Question

Bag A contains 1 white, 2 blue, and 3 red balls. Bag B contains 3 white, 3 blue, and 2 red balls. Bag C contains 2 white, 3 blue, and 4 red balls. One bag is selected at random and then two balls are drawn from the selected bag. Find the probability that the balls drawn are white and red.

Answer 1

Hint: Let us consider one of the bags given in the question. In bag A, the total number of balls is equal to 6. If we draw a random ball from the bag, the chances of the ball being blue are 2 out of 6. Therefore the probability of drawing a blue ball from bag A is $P(B)=\dfrac{2}{6}=\dfrac{1}{3}$.
In theory, probability deals with finding the possibilities of achieving the desired result.

Complete Step by Step Solution:
First, calculate the probability of selecting a bag out of the three bags A, B, and C. The total number of bags is equal to 3 and the number of bags drawn at a time is one.
The probability of drawing a bag out of 3 bags is equal to $\dfrac{1}{3}$
Now let us find out the probability of drawing a white ball and a red ball out of the available balls in each bag.
Bag A has 6 balls. The sample space of the balls in the bag drawn 2 at a time will have ${}^{6}{{C}_{2}}$ combinations.
$\Rightarrow {}^{6}{{C}_{2}}=\dfrac{6!}{2!\left( 6-2 \right)!}=15$
The number of white and red balls in this bag is 1 and 3 respectively. Out of the two balls drawn, the number of combinations the white and red balls will have is equal to $1\times 3=3$.
Therefore the probability that two balls that are drawn from bag A are white and red is $\Rightarrow P(A)=\dfrac{3}{15}=\dfrac{1}{5}$
Bag B has 8 balls in total. Therefore the sample space of the ball in the bag drawn two at a random draw have ${}^{8}{{C}_{2}}$ combinations.
$\Rightarrow {}^{8}{{C}_{2}}=\dfrac{8!}{2!\left( 8-2 \right)!}=28$
And the number of combinations formed by the white and red balls in bag B is $3\times 2=6$. And so the probability of drawing a white ball and a red ball took two at a time from bag B is
$\Rightarrow P\left( B \right)=\dfrac{6}{28}=\dfrac{3}{14}$
In bag C, the sample space will have ${}^{9}{{C}_{2}}$ combinations.
$\Rightarrow {}^{9}{{C}_{2}}=\dfrac{9!}{2!\left( 9-2 \right)!}=36$
And the number of combinations formed by the white and red ball in bag C is $2\times 4=8$. And so the probability of drawing a white ball and a red ball from bag C is
$\Rightarrow P\left( C \right)=\dfrac{8}{36}=\dfrac{2}{9}$
The probability of selecting one bag out of the three and drawing two balls which is a combination of white and red is
$\Rightarrow P=\dfrac{1}{3}\left( P\left( A \right)+P\left( B \right)+P\left( C \right) \right)=\dfrac{1}{3}\left( \dfrac{1}{5} \right)+\dfrac{1}{3}\left( \dfrac{3}{14} \right)+\dfrac{1}{3}\left( \dfrac{2}{9} \right)$
$\Rightarrow P=\dfrac{1}{15}+\dfrac{3}{42}+\dfrac{2}{27}=\dfrac{401}{1890}$

Hence in every 1890 draws, 401 draws will be a combination of red and white balls taken two at a time.

Note:
Probability can be expressed in percentages also. This is one of the ways where the probability can be easily understood by just looking at it. The probability $P=\dfrac{401}{1890}$ can be expressed in percentage as $\dfrac{401}{1890}\times 100=21.21%$. It can be stated that there is a 21.21 percentage possibility of finding our desired result.