Similar question All parts showing An experiment consists of tossing 6 fair (not weighted) coins, except one of the 6 coins has a head on both sides. Compute the probability of obtaining exactly 5 heads. The probability of obtaining exactly 5 heads is $\square$. (Type an integer or a simplified fraction.)

Step 1 ：Consider two cases: Case 1 where the two-headed coin is one of the 5 heads, and Case 2 where the two-headed coin is not one of the 5 heads.

Step 2 ：For Case 1, we need to get 4 heads from the remaining 5 coins. The probability of getting a head from a fair coin is \(\frac{1}{2}\). So, the probability of getting 4 heads from 5 coins is given by the binomial probability formula: \(P(4 heads from 5 coins) = C(5, 4) * (\frac{1}{2})^4 * (\frac{1}{2})^1 = 5 * (\frac{1}{16}) * (\frac{1}{2}) = \frac{5}{32}\)

Step 3 ：For Case 2, we need to get 5 heads from the remaining 5 coins. The probability of getting a head from a fair coin is \(\frac{1}{2}\). So, the probability of getting 5 heads from 5 coins is given by the binomial probability formula: \(P(5 heads from 5 coins) = C(5, 5) * (\frac{1}{2})^5 * (\frac{1}{2})^0 = 1 * (\frac{1}{32}) * 1 = \frac{1}{32}\)

Step 4 ：Adding the probabilities from the two cases gives the total probability of getting exactly 5 heads: \(P(5 heads) = P(Case 1) + P(Case 2) = \frac{5}{32} + \frac{1}{32} = \frac{6}{32} = \frac{3}{16}\)

Step 5 ：\(\boxed{\text{So, the probability of obtaining exactly 5 heads is } \frac{3}{16}}\)