The probability of a boy winning a game is 32. 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) 32 (C) 2 (D) 0.81 Choose the correct answer from the options given below:
Finding n such that P(winning at least once)>0.9. The probability of losing a single game is: P(loss)=1−32=31 The probability of losing all n games is: P(losingalln games)=(31)n The probability of winning at least once is: P(winningatleastonce)=1−P(losingallngames)=1−(31)n We require: 1−(31)n>0.9 (31)n<0.1 Taking the natural logarithm on both sides: nln(31)<ln(0.1) Since ln(31)<0, dividing by it reverses the inequality: n>ln(31)ln(0.1) Using approximate values: ln(0.1)≈−2.3026, ln(31)≈−1.0986 n>−1.0986−2.3026≈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=32. The mean, variance, and standard deviation are given by: - Mean: Mean =n⋅p=3⋅32=2 - Variance: Variance =n⋅p⋅(1−p)=3⋅32⋅31=32≈0.67 - Standard deviation: Standard Deviation =Variance=32≈0.816 Matching options. From the above calculations: n=3 corresponds to (A). Mean =2 corresponds to (C). Variance =32 corresponds to (B). Standard deviation =0.816 corresponds to (D). Thus, the correct sequence is: (A), (C), (B), (D)