Question

A bag contains \[4\] white balls \[3\] black balls. Two balls are drawn at random. Find the probability that they are of the same colour.

Answer 1

Hint: At first, we will find the ways the white balls and black balls can be drawn separately. Then we will find the ways the two can be drawn from the total number balls. From there we can find our required probability.

Complete step-by-step answer:
It is given that a bag contains \[4\] white balls \[3\] black balls
Two balls are drawn at random.
We have to find the probability that they are of same colour.
We know that, if \[n\]be the total number of outcomes and \[m\] be the outcomes of an event.
So, the probability of any event is\[\dfrac{m}{n}\].
As the bag contains \[4\] white balls, we can draw \[2\] white balls at a time.
So, \[2\] white balls can be drawn in \[^4{C_2}\] ways.
Again, the bag contains 3 blue balls, we can draw \[2\] blue balls at a time.
So, \[2\] blue balls can be drawn in \[^3{C_2}\] ways.
Now, total number of balls in the bag is \[4 + 3 = 7\]
We can draw \[2\] balls at a time from \[7\] balls.
So, \[2\] balls at a time from \[7\] balls in \[^7{C_2}\] ways.
Therefore, the required probability is \[ = \dfrac{{^4{C_2}{ + ^3}{C_2}}}{{^7{C_2}}}\]
On substituting the formula of combinations we get,
Probability \[ = \dfrac{{\dfrac{{4!}}{{2!2!}} + \dfrac{{3!}}{{2!1!}}}}{{\dfrac{{7!}}{{2!5!}}}}\]
Let us simplify again using factorials we get,
\[ \Rightarrow \dfrac{{\dfrac{{4!}}{{2!2!}} + \dfrac{{3!}}{{2!1!}}}}{{\dfrac{{7!}}{{2!5!}}}} = \dfrac{{\dfrac{{4 \times 3 \times 2}}{{2 \times 2}} + \dfrac{{3 \times 2}}{2}}}{{\dfrac{{7 \times 6 \times 5!}}{{2 \times 5!}}}}\]
\[ \Rightarrow \dfrac{{\dfrac{{3 \times 2}}{1} + \dfrac{3}{1}}}{{\dfrac{{7 \times 3}}{1}}} = \dfrac{{2 \times 3 + 3}}{{7 \times 3}}\]
Probability that the drawn balls have same colour is \[\dfrac{{2 \times 3 + 3}}{{7 \times 3}} = \dfrac{9}{{21}} = \dfrac{3}{7}\]

Hence, we have found the probability that they are of the same colour as \[\dfrac{3}{7}\].

Note: We know that, if \[n\] be the total number of outcomes and \[m\] be the outcomes of an event. So, the probability of any event is \[\dfrac{m}{n}\].
Since the balls are taken randomly we use combinations to solve the problem.