the correct answer from the options given below: 

Updated On: Mar 12, 2025
  • (A), (C), (B), (D)

  • (A), (B), (C), (D)

  • (B), (A), (D), (C)
  • (C), (A), (D), (B)
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The Correct Option is A

Solution and Explanation

Finding nn such that P(winning at least once)>0.9P(\text{winning at least once}) > 0.9. The probability of losing a single game is:
P(loss)=123=13P(loss) = 1 - \frac{2}{3} = \frac{1}{3}
The probability of losing all nn games is:
P(losingall n games)=(13)nP(losing all  n \text{ games}) = \left(\frac{1}{3}\right)^n
The probability of winning at least once is:
P(winningatleastonce)=1P(losingall n games)=1(13)nP(winning at least once) = 1 - P(losing all  n  games) = 1 - \left(\frac{1}{3}\right)^n
We require:
1(13)n>0.91 - \left(\frac{1}{3}\right)^n > 0.9
(13)n<0.1\left(\frac{1}{3}\right)^n < 0.1
Taking the natural logarithm on both sides:
nln(13)<ln(0.1)n \ln\left(\frac{1}{3}\right) < \ln(0.1)
Since ln(13)<0\ln\left(\frac{1}{3}\right) < 0, dividing by it reverses the inequality:
n>ln(0.1)ln(13)n > \frac{\ln(0.1)}{\ln\left(\frac{1}{3}\right)}
Using approximate values:
ln(0.1)2.3026\ln(0.1) \approx -2.3026, ln(13)1.0986\ln\left(\frac{1}{3}\right) \approx -1.0986
n>2.30261.09862.1n > \frac{-2.3026}{-1.0986} \approx 2.1
Thus, the least integer nn is:
n=3n = 3
Mean, variance, and standard deviation of XX. The number of wins XX follows a Binomial distribution with parameters n=3n = 3 and p=23p = \frac{2}{3}. The mean, variance, and standard deviation are given by:
- Mean:
Mean =np=323=2= n \cdot p = 3 \cdot \frac{2}{3} = 2
- Variance:
Variance =np(1p)=32313=230.67= 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 =Variance=230.816= \sqrt{\text{Variance}} = \sqrt{\frac{2}{3}} \approx 0.816
Matching options. From the above calculations:
n=3n = 3 corresponds to (A).
Mean =2= 2 corresponds to (C).
Variance =23= \frac{2}{3} corresponds to (B).
Standard deviation =0.816= 0.816 corresponds to (D).
Thus, the correct sequence is:
(A), (C), (B), (D)

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