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 $K$ 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 $\frac{45}{1024}$.

Answer 2

The number of tosses \( K \) satisfies the condition:

\[ P(3 \text{ heads}) = P(7 \text{ heads}). \]

Step 1: Probability of \( r \) heads in \( K \) tosses:

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

Step 2: Equate \( P(3) = P(7) \):

From the given condition \( P(3) = P(7) \), we have: \[ {K \choose 3} = {K \choose 7}. \]

Step 3: Solve the binomial equation:

Using the symmetry of binomial coefficients \( {K \choose r} = {K \choose K-r} \), we get: \[ {K \choose 3} = {K \choose K-3}. \] Therefore, \( K - 3 = 7 \), which gives: \[ K = 10. \]

Step 4: Calculate the probability of getting 8 tails (or 2 heads):

The number of tails is 8, which means the number of heads is \( 2 \). The probability of getting 8 tails (or 2 heads) is: \[ P(8 \text{ tails}) = {10 \choose 2} \left( \frac{1}{2} \right)^{10}. \]

Step 5: Simplify the calculation:

The binomial coefficient \( {10 \choose 2} \) is calculated as: \[ {10 \choose 2} = \frac{10 \cdot 9}{2} = 45. \] Now, the probability is: \[ P(8 \text{ tails}) = 45 \cdot \frac{1}{1024} = \frac{45}{1024}. \]

Conclusion: The probability of getting 8 tails (or 2 heads) is \( \frac{45}{1024} \).