Question

An unbiased coin is tossed \(n\) times. If the probability of getting 5 heads is equal to the probability of getting 6 heads, then the probability of getting 3 heads is:

• A. \(^{11}C_5 \left(\frac{1}{2}\right)^5\)
• B. \(^{11}C_6 \left(\frac{1}{2}\right)^6\)
• C. \(^{11}C_3 \left(\frac{1}{2}\right)^{11}\)
• D. \( \frac{11}{1024} \)

Hint:

Equal binomial probabilities occur when \(r\) and \(n-r\) are equal → use symmetry.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
This problem follows a Binomial Distribution \(B(n, p)\). For an unbiased coin, the probability of success (heads) is \(p = 1/2\) and failure (tails) is \(q = 1 - p = 1/2\).
Step 2: Key Formula or Approach:
\(P(X = r) = \binom{n}{r} p^{r} q^{n-r}\).
For \(p = q = 1/2\), this simplifies to \(P(X = r) = \binom{n}{r} (1/2)^{n}\).
Step 3: Detailed Explanation:
1. Given: \(P(X = 5) = P(X = 6)\).
\[ \binom{n}{5} (1/2)^{n} = \binom{n}{6} (1/2)^{n} \]
\[ \binom{n}{5} = \binom{n}{6} \]
2. Using the property \(\binom{n}{x} = \binom{n}{y} \implies x + y = n\):
\[ 5 + 6 = n \implies n = 11 \]
3. Now calculate the probability of getting 3 heads:
\[ P(X = 3) = \binom{11}{3} (1/2)^{3} (1/2)^{11-3} \]
\[ P(X = 3) = \binom{11}{3} (1/2)^{11} \]
Step 4: Final Answer:
The probability is \({}^{11}C_{3} (1/2)^{11}\).