Finding probability at least one male child A family has three children. what are the probabilities that it has no male child and at least one male child if we consider the order of birth. 
My Try: 
I didn't understand the complete question. It's confusing me while solving.
 A: There are different ways to solve the problem. A simple, straight forward way is to enumerate the possible options (since the problem size is small). Since we consider the order of birth to important, the possible options are:
\begin{align}
\color{blue}F \color{blue}F \color{blue}F\\
\color{red}M \color{blue}F \color{blue}F\\
\color{blue}F \color{red}M \color{blue}F\\
\color{blue}F \color{blue}F \color{red}M\\
\color{red}M \color{red}M \color{blue}F\\
\color{red}M \color{blue}F \color{red}M\\
\color{blue}F \color{red}M \color{red}M\\
\color{red}M \color{red}M \color{red}M
\end{align}
As you can see there are $8$ possible options in total. Hence, the probability that there is no male child is $\dfrac18$ and the probability that there is at-least one male child is $\dfrac78$.
A: Probability with no male child, which is the same as all female childen, is $$\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}=\frac{1}{8}$$
The probability with at least on male child can be obtained by subtracting the probability of all children being female from $1$. Thus:
$$1-\frac{1}{8}=\frac{7}{8}$$
Or you can obtain the answer by adding up the possible ways of at least $1$ male child, $2$ male children and $3$:
$${3\choose1}+{3\choose2}+{3\choose3}=3+3+1=7\implies\frac{7}{8}$$
A: Presuming the standard probability of 50% chance of male child and 50% of female child, with each child being an independent event, here is some background on each case:
No male child, implies all 3 children are female which would be $(\frac{1}{2})^3 = \frac{1}{8} = .125$ or $12.5$%
At least one male child would be any other possibility which could be computed as $1-.125=.875$ which is 87.5%
Order of birth would be a red herring here as it has no bearing on the computation.
