Question

probability of getting 7 heads out of n coins is equal to probability of getting 8 heads out of n coins. Find probability of getting 3 heads out of these n coins

• A. 455/2^15
• B. 455/2^8
• C. 455/2^7
• D. 1/2^15

Answer 1

Answer: (A)

455/2^15

Answer 2

Given that the probability of getting 7 heads out of $n$ coins is equal to the probability of getting 8 heads out of $n$ coins. For a fair coin, the probability of getting $k$ heads in $n$ tosses is given by the binomial probability formula: $P(k \text{ heads}) = \binom{n}{k} p^k (1-p)^{n-k}$ Assuming a fair coin, $p = 1/2$, so $P(k \text{ heads}) = \binom{n}{k} (1/2)^k (1/2)^{n-k} = \binom{n}{k} (1/2)^n$.

We are given: $P(7 \text{ heads}) = P(8 \text{ heads})$ $\binom{n}{7} (1/2)^n = \binom{n}{8} (1/2)^n$

Since $(1/2)^n$ is a common factor and is non-zero, we can cancel it out: $\binom{n}{7} = \binom{n}{8}$

Using the property of binomial coefficients that if $\binom{n}{a} = \binom{n}{b}$ and $a \neq b$, then $n = a+b$. Here, $a=7$ and $b=8$. So, $n = 7 + 8 = 15$.

Now we need to find the probability of getting 3 heads out of these $n=15$ coins. $P(3 \text{ heads}) = \binom{15}{3} (1/2)^{15}$

Calculate the binomial coefficient $\binom{15}{3}$: $\binom{15}{3} = \frac{15!}{3!(15-3)!} = \frac{15!}{3!12!} = \frac{15 \times 14 \times 13}{3 \times 2 \times 1}$ $\binom{15}{3} = \frac{15}{3} \times \frac{14}{2} \times 13 = 5 \times 7 \times 13 = 455$.

Therefore, the probability of getting 3 heads out of 15 coins is: $P(3 \text{ heads}) = 455 \times (1/2)^{15} = \frac{455}{2^{15}}$.