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# There are two bags A and B. Bag A contains 3 white and 4 red balls whereas bag B contains 4 white and 3 red balls. Three balls are drawn at random (without replacement) from one of the bags and are found to be two white and one red. Find the probability that these were drawn from bag B. (a) $\dfrac{7}{11}$ (b) $\dfrac{5}{11}$ (c) $\dfrac{2}{5}$ (d) $\dfrac{3}{5}$  Hint: Here, we will use the concept of conditional probability. The probability of an event A when another event B has already occurred is known as conditional property and it is given as $P\left( \dfrac{A}{B} \right)=\dfrac{P\left( A\cap B \right)}{P\left( B \right)}$ .

Complete step-by-step solution -
Let us assume that the event of drawing two white and one red ball is E.
Let B be the event of drawing the balls from bag B and A be the event of drawing the balls from bag A.
Since, we have to find that if two white and one red ball is drawn, then what is the probability of them being from bag B, that is we have to find the value of $P\left( \dfrac{B}{E} \right)$.
Since, $P\left( \dfrac{B}{E} \right)=\dfrac{P\left( B\cap E \right)}{P\left( E \right)}.........\left( 1 \right)$
Let us find the value of P(E).
We know that 2 white and 1 red ball can be drawn either from bag A or from bag B. So, the probability of drawing them is equal to the sum of the probability of them being drawn from bag A and bag B.
So, $P\left( E \right)=P\left( \dfrac{E}{A} \right)\times P\left( A \right)+P\left( \dfrac{E}{B} \right)\times P\left( B \right)$
$P\left( \dfrac{E}{A} \right)$ = Probability of drawing 2 white and 1 red ball from bag A.
Since the number of white balls in bag A = 3.
So, the probability of drawing 1st white ball = $\dfrac{3}{7}$
Now, two white balls are remaining. So, the probability of drawing 2nd white ball from a total of 6 balls = $\dfrac{2}{6}$.
And, the probability of selecting 1 red ball from a total of 5 remaining balls = $\dfrac{4}{5}$.
The probability of selecting bag A = $\dfrac{1}{2}$
So, $P\left( \dfrac{E}{A} \right)\times P\left( A \right)=\left( \dfrac{3}{7}\times \dfrac{2}{6}\times \dfrac{4}{5}\times \dfrac{1}{2} \right)=\dfrac{24}{420}=\dfrac{2}{35}$
Now, probability of drawing 1st white ball from bag B = $\dfrac{4}{7}$
Probability of drawing 2nd white ball from a total of 6 balls = $\dfrac{3}{6}$
Probability of drawing 1 white ball from a total of 5 remaining balls = $\dfrac{3}{5}$
Probability of selecting bag B = $\dfrac{1}{2}$.
Therefore, $P\left( \dfrac{E}{B} \right)\times P\left( B \right)=\dfrac{4}{7}\times \dfrac{3}{6}\times \dfrac{3}{5}\times \dfrac{1}{2}=\dfrac{36}{420}=\dfrac{3}{35}$
And, $P\left( B\cap E \right)=P\left( \dfrac{E}{B} \right)\times P\left( B \right)$ {from equation (1)}
= $\dfrac{3}{35}$
So, $P\left( E \right)=\dfrac{2}{35}+\dfrac{3}{35}=\dfrac{5}{35}=\dfrac{1}{7}$
Therefore, from equation (1), $P\left( \dfrac{B}{E} \right)=\dfrac{\left( \dfrac{3}{35} \right)}{\left( \dfrac{1}{7} \right)}=\dfrac{3\times 7}{35}=\dfrac{3}{5}$
Hence, option (d) is the correct answer.

Note: Students should note here that after selecting one white ball from any of the bags, the total no of balls get reduced by one. So, while selecting the second white ball, the total no of balls is one less than that in the first case. Again while choosing the third ball, the total number of balls is one less than the total number of balls in the second case.

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