Explanation
Solution
1. Identify the Possible Cases
Let Ei be the event that the bag contains i black balls. Since we already drew 2 black balls, the bag must have contained at least 2 black balls. Therefore, the possible number of black balls in the bag originally could be i∈{2,3,4,5,6}. Assuming all these compositions are equally likely, P(Ei)=51.
2. Define the Observed Event
Let A be the event that 2 balls drawn at random are black. The probability of drawing 2 black balls given that there are i black balls in a bag of 6 is:
P(A∣Ei)=(26)(2i)=6×5i(i−1)=30i(i−1)
3. Apply Bayes' Theorem
We want to find the probability that the bag contains at least 5 black balls, which is P(E5∪E6∣A).
P(E5∪E6∣A)=∑i=26P(A∣Ei)P(Ei)P(A∣E5)P(E5)+P(A∣E6)P(E6)
Since P(Ei) is constant for all i, they cancel out:
P(at least 5 black)=(22)+(23)+(24)+(25)+(26)(25)+(26)
4. Calculation
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(22)=1
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(23)=3
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(24)=6
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(25)=10
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(26)=15
Sum of all combinations =1+3+6+10+15=35.
Sum for i≥5 is 10+15=25.
Correct Option: (2) 75
