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Question: A bag contains 3 white and 3 red balls, pairs of balls are drawn without replacement until the bag i...

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. 25\dfrac{2}{5}
b. 410\dfrac{4}{{10}}
c. 510\dfrac{5}{{10}}
d. 610\dfrac{6}{{10}}

Explanation

Solution

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 6C2{}^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 3C1{}^3{C_1}
And as there are 3 red balls, so the number of ways of selecting 1 red ball is 3C1{}^3{C_1}
Hence, In first draw, probability =3C1×3C16C2 = \dfrac{{{}^3{C_1} \times {}^3{C_1}}}{{{}^6{C_2}}}
As nCn−1=n{}^n{C_{n - 1}} = n and nCr=n!r!(n−r)!{}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}} , we get,
=3×36!2!4!= \dfrac{{3 \times 3}}{{\dfrac{{6!}}{{2!4!}}}}
On simplification we get,
=96×52= \dfrac{9}{{\dfrac{{6 \times 5}}{2}}}
On simplifying the denominator we get,
=915= \dfrac{9}{{15}}
On dividing the numerator and denominator by 3 we get,
=35= \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 2C1{}^2{C_1}
And, the number of ways of selecting 1 ball from 2 red balls are 2C1{}^2{C_1}
The number of ways of selecting 2 balls from 4 balls are 4C2{}^4{C_2}
Hence, In second draw, probability =2C1×2C14C2 = \dfrac{{{}^2{C_1} \times {}^2{C_1}}}{{{}^4{C_2}}}
As nCn−1=n{}^n{C_{n - 1}} = n and nCr=n!r!(n−r)!{}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}} , we get,
=2×24!2!2!= \dfrac{{2 \times 2}}{{\dfrac{{4!}}{{2!2!}}}}
On simplification we get,
=2×24×32= \dfrac{{2 \times 2}}{{\dfrac{{4 \times 3}}{2}}}
On simplifying the denominator we get,
=46= \dfrac{4}{6}
On dividing the numerator and denominator by 2 we get,
=23= \dfrac{2}{3}
Now there are only 2 balls in the bag
So number of ways of selecting 2 balls from 2 balls are 2C2{}^2{C_2}
Now the number of ways of selecting 1 ball from 1 white ball is 1C1{}^1{C_1}
Now the number of ways of selecting 1 ball from 1 red ball is 1C1{}^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.