Question

the probability of getting exactly 2 heads

Explanation:

Step1: Identify the problem type

This is a probability problem, likely involving coin tosses (getting exactly 2 heads). We'll use the binomial probability formula. The binomial probability formula is \( P(X = k) = \binom{n}{k} p^k (1 - p)^{n - k} \), where \( n \) is the number of trials, \( k \) is the number of successful trials, \( p \) is the probability of success on a single trial. For a fair coin, \( p = 0.5 \), and we assume \( n = 3 \) (common for exactly 2 heads, but need to confirm). Wait, maybe the problem is about 3 coins? Let's assume \( n = 3 \), \( k = 2 \), \( p = 0.5 \).

Step2: Calculate combinations

First, calculate \( \binom{n}{k} = \binom{3}{2} \). The combination formula is \( \binom{n}{k} = \frac{n!}{k!(n - k)!} \). So \( \binom{3}{2} = \frac{3!}{2!(3 - 2)!} = \frac{3 \times 2!}{2! \times 1!} = 3 \).

Step3: Apply binomial formula

Now, \( p = 0.5 \), \( k = 2 \), \( n = 3 \). So \( P(X = 2) = 3 \times (0.5)^2 \times (1 - 0.5)^{3 - 2} = 3 \times 0.25 \times 0.5 = 3 \times 0.125 = 0.375 \) or \( \frac{3}{8} \). Wait, maybe the problem has different numbers? Wait, the handwritten part is hard to read, but assuming it's 3 coins, the probability of exactly 2 heads is \( \frac{3}{8} \).

Answer:

If the problem is about 3 fair coin tosses, the probability of exactly 2 heads is \(\frac{3}{8}\) (or 0.375).