Question

In a family with 4 children, the probability that there are at least two girls is:
A) \[\dfrac{1}{2}\]
B) $\dfrac{9}{{16}}$
C) $\dfrac{3}{4}$
D) $\dfrac{{11}}{{16}}$

Answer 1

Hint: First we will find the arrangements for how many can be boys and how many can be girls out 4 children keeping in mind that there should be at least two girls. In this problem we will use a binomial distribution method for finding the probability of any random variable, where binomial distribution is used to find the probability of a success or failure.

Complete step-by-step answer:
Given: we have given that in a family there are 4 children and we have to find the probability of having at least 2 girls out of 4 children in the family.
First, we will let the probability of having girl is
\[p\]and probability of having a boy is $q$, in the family.
Now, we will solve for the probability of \[p\]
Which is$ = \dfrac{{{\text{chances of having girl}}}}{{{\text{total chances}}}}$
Mathematically, $p = \dfrac{2}{4} = \dfrac{1}{2}$
So, the probability of having boys will be \[1 - p\]
Mathematically, $q = 1 - p$
Substitute the obtained value of \[p\]to find the value of \[q\]
$
  q = 1 - \dfrac{1}{2} \\
   = \dfrac{1}{2} \\
$
Now, we have the probability of having 2 girls and 2 boys out of 4 children.
And now, we will find out all the cases having at least two girls.
Which is, Probability of having two girls and two boys $ + $Probability of having three girls and one boy$ + $probability of four girls.
Mathematically, $P\left( {2G,2B} \right) + P\left( {3G,1B} \right) + P\left( {4G} \right)$
Where, $P$represents the probability, $G$represents girls and $B$represents boys.
So, to solve this we will use a binomial distribution method.
So, let $x$be the number of girls.
Then the probability of random variable $x$is $P\left( x \right) = {}^n{C_x}{p^x}{q^{n - x}}$
So, our required probability is
$P\left( {2G} \right) + P\left( {3G} \right) + P\left( {4G} \right)$
Now, put the formula of probability of random variable according to the required probability.
${}^4{C_2}{p^2}{q^{4 - 2}} + {}^4{C_3}{p^3}{q^{4 - 3}} + {}^4{C_4}{p^4}$
Now, substitute the values of \[p\]and $q$ which is equal to $\dfrac{1}{2}$
We will get,
${}^4{C_2}{\left( {\dfrac{1}{2}} \right)^2}{\left( {\dfrac{1}{2}} \right)^{4 - 2}} + {}^4{C_3}{\left( {\dfrac{1}{2}} \right)^3}{\left( {\dfrac{1}{2}} \right)^{4 - 3}} + {}^4{C_4}{\left( {\dfrac{1}{2}} \right)^4}$
Now, solve this equation further,
$
   \Rightarrow 6 \times \dfrac{1}{{{2^4}}} + 4 \times \dfrac{1}{{{2^4}}} + 1 \times \dfrac{1}{{{2^4}}} \\
   = \dfrac{6}{{16}} + \dfrac{4}{{16}} + \dfrac{1}{{16}} \\
   = \dfrac{{11}}{{16}} \\
$
So, the probability of having at least two girls out of 4 children is $\dfrac{{11}}{{16}}$.
Hence, the option D is correct.

Note:
If any one of the combination of girls and boys is skipped then we can get the wrong answer, so be careful while making the combinations of girls and boys out of 4 children.