If X has binomial distribution with mean np and variance npq, then `(P(X = r))/(P(X = r - 1))` is equal to

Options

A. `(n-r)/(r-1)p/q`
B. `(n-r+1)/rp/q`
C. `(n+1)/rp/q`
D. `(n-1)/(r+1)p/q`

Answer

To solve the problem, we need to find the ratio \( \frac{P(X = r)}{P(X = r - 1)} \) for a binomial distribution. The probability mass function for a binomial distribution is given by:

\[ P(X = r) = \binom{n}{r} p^r q^{n-r} \]

where \( n \) is the number of trials, \( p \) is the probability of success, \( q = 1 - p \) is the probability of failure, and \( \binom{n}{r} \) is the binomial coefficient.

### Step 1: Write down the probabilities
We start by writing the expressions for \( P(X = r) \) and \( P(X = r - 1) \):

\[ P(X = r) = \binom{n}{r} p^r q^{n-r} \]
\[ P(X = r - 1) = \binom{n}{r-1} p^{r-1} q^{n-(r-1)} = \binom{n}{r-1} p^{r-1} q^{n-r+1} \]

### Step 2: Set up the ratio
Now we can set up the ratio:

\[ \frac{P(X = r)}{P(X = r - 1)} = \frac{\binom{n}{r} p^r q^{n-r}}{\binom{n}{r-1} p^{r-1} q^{n-r+1}} \]

### Step 3: Simplify the ratio
We can simplify this expression step by step:

1. **Cancel \( p^{r-1} \)**:
\[ = \frac{\binom{n}{r} p}{\binom{n}{r-1} q} \]

2. **Use the relationship between binomial coefficients**:
\[ \binom{n}{r} = \frac{n!}{r!(n-r)!}, \quad \binom{n}{r-1} = \frac{n!}{(r-1)!(n-r+1)!} \]
Thus,
\[ \frac{\binom{n}{r}}{\binom{n}{r-1}} = \frac{(r-1)!(n-r+1)!}{r!(n-r)!} = \frac{n-r+1}{r} \]

3. **Substituting back**:
\[ \frac{P(X = r)}{P(X = r - 1)} = \frac{n-r+1}{r} \cdot \frac{p}{q} \]

### Final Result
Therefore, we have:

\[ \frac{P(X = r)}{P(X = r - 1)} = \frac{(n - r + 1)p}{rq} \]