Step 1: Understanding the Question:
\(X\) follows a binomial distribution with parameters \(n = 35\) and \(p\).
We are given a relation between \(P(X=0)\) and \(P(X=1)\) and need to find a ratio of probabilities.
Step 2: Key Formula or Approach:
The binomial probability mass function is \(P(X = k) = \binom{n}{k} p^k q^{n-k}\), where \(q = 1-p\).
Step 3: Detailed Explanation:
Given \(n = 35\) and \(7P(X=0) = P(X=1)\):
\[ 7 \times \binom{35}{0} p^0 q^{35} = \binom{35}{1} p^1 q^{34} \]
\[ 7 \times 1 \times 1 \times q^{35} = 35 \times p \times q^{34} \]
Divide by \(q^{34}\) (assuming \(q \neq 0\)):
\[ 7q = 35p \implies q = 5p \]
Since \(p + q = 1\), substitute \(q = 5p\):
\[ p + 5p = 1 \implies 6p = 1 \implies p = \frac{1}{6} \]
Then, \(q = 1 - \frac{1}{6} = \frac{5}{6}\).
Now find the ratio \(\frac{P(X=15)}{P(X=20)}\):
\[ \frac{P(X=15)}{P(X=20)} = \frac{\binom{35}{15} p^{15} q^{20}}{\binom{35}{20} p^{20} q^{15}} \]
Since \(\binom{n}{r} = \binom{n}{n-r}\), \(\binom{35}{15} = \binom{35}{20}\). They cancel out.
\[ \text{Ratio} = \frac{p^{15}}{p^{20}} \times \frac{q^{20}}{q^{15}} = \frac{q^5}{p^5} = \left(\frac{q}{p}\right)^5 \]
Substitute \(q/p = 5\):
\[ \text{Ratio} = (5)^5 = 3125 \]
Step 4: Final Answer:
The value is \(3125\).