Bag A contains 6 white balls and 4 black balls and bag B contains 3 wh

nick1816
I posted the solution incomplete the combination part along each condition got left out my bad.. ?



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Solution

Given

• Bag A contains 6 white balls and 4 black balls and bag B contains 3 white balls and 2 black balls.
• A white ball is picked from bag A and put into bag B.

o Then, three balls are picked from bag B and put into bag A.

To find

• The probability that a ball picked now from bag A is black.

Approach and Working out

• Bag A contains 6 white balls and 4 black balls and bag B contains 3 white balls and 2 black balls
• A white ball is picked from bag A and put into bag B

o So, Bag A - 5 white balls and 4 black balls
o Bag B - 4 white and 2 black balls.
• Then, three balls are picked from bag B and put into bag A

o There will be 3 cases in which the balls from bag B can be picked.

 2 black and 1 white

• Bag A – 6 White and 6 black
• Probability = \(\frac{4c1 * 2c2 }{6c3} * \frac{6}{12}\) = \(\frac{1}{10}\)
 1 black and 2 white

• Bag A – 7 White and 5 black
• Probability = \(\frac{4c2 * 2c1 }{6c3} * \frac{6}{12}\) = \(\frac{1}{4}\)
 3 white

• Bag A – 8 White and 4 black
• Probability = \(\frac{4c3 }{6c3} * \frac{6}{12}\) = \(\frac{1}{15}\)
• Total probability= 1/10 +1 /4 + 1/15 =5/12

Thus, option D is the correct answer.
Correct Answer: Option D

Can someone explain why 4c3/6c3? I dont understand the need of combinations here?

AbenGeorge

Three balls are picked from bag B and put into bag A and a ball picked now from bag A is black are dependent events [ the outcome of the first event influences the outcome of the second event]. Hence, The probability of two dependent events is the product of the probability of first event and that of second event.

Case 1- when all 3 picks from bag B are white, then probability of a ball picked now from bag A is black.

Number of balls we have after First transfer
Bag B = 4 white and 2 black

Probability of picking 3 white balls from bag B = Number of ways to pick 3 white balls from bag B/Number of ways to pick any 3 balls from bag B = \(\frac{4C3}{6C3}\)

Number of balls we have after Second transfer
Bag A = 8 white and 4 black

probability of a ball picked now from bag A is black= \(\frac{4C1}{12C1}\)

Total probability = \(\frac{4C3}{6C3} * \frac{4C1}{12C1}\)

Similarly you can find probability for other 2 cases.

me too.... isnt the probablility enough for the solution
in the first case isnt the answer 4/12=1/3 enough.
Why are we bringing combinations here ?
Can someone help? Bunuel

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