Question:
If the mean and variance of a random variable X having binomial distribution are 4 and 2 respectively, then \( P(X=2)= \)
If the mean and variance of a random variable X having binomial distribution are 4 and 2 respectively, then \( P(X=2)= \)
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Whenever the success and failure probabilities are equal (\( p = q = 1/2 \)), the component term \( p^r q^{n-r} \) simplifies beautifully to just \( \frac{1}{2^n} \), regardless of the value of \( r \). This lets you compute the final fraction much faster!
Updated On: Jun 8, 2026
- \( \frac{7}{64} \)
- \( \frac{15}{64} \)
- \( \frac{21}{64} \)
- \( \frac{39}{64} \)
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The Correct Option is A
Solution and Explanation
Concept:
For a binomial distribution:
• \( \text{Mean} = np = 4 \)
• \( \text{Variance} = npq = 2 \)
We can use these two properties to find the individual parameters \( n, p, \) and \( q \), and then use the general binomial formula \( P(X=r) = \binom{n}{r} p^r q^{n-r} \) to compute the requested probability.
Step 1: Solving for parameter values \( q, p, \) and \( n \).
Divide the variance by the mean: \[ \frac{npq}{np} = \frac{2}{4} \implies q = \frac{1}{2} \] Since \( p + q = 1 \), we have: \[ p = 1 - \frac{1}{2} = \frac{1}{2} \] Now, substitute \( p = \frac{1}{2} \) back into the mean equation to find \( n \): \[ n \left(\frac{1}{2}\right) = 4 \implies n = 8 \]
Step 2: Evaluating the probability \( P(X = 2) \).
Using the binomial formula with \( n = 8 \), \( p = \frac{1}{2} \), \( q = \frac{1}{2} \), and \( r = 2 \): \[ P(X = 2) = \binom{8}{2} \left(\frac{1}{2}\right)^2 \left(\frac{1}{2}\right)^{8-2} = \binom{8}{2} \left(\frac{1}{2}\right)^8 \] Calculate the combinations and powers: \[ \binom{8}{2} = \frac{8 \times 7}{2} = 28 \] \[ \left(\frac{1}{2}\right)^8 = \frac{1}{256} \] Multiply the values together: \[ P(X = 2) = \frac{28}{256} \] Dividing both numerator and denominator by 4 to simplify the fraction: \[ P(X = 2) = \frac{7}{64} \]
• \( \text{Mean} = np = 4 \)
• \( \text{Variance} = npq = 2 \)
We can use these two properties to find the individual parameters \( n, p, \) and \( q \), and then use the general binomial formula \( P(X=r) = \binom{n}{r} p^r q^{n-r} \) to compute the requested probability.
Step 1: Solving for parameter values \( q, p, \) and \( n \).
Divide the variance by the mean: \[ \frac{npq}{np} = \frac{2}{4} \implies q = \frac{1}{2} \] Since \( p + q = 1 \), we have: \[ p = 1 - \frac{1}{2} = \frac{1}{2} \] Now, substitute \( p = \frac{1}{2} \) back into the mean equation to find \( n \): \[ n \left(\frac{1}{2}\right) = 4 \implies n = 8 \]
Step 2: Evaluating the probability \( P(X = 2) \).
Using the binomial formula with \( n = 8 \), \( p = \frac{1}{2} \), \( q = \frac{1}{2} \), and \( r = 2 \): \[ P(X = 2) = \binom{8}{2} \left(\frac{1}{2}\right)^2 \left(\frac{1}{2}\right)^{8-2} = \binom{8}{2} \left(\frac{1}{2}\right)^8 \] Calculate the combinations and powers: \[ \binom{8}{2} = \frac{8 \times 7}{2} = 28 \] \[ \left(\frac{1}{2}\right)^8 = \frac{1}{256} \] Multiply the values together: \[ P(X = 2) = \frac{28}{256} \] Dividing both numerator and denominator by 4 to simplify the fraction: \[ P(X = 2) = \frac{7}{64} \]
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