Question

It is given that the discrete random variable \[ X\sim B(n,p) \] and \[ P(X=2)=P(X=3). \] Then mean of \(X\) is

• A. \(2-p\)
• B. \(3-p\)
• C. \(p-2\)
• D. \(p-3\)

Hint:

For binomial distribution, \[ P(X=r)={}^{n}C_r p^r(1-p)^{n-r} \] and mean is always \[ np. \] Equating two probabilities often helps in directly obtaining \(np\).

Answer 1

The Correct Option is B

Step 1: Write probabilities using binomial distribution.
For \[ X\sim B(n,p), \] the probability mass function is \[ P(X=r)={}^{n}C_r p^r(1-p)^{n-r} \] Given, \[ P(X=2)=P(X=3) \] Therefore, \[ {}^{n}C_2 p^2(1-p)^{n-2} = {}^{n}C_3 p^3(1-p)^{n-3} \]

Step 2: Simplify the equation.
Dividing both sides by \[ p^2(1-p)^{n-3}, \] we get \[ {}^{n}C_2(1-p) = {}^{n}C_3 p \] Using \[ {}^{n}C_2=\frac{n(n-1)}{2} \] and \[ {}^{n}C_3=\frac{n(n-1)(n-2)}{6}, \] we get \[ \frac{n(n-1)}{2}(1-p) = \frac{n(n-1)(n-2)}{6}p \] Cancelling \(n(n-1)\), \[ \frac{1-p}{2} = \frac{(n-2)p}{6} \] Multiplying by \(6\), \[ 3(1-p)=(n-2)p \] \[ 3-3p=np-2p \] \[ 3-p=np \]

Step 3: Find the mean.
Mean of a binomial distribution is \[ np \] From above, \[ np=3-p \] Hence, mean of \(X\) is \[ \boxed{3-p} \]

Step 4: Final conclusion.
Therefore, \[ \boxed{3-p} \]