Question

A coin is tossed K times. If the probability of getting 3 heads is equal to the probability of getting 7 heads, then the probability of getting 8 tails is:

• A. $\frac{5}{512}$
• B. $\frac{45}{2^{21}}$
• C. $\frac{45}{1024}$
• D. $\frac{210}{2^{21}}$

Answer 1

Answer: (C)

$\frac{45}{1024}$

Answer 2

The number of tosses $K$ satisfies the condition:
$P(3 \text{ heads}) = P(7 \text{ heads})$.

The probability of $r$ heads in $K$ tosses is:
$P(r) = {K \choose r} \left(\frac{1}{2}\right)^K$.

Equating $P(3) = P(7)$:
${K \choose 3} = {K \choose 7}$.

$84 = {K \choose 7}$.

From symmetry of binomial coefficients:
${K \choose 3} = {K \choose K-3} \Rightarrow K-3 = 7 \Rightarrow K = 10$.

The probability of getting 8 tails (or 2 heads) is:
$P(8 \text{ tails}) = {10 \choose 2} \left(\frac{1}{2}\right)^{10}$.

$P(8 \text{ tails}) = \frac{10 \cdot 9}{2} \cdot \frac{1}{1024} = \frac{45}{1024}$.

Thus, the probability is $\frac{45}{1024}$.