Bag I contains 3 red, 4 black and 3 white balls and Bag II contains 2 red, 5 black and 2 white balls. One balls is transferred from Bag I to Bag II and then a ball is drawn from Bag II. The ball so drawn is found to be black in colour. Then the probability, that the transferred is red, is:
Explanation
1. Define the Events:
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E1: Transferred ball is Red.
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E2: Transferred ball is Black.
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E3: Transferred ball is White.
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A: Ball drawn from Bag II is Black.
2. Probabilities of Transferring Balls from Bag I:
Total balls in Bag I = 3+4+3=10.
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P(E1)=103
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P(E2)=104
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P(E3)=103
3. Conditional Probabilities of Drawing a Black Ball from Bag II:
Bag II initially has 2+5+2=9 balls. After transfer, it has 10 balls.
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If Red is transferred (E1), Bag II black balls remain 5: P(A∣E1)=105
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If Black is transferred (E2), Bag II black balls become 6: P(A∣E2)=106
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If White is transferred (E3), Bag II black balls remain 5: P(A∣E3)=105
4. Using Bayes' Theorem:
We need to find P(E1∣A):
P(E1∣A)=P(E1)⋅P(A∣E1)+P(E2)⋅P(A∣E2)+P(E3)⋅P(A∣E3)P(E1)⋅P(A∣E1)
5. Substitution:
P(E1∣A)=(103×105)+(104×106)+(103×105)103×105
P(E1∣A)=10015+10024+1001510015
6. Simplification:
Dividing both numerator and denominator by 3:
