Statistics and Probability: Mathematics | JEE Main
A bag contains 6 balls. Two balls are drawn from it at random and both are found to be black. The probability that the bag contains at least 5 black balls is
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Hint 1 of 3
Assuming the bag initially contains between 2 and 6 black balls with equal prior probability, how is the conditional probability computed via Bayes' theorem?
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Step-by-step solutionView
Correct answer
The probability that the bag contains at least 5 black balls given that two drawn balls are black is 5/7.
Option analysis
Why each option works or fails
A · 73
The student likely computes the probability of having exactly 5 black balls instead of at least 5, or makes an arithmetic error summing the favorable likelihoods. Include both favorable hypotheses (5 black balls and 6 black balls) when summing the favorable numerator values.
B · 75
This is the correct option. Using Bayes' theorem with an equally likely prior for the number of black balls (2 to 6), the sum of favorable weights for 5 and 6 black balls divided by the total weight gives (10 + 15) / (1 + 3 + 6 + 10 + 15) = 25/35 = 5/7. Correctly applied Bayes' theorem to find the posterior probability.
C · 65
The student confuses the prior ratio of black ball counts (5 out of 6) directly with the posterior probability, ignoring the conditional likelihood given by drawing two black balls. Weight each possible number of black balls by the conditional probability of drawing two black balls from it.
D · 72
The student computes the probability of having at most 4 black balls (the complement) or only considers the event of having exactly 6 black balls while making an arithmetic error. Ensure the target event represents having 5 or 6 black balls, not the complementary cases.
Reviewed route
Solution
StepWorking
01given
A bag contains 6 balls of unknown colors. Two balls drawn at random are both black. Let E be the event that 2 drawn balls are black. The number of black balls in the bag can initially be any integer from 2 to 6 (since at least 2 black balls exist), each composition being equally likely.
02goal
Find the posterior probability that the bag contains at least 5 black balls, i.e., P(B5∪B6∣E).
03approach
Apply Bayes' theorem. Let Bi denote the hypothesis that the bag contains i black balls (i∈{2,3,4,5,6}). Assuming equal prior probabilities P(Bi)=51, the conditional probability P(E∣Bi)=(26)(2i). Then P(B5∪B6∣E)=∑i=26(2i)(25)+(26).
04execute
Calculate the combinations: (22)=1, (23)=3, (24)=6, (25)=10, (26)=15. The numerator is 10+15=25 and the denominator is 1+3+6+10+15=35. Thus, P=3525=75.
✓verify
The sum ∑i=26(2i)=(37)=35 by the hockey-stick identity. Numerator is 10+15=25. Ratio is 3525=75.
Hints that build this answer step by step
Assuming the bag initially contains between 2 and 6 black balls with equal prior probability, how is the conditional probability computed via Bayes' theorem?
P(≥5∣B2)=∑k=26P(B2∣k)P(B2∣5)+P(B2∣6)
What is the likelihood (2k) proportional to P(B2∣k) for each k∈{2,3,4,5,6}?
1,3,6,10,15
What is the ratio of the favorable sum to the total sum of these likelihood weights?
Why are hypotheses B0 and B1 omitted in the denominator?
Because if the bag had 0 or 1 black balls, the probability of drawing 2 black balls is 0 (i.e., (20)=(21)=0), so they contribute 0 to the total probability.