Step 1: Understanding the Concept:
The random variable \( X \) follows a Binomial distribution with parameters \( n \) (number of trials) and \( p \) (probability of success).
The probability mass function is given by \( P(X = r) = \binom{n}{r} p^r q^{n-r} \), where \( q = 1 - p \).
We need to find the ratio of two consecutive probabilities. Step 2: Key Formula or Approach:
Write down the expressions for \( P(X=k) \) and \( P(X=k-1) \) and simplify their ratio.
Recall that \( \binom{n}{r} = \frac{n!}{r!(n-r)!} \). Step 3: Detailed Explanation:
The probability of \( k \) successes is:
\[ P(X=k) = \binom{n}{k} p^k q^{n-k} = \frac{n!}{k!(n-k)!} p^k q^{n-k} \]
The probability of \( k-1 \) successes is:
\[ P(X=k-1) = \binom{n}{k-1} p^{k-1} q^{n-(k-1)} = \frac{n!}{(k-1)!(n-k+1)!} p^{k-1} q^{n-k+1} \]
Now, take the ratio:
\[ \frac{P(X=k)}{P(X=k-1)} = \frac{\frac{n!}{k!(n-k)!} p^k q^{n-k}}{\frac{n!}{(k-1)!(n-k+1)!} p^{k-1} q^{n-k+1}} \]
Rearrange the terms:
\[ = \frac{n!}{n!} \cdot \frac{(k-1)!}{k!} \cdot \frac{(n-k+1)!}{(n-k)!} \cdot \frac{p^k}{p^{k-1}} \cdot \frac{q^{n-k}}{q^{n-k+1}} \]
Simplify the factorials and powers:
\( \frac{(k-1)!}{k!} = \frac{(k-1)!}{k \cdot (k-1)!} = \frac{1}{k} \)
\( \frac{(n-k+1)!}{(n-k)!} = \frac{(n-k+1) \cdot (n-k)!}{(n-k)!} = n-k+1 \)
\( \frac{p^k}{p^{k-1}} = p \)
\( \frac{q^{n-k}}{q^{n-k+1}} = \frac{1}{q} \)
Substitute these back into the ratio:
\[ \frac{P(X=k)}{P(X=k-1)} = 1 \cdot \frac{1}{k} \cdot (n-k+1) \cdot p \cdot \frac{1}{q} \]
\[ = \frac{n-k+1}{k} \cdot \frac{p}{q} \]
Step 4: Final Answer:
The ratio is \( \frac{n-k+1}{k} \cdot \frac{p}{q} \).