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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.

Last updated date: 20th Jul 2024
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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.