Question

A bag contains 3 white and 3 red balls, pairs of balls are drawn without replacement until the bag is empty. The probability that each pair consist of one white and one red ball is-
a. \[\dfrac{2}{5}\]
b. \[\dfrac{4}{{10}}\]
c. \[\dfrac{5}{{10}}\]
d. \[\dfrac{6}{{10}}\]

Answer 1

Hint:
It is given that we have 3 red and 3 white balls from which we need to take out a pair of 1 white and red ball, without replacement until the bag is empty. So we will start choosing one ball from 3 white and one ball from 3 red balls, and find the number of ways in which we can do so, and divide it by the total number of ways of selecting 2 balls from 6 balls, then for the second time we have 2 red and 2 white balls remaining, we again follow same steps until the bag is empty. The product of all the probabilities will be the required answer.

Complete step by step solution:
Given A bag contains 3 white and 3 red balls
The total number of ways of selecting 2 balls from 6 balls is \[{}^6{C_2}\].
Now firstly we select one white ball and one red ball from the bag.
As there are 3 white balls, so the number of ways of selecting 1 white ball is \[{}^3{C_1}\]
And as there are 3 red balls, so the number of ways of selecting 1 red ball is \[{}^3{C_1}\]
Hence, In first draw, probability \[ = \dfrac{{{}^3{C_1} \times {}^3{C_1}}}{{{}^6{C_2}}}\]
As \[{}^n{C_{n - 1}} = n\] and \[{}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}\] , we get,
\[ = \dfrac{{3 \times 3}}{{\dfrac{{6!}}{{2!4!}}}}\]
On simplification we get,
\[ = \dfrac{9}{{\dfrac{{6 \times 5}}{2}}}\]
On simplifying the denominator we get,
\[ = \dfrac{9}{{15}}\]
On dividing the numerator and denominator by 3 we get,
\[ = \dfrac{3}{5}\]
Now in the bag there are 2 white balls and 2 red balls,
So, the number of ways of selecting 1 ball from 2 white balls are \[{}^2{C_1}\]
And, the number of ways of selecting 1 ball from 2 red balls are \[{}^2{C_1}\]
The number of ways of selecting 2 balls from 4 balls are \[{}^4{C_2}\]
Hence, In second draw, probability \[ = \dfrac{{{}^2{C_1} \times {}^2{C_1}}}{{{}^4{C_2}}}\]
As \[{}^n{C_{n - 1}} = n\] and \[{}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}\] , we get,
\[ = \dfrac{{2 \times 2}}{{\dfrac{{4!}}{{2!2!}}}}\]
On simplification we get,
\[ = \dfrac{{2 \times 2}}{{\dfrac{{4 \times 3}}{2}}}\]
On simplifying the denominator we get,
\[ = \dfrac{4}{6}\]
On dividing the numerator and denominator by 2 we get,
\[ = \dfrac{2}{3}\]
Now there are only 2 balls in the bag
So number of ways of selecting 2 balls from 2 balls are \[{}^2{C_2}\]
Now the number of ways of selecting 1 ball from 1 white ball is \[{}^1{C_1}\]
Now the number of ways of selecting 1 ball from 1 red ball is \[{}^1{C_1}\]
In third draw, probability \[ = \dfrac{{{}^1{C_1} \times {}^1{C_1}}}{{{}^2{C_2}}}\]\[ = 1\]
So, the total probability \[ = \dfrac{3}{5} \times \dfrac{2}{3} \times 1\]\[ = \dfrac{2}{5}\]

So, we have our solution as, \[\dfrac{4}{{10}}\]\[ = \dfrac{2}{5}\] as our option (a) and (b).

Note:
Here in this type of general probability problems once we get the results of how to arrange the objects, we usually make a mistake of arranging them among themselves. The value in which way it can be arranged among themselves is to be divided in the probability term.