Question

A bag contains 6 white and 4 red balls. Three balls are drawn at random. What is the probability that one ball is red and the other two are white?
$(a){\text{ }}\dfrac{1}{2} \\\ (b){\text{ }}\dfrac{1}{{12}} \\\ (c){\text{ }}\dfrac{3}{{10}} \\\ (d){\text{ }}\dfrac{7}{{12}} \\\$

Answer 1

Hint – In this problem use the basic concept of probability which is the ratio of favorable outcome and total possible outcome. Find the total possible outcomes as the bags contain in total 10 balls. Then find the ways of selecting 1 red ball from 4 red balls and ways of selecting 2 white balls from 6 white balls. This will help getting the right option.

Complete step-by-step answer:
Given data
White balls = 6
Red balls = 4
So the total number of balls = 6 + 4 = 10.
Now three balls are drawn random so the number of ways of drawing 3 red balls from 16 balls.
$= {}^{10}{C_3}$
So the total number of outcomes $= {}^{10}{C_3}$.
Now there are 10 red balls so the total number of ways to draw 1 red ball out of 4 red balls are
$= {}^4{C_1}$
Now there are 6 white balls so the total number of ways to draw 2 white balls out of 6 white balls are
$= {}^6{C_2}$
So the total number of ways to draw 1 red ball and 2 white balls are
$= {}^4{C_1} \times {}^6{C_2}$
So the favorable number of outcomes $= {}^4{C_1} \times {}^6{C_2}$.
Now as we know that the probability (P) is the ratio of favorable number of outcomes to the total number of outcomes.
$\Rightarrow P = \dfrac{{{}^4{C_1} \times {}^6{C_2}}}{{{}^{10}{C_3}}}$
Now as we know that $\left[ {{}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}} \right]$ so use this property we have,
$\Rightarrow P = \dfrac{{{}^4{C_1} \times {}^6{C_2}}}{{{}^{10}{C_3}}} = \dfrac{{\dfrac{{4!}}{{1!\left( {4 - 1} \right)!}} \times \dfrac{{6!}}{{2!\left( {6 - 2} \right)!}}}}{{\dfrac{{10!}}{{3!\left( {10 - 3} \right)!}}}}$
Now simplify the above equation we have,
$\Rightarrow P = \dfrac{{\dfrac{{4!}}{{1!\left( {4 - 1} \right)!}} \times \dfrac{{6!}}{{2!\left( {6 - 2} \right)!}}}}{{\dfrac{{10!}}{{3!\left( {10 - 3} \right)!}}}} = \dfrac{{4!.6!.3!.7!}}{{1!.3!.2!.4!.10!}}$
Now as we know that $n! = n\left( {n - 1} \right)\left( {n - 2} \right)\left( {n - 3} \right)............$ so we have,
$\Rightarrow P = \dfrac{{4!.6!.3!.7!}}{{1!.3!.2!.4!.10!}} = \dfrac{{6!.7!}}{{2!.10!}} = \dfrac{{6 \times 5 \times 4 \times 3 \times 2! \times 7!}}{{2! \times 10 \times 9 \times 8 \times 7!}}$
$\Rightarrow P = \dfrac{{6 \times 5 \times 4 \times 3}}{{10 \times 9 \times 8}} = \dfrac{1}{2}$
So this is the required probability.
Hence option (A) is correct.

Note – The key point here is that we have multiplied both the possible ways so selecting red balls and white balls from the given number of balls and that constitute the favorable outcome. This happens because we need to find the probability of one red and other two white, this and word is the key word here as it resembles the same just like the intersection of two events.