Question

A father has three children with at least one boy. The probability that he has 2 boys and 1 girl is
$
  {\text{A}}{\text{. }}\dfrac{1}{4} \\
  {\text{B}}{\text{. }}\dfrac{1}{3} \\
  {\text{C}}{\text{. }}\dfrac{2}{3} \\
  {\text{D}}{\text{. None of these}} \\
$

Answer 1

Hint: To solve the question of probability we have to find the total number of outcomes in sample space and the total number of favorable outcomes of our event. If we can find these two things now just use the formula of probability and get the answer.

Complete step-by-step answer:
Let's represent boy with B and girl with G then possible outcomes are BBB,BBG,BGG.
That means we have a total number of outcomes in sample space is 3.
And the favorable outcome is BBG
That means the number of favorable outcomes is 1.
Let's call this event to be E
Then
$P\left( E \right) = \dfrac{{{\text{Number of favorable outcomes}}}}{{{\text{Total number of outcomes}}}}$
$ \Rightarrow P\left( E \right) = \dfrac{1}{3}$
Hence option B is the correct option.

Note: Whenever we get this type of question the key concept of solving is we have to understand what is given in question and from which we will try to find total outcomes of sample space and also favorable outcomes of our event. One thing is important here that you will not always be able to find all the terms of sample space so in that condition we have to get more and more knowledge about sample space.