Question

In a binomial distribution \[ P(X=2)\div P(X=23)=\left(\frac{2}{3}\right)^{21}, \] then the mean of the binomial distribution is

• A. \(10\)
• B. \(15\)
• C. \(20\)
• D. \(25\) \bigskip

Hint:

For binomial distributions, whenever probabilities of two terms are compared, first use \[ P(X=r)=\binom{n}{r}p^rq^{\,n-r}. \] Look for complementary indices since \[ \binom{n}{r}=\binom{n}{n-r}, \] which often helps determine \(n\) immediately.

Answer 1

The Correct Option is B

Concept: For a binomial distribution \[ X\sim B(n,p), \] \[ P(X=r)=\binom{n}{r}p^r q^{\,n-r}, \] where \[ q=1-p. \] Also, the mean of the distribution is \[ \mu=np. \]

Step 1: Write the ratio using the binomial probability formula. Given \[ \frac{P(X=2)}{P(X=23)} = \left(\frac23\right)^{21}. \] Using \[ P(X=r)=\binom{n}{r}p^r q^{n-r}, \] we get \[ \frac{\binom{n}{2}p^2q^{\,n-2}} {\binom{n}{23}p^{23}q^{\,n-23}} = \left(\frac23\right)^{21}. \] \[ \frac{\binom{n}{2}} {\binom{n}{23}} \left(\frac{q}{p}\right)^{21} = \left(\frac23\right)^{21}. \]

Step 2: Determine \(n\). Since \[ \binom{n}{r} = \binom{n}{n-r}, \] for the combination terms to cancel, we require \[ 23=n-2. \] Thus, \[ n=25. \] Hence, \[ \binom{25}{2} = \binom{25}{23}. \] Therefore, \[ \left(\frac{q}{p}\right)^{21} = \left(\frac23\right)^{21}. \] \[ \frac{q}{p} = \frac23. \]

Step 3: Find \(p\). Since \[ p+q=1, \] and \[ q=\frac23p, \] we have \[ p+\frac23p=1. \] \[ \frac53p=1. \] \[ p=\frac35. \]

Step 4: Find the mean. The mean of the binomial distribution is \[ np. \] Therefore, \[ np = 25\left(\frac35\right). \] \[ =15. \]

Step 5: Write the final answer. \[ \boxed{15} \]