Question

7 coins are tossed simultaneously and the number of heads turned up is denoted by the random variable \( X \). If \( \mu \) is the mean and \( \sigma^2 \) is the variance of \( X \), then \( \frac{\mu^2}{P(X = 3)} \) is:

• A. \( \frac{56}{5} \)
• B. \( \frac{84}{5} \)
• C. \( \frac{112}{5} \)
• D. \( \frac{224}{5} \)

Hint:

For a binomial distribution, the mean is \( \mu = np \) and the variance is \( \sigma^2 = np(1-p) \). To find the probability for a specific number of successes, use the binomial probability formula.

Answer 1

The Correct Option is C

We are given that 7 coins are tossed simultaneously, and the random variable \( X \) represents the number of heads. We are asked to find \( \frac{\mu^2}{P(X = 3)} \), where \( \mu \) is the mean and \( \sigma^2 \) is the variance of \( X \).
Step 1: Since we are tossing 7 coins, the number of heads \( X \) follows a binomial distribution with parameters \( n = 7 \) and \( p = \frac{1}{2} \). The probability mass function for a binomial random variable is given by: \[ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \] Substitute \( n = 7 \) and \( p = \frac{1}{2} \) to find the probabilities. 
Step 2: The mean \( \mu \) of a binomial distribution is given by: \[ \mu = np = 7 \times \frac{1}{2} = \frac{7}{2} \] The variance \( \sigma^2 \) of a binomial distribution is given by: \[ \sigma^2 = np(1-p) = 7 \times \frac{1}{2} \times \frac{1}{2} = \frac{7}{4} \] Step 3: Now, calculate \( P(X = 3) \), which is the probability of getting exactly 3 heads: \[ P(X = 3) = \binom{7}{3} \left( \frac{1}{2} \right)^3 \left( \frac{1}{2} \right)^{7-3} = \binom{7}{3} \left( \frac{1}{2} \right)^7 \] First, calculate the binomial coefficient: \[ \binom{7}{3} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35 \] Thus, \[ P(X = 3) = 35 \times \frac{1}{128} = \frac{35}{128} \] Step 4: We are asked to find \( \frac{\mu^2}{P(X = 3)} \). Substituting the values of \( \mu \) and \( P(X = 3) \): \[ \frac{\mu^2}{P(X = 3)} = \frac{\left( \frac{7}{2} \right)^2}{\frac{35}{128}} = \frac{\frac{49}{4}}{\frac{35}{128}} = \frac{49}{4} \times \frac{128}{35} = \frac{49 \times 128}{4 \times 35} = \frac{6272}{140} = \frac{112}{5} \] Thus, \( \frac{\mu^2}{P(X = 3)} = \frac{112}{5} \).