Question

A fair coin is tossed five times. The probability of obtaining at most two “HEAD” (correct up to 1 decimal place) is ________________.

Hint:

The binomial distribution can be used to find the probability of getting a certain number of heads in a fixed number of coin tosses.

Answer 1

Correct Answer: 0.4

The problem involves calculating the probability of obtaining at most two heads in five coin tosses. This follows a binomial distribution. The probability of getting exactly \(k\) heads in \(n\) tosses is given by the binomial formula: \[ P(X = k) = \binom{n}{k} \cdot p^k \cdot (1-p)^{n-k}, \] where \( p = 0.5 \) (probability of heads), \(n = 5\) (number of tosses), and \(k\) can be 0, 1, or 2 for "at most two heads". We calculate the probabilities for \( k = 0 \), \( k = 1 \), and \( k = 2 \): \[ P(X = 0) = \binom{5}{0} \cdot (0.5)^0 \cdot (0.5)^5 = \frac{1}{32}, \] \[ P(X = 1) = \binom{5}{1} \cdot (0.5)^1 \cdot (0.5)^4 = \frac{5}{32}, \] \[ P(X = 2) = \binom{5}{2} \cdot (0.5)^2 \cdot (0.5)^3 = \frac{10}{32}. \] Thus, the total probability of obtaining at most two heads is: \[ P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) = \frac{1}{32} + \frac{5}{32} + \frac{10}{32} = \frac{16}{32} = 0.5. \] Final Answer: \[ \boxed{0.5}. \]