Question

Let 𝑋₁~𝑁(2, 1), 𝑋₂~𝑁(−1,4) and 𝑋₃~𝑁(0, 1) be mutually independent random variables. Then, the probability that exactly two of these three random variables are less than 1, equals____(round off to two decimal places)

Answer 1

Correct Answer: 0.63

To solve for the probability that exactly two of the three independent normal random variables are less than 1, we first determine the probability for each random variable to be under 1 using the standard normal distribution.

1. Determine individual probabilities:

• For \(X_1 \sim N(2, 1)\), calculate \(P(X_1 < 1)\). Convert to standard normal: \[ Z = \frac{1 - 2}{1} = -1. \quad \text{Thus,} \quad P(X_1 < 1) = P(Z < -1) \approx 0.1587. \]
• For \(X_2 \sim N(-1, 4)\), calculate \(P(X_2 < 1)\). Convert to standard normal: \[ Z = \frac{1 - (-1)}{2} = 1. \quad \text{Thus,} \quad P(X_2 < 1) = P(Z < 1) \approx 0.8413. \]
• For \(X_3 \sim N(0, 1)\), calculate \(P(X_3 < 1)\). Convert to standard normal: \[ Z = \frac{1 - 0}{1} = 1. \quad \text{Thus,} \quad P(X_3 < 1) = P(Z < 1) \approx 0.8413. \]

2. Calculate the probability for exactly two random variables being less than 1:

We use the formula for the probability of exactly \(k\) successes (being less than 1) in \(n\) trials (3 random variables) where each trial is independent. The trials where \(X_i < 1\) are considered successes:

Compute for different combinations:

Combination	Probability
\(X_1 < 1, X_2 < 1, X_3 \geq 1\)	\(0.1587 \times 0.8413 \times (1 - 0.8413) = 0.0213\)
\(X_1 < 1, X_2 \geq 1, X_3 < 1\)	\(0.1587 \times (1 - 0.8413) \times 0.8413 = 0.0213\)
\(X_1 \geq 1, X_2 < 1, X_3 < 1\)	\((1 - 0.1587) \times 0.8413 \times 0.8413 = 0.5898\)

Sum these probabilities for exactly two being less than 1:

\[ 0.0213 + 0.0213 + 0.5898 = 0.6324. \]

3. Validation:

The calculated probability is 0.63 when rounded to two decimal places. It fits within the given range [0.63, 0.63].

Conclusion:

The probability that exactly two of the three random variables are less than 1 is \( \boxed{0.63} \).