To solve this problem, we need to consider the possible gender combinations for a family with 4 children, where each child has an independent probability of 0.5 of being male (M) or female (F).
(a) We list all possible combinations of M and F for 4 children. Since each child can be either M or F, there are \(2^4 = 16\) possible combinations.
(b) To find the probability that all 4 children are male, we calculate the probability of the event "MMMM". Since each child being male is independent and has a probability of 0.5, the probability of all 4 being male is \(0.5^4\).
(c) The probability that at least one child is female is the complement of the probability that all children are male. We subtract the probability of all being male from 1.
To determine the possible gender combinations for a family with 4 children, where each child can be either male (M) or female (F), we calculate all permutations of M and F for 4 positions. Since each child has 2 possible outcomes, there are \(2^4 = 16\) combinations. These combinations are:
\[
\begin{align_}
&\text{MMMM}, \text{MMMF}, \text{MMFM}, \text{MMFF}, \text{MFMM}, \text{MFMF}, \text{MFFM}, \text{MFFF}, \\
&\text{FMMM}, \text{FMMF}, \text{FMFM}, \text{FMFF}, \text{FFMM}, \text{FFMF}, \text{FFFM}, \text{FFFF}
\end{align_}
\]
The probability that each child is male is \(0.5\). Since the events are independent, the probability that all 4 children are male (MMMM) is:
\[
P(\text{MMMM}) = 0.5^4 = \frac{1}{16}
\]
The probability that at least one child is female is the complement of the probability that all children are male. Therefore, we subtract the probability of all being male from 1:
\[
P(\text{at least one F}) = 1 - P(\text{MMMM}) = 1 - \frac{1}{16} = \frac{15}{16}
\]
- The equally likely events for the gender of the 4 children are: \(\text{MMMM}, \text{MMMF}, \text{MMFM}, \text{MMFF}, \text{MFMM}, \text{MFMF}, \text{MFFM}, \text{MFFF}, \text{FMMM}, \text{FMMF}, \text{FMFM}, \text{FMFF}, \text{FFMM}, \text{FFMF}, \text{FFFM}, \text{FFFF}\).
- The probability that all 4 children are male is \(\boxed{\frac{1}{16}}\).
- The probability that at least one child is female is \(\boxed{\frac{15}{16}}\).