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. \(\frac{210}{2^{21}}\)

Hint:

When solving problems involving binomial probabilities, the key is recognizing the symmetry in binomial coefficients. \( {K \choose r} = {K \choose K - r} \), which helps simplify problems like this one. Always pay attention to conditions that relate different probabilities (such as equal probabilities for 3 heads and 7 heads in this case), and use the symmetry to find the total number of tosses. Once you have the number of tosses, the rest is just applying the binomial formula!

Answer 1

The Correct Option is C

The number of tosses, denoted by $K$, is determined by the condition $P(3 \text{ heads}) = P(7 \text{ heads})$.

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

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

This equation simplifies to $84 = {K \choose 7}$.

Utilizing the symmetry property of binomial coefficients, where ${K \choose n} = {K \choose K-n}$, we have ${K \choose 3} = {K \choose K-3}$. Therefore, $K-3 = 7$, which implies $K = 10$.

The probability of observing 8 tails, which is equivalent to 2 heads in 10 tosses, is calculated as: $P(8 \text{ tails}) = P(2 \text{ heads}) = {10 \choose 2} \left(\frac{1}{2}\right)^{10}$.

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

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