Question

A fair coin is tossed a fixed number of times. If the probability of getting 5 heads is equal to the probability of getting 4 heads, then the probability of getting 6 heads is:

• A. \(\frac{7}{64}\)
• B. \(\frac{9}{32}\)
• C. \(\frac{21}{128}\)
• D. \(\frac{35}{256}\)

Hint:

In binomial distributions, symmetry exists between heads and tails, i.e., \(\binom{n}{r} = \binom{n}{n-r}\).
- When probabilities for different outcomes are equal, it often involves solving for the number of trials \(n\).

Answer 1

The Correct Option is C

Step 1: Let the number of tosses be \(n\).
The number of heads follows a binomial distribution, so the probability of getting \(r\) heads in \(n\) tosses is: \[ P(\text{r heads}) = \binom{n}{r} \left(\frac{1}{2}\right)^n. \] Given that the probability of getting 5 heads is equal to the probability of getting 4 heads, we have: \[ P(5 \text{ heads}) = P(4 \text{ heads}), \] which simplifies to: \[ \binom{n}{5} \left(\frac{1}{2}\right)^n = \binom{n}{4} \left(\frac{1}{2}\right)^n. \] Canceling out \(\left(\frac{1}{2}\right)^n\) from both sides, we get: \[ \binom{n}{5} = \binom{n}{4}. \] From the property of binomial coefficients, we know: \[ \binom{n}{5} = \binom{n}{n-5}, \] so the equation becomes: \[ \binom{n}{5} = \binom{n}{4} \quad \Rightarrow \quad n = 9. \] Step 2: Find the probability of getting 6 heads.
Using the binomial distribution for \(n = 9\), the probability of getting 6 heads is: \[ P(6 \text{ heads}) = \binom{9}{6} \left(\frac{1}{2}\right)^9. \] Using the binomial coefficient \(\binom{9}{6} = \binom{9}{3} = 84\), we find: \[ P(6 \text{ heads}) = 84 \times \frac{1}{2^9} = \frac{84}{512} = \frac{21}{128}. \] Thus, the probability of getting 6 heads is \(\boxed{\frac{21}{128}}\).