archit Always remember: probability can never be greater than 1 (at least on Earth
It's a conditional probability question.
Number of balls we have after First transfer
Bag A = 5 white and 4 black
Bag B = 4 white and 2 black
Now there are 3 cases possible for 2nd transfer (as you mentioned)
1. when all 3 picks from bag B are white, then probability of a ball picked now from bag A is black.
\(\frac{4C3}{6C3} * \frac{4}{12}= \frac{1}{15}\)
2. when 2 picks from bag B are white and 1 pick is black, then probability of a ball picked now from bag A is black.
\(\frac{4C2*2C1}{6C3} * \frac{5}{12}= \frac{1}{4}\)
3. when 1 pick from bag B is white and 2 picks are black, then probability of a ball picked now from bag A is black.
\(\frac{4C1*2C2}{6C3} * \frac{6}{12}= \frac{1}{10}\)
Total probability= \(\frac{1}{15}+\frac{1}{4}+\frac{1}{10}= \frac{4+15+6}{60}= \frac{5}{12}\)
Archit3110 wrote:
Bag A = 6 white and 4 black = 10 total
Bag B = 3 white and 2 black = 5 total
now
Bag A = 5 white and 4 black = 9 total
Bag B = 4 white and 2 black = 6 total
three balls are picked from bag B and put into bag A Find the probability that a ball picked now from bag A is black.
case 1 ; all three white = Bag A ; 8 white + 4 black ; 4/12
case 2 ; 2 white and 1 black ; Bag A ; 7 white + 5 black ; 5/12
case 3 ; 1 white and 2 black ; Bag A; 6 white + 6 black ; 6/12
total probability that a ball picked now from bag A is black = 4/12 +5/12+6/12 ; 15/12 ; 5/4
Bunuel wrote:
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. Then, three balls are picked from bag B and put into bag A. Find the probability that a ball picked now from bag A is black.
A. 1/4
B. 1/3
C. 7/12
D. 5/12
E. 11/24
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