Question

A bag contains 4 brown and 5 white socks. A man pulls two socks at random without
replacement. The probability that the man gets both the socks of the same colour is
A. $\dfrac{5}{{108}}$
B.$\dfrac{1}{6}$
C. $\dfrac{5}{{18}}$
D. $\dfrac{4}{9}$

Answer 1

Hint: In this problem, first we have to find the total number of socks present in the bag by adding
the total number of brown socks and the total number of white socks in the bag are given. Since
the man pulls two socks randomly without replacement, so we need to divide each number of
socks with the total number. This will give the probability of pulling socks on first draw. After
that we need to reduce one socks from each colour and divide it with the total number of socks.
This will give the probability of pulling a socks in the second draw. Finally, we have to multiply
the corresponding colours. By adding them we get the required probability.

Complete step-by-step answer:
It is given that total number of brown socks in the bag = 4
Total number of white socks in the bag = 5
Step I
Adding the total number of brown socks and white socks, we get the total number of socks.
Thus Total number of socks
$\begin{array}{l} = 4 + 5\\ = 9\end{array}$
Step II
Given that the man pulls two socks at random without replacement.
Now, the probability of pulling a brown socks in first draw $P\left( {{\rm{brow}}{{\rm{n}}_1}} \right) =
\dfrac{4}{9}$
Since, the socks are not replaced so after pulling one brown socks, there are 3 brown socks left in the
bag.
Probability of pulling a brown socks in second draw $P\left( {{\rm{brow}}{{\rm{n}}_2}} \right) =
\dfrac{3}{8}$
Similarly, for white socks
Probability of pulling a white socks in first draw $P\left( {{\rm{whit}}{{\rm{e}}_1}} \right) =
\dfrac{5}{9}$
After pulling one white socks, there are 4 white socks left in the bag.
Thus the probability of pulling a white socks in second draw $P\left( {{\rm{whit}}{{\rm{e}}_2}} \right) =
\dfrac{4}{8}$
Step III
We have to find the probability that the man gets both the socks of the same colour.
Required probability
 $\begin{array}{l} = P\left( {{\rm{brow}}{{\rm{n}}_1}} \right)P\left(
{{\rm{brow}}{{\rm{n}}_2}} \right) + P\left( {{\rm{whit}}{{\rm{e}}_1}} \right)P\left(
{{\rm{whit}}{{\rm{e}}_2}} \right)\\ = \dfrac{4}{9} \times \dfrac{3}{8} + \dfrac{5}{9} \times \dfrac{4}{8}\\ =
\dfrac{1}{6} + \dfrac{5}{{18}}\\ = \dfrac{3+5}{18}\end{array}$
Thus, the probability that the man gets both the socks of the same colour is $\dfrac{4}{9}$.
Hence, the correct option is D.

Note: Probability explains the chances of happening an even Here we have to determine the
probability that the man gets both the socks of the same colour. In step I, the number of brown
socks and white socks are given, so we can easily calculate the total number of socks present in
the bag. In step II, we calculate the probability of selecting the same coloured socks for each
case. Finally, in step III we find the probability of selecting the both the socks of the same
colour, using the data from step I and step II.