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Bag 1 contains 4 white and 6 black balls while another Bag 2 contains 4 white and 3 black balls. One ball is drawn at random from one of the bags and it is found to be black. Find the probability that it was drawn from Bag 1.
- a)12/13
- b)5/12
- c)7/11
- d)7/12
Correct answer is option 'D'. Can you explain this answer?
Most Upvoted Answer
Bag 1 contains 4 white and 6 black balls while another Bag 2 contains ...
Solution:
We are given that there are two bags, Bag 1 and Bag 2.
Bag 1 contains 4 white and 6 black balls.
Bag 2 contains 4 white and 3 black balls.
One ball is drawn at random from one of the bags and it is found to be black.
We need to find the probability that it was drawn from Bag 1.
Let A be the event that the ball is drawn from Bag 1.
Let B be the event that the ball is black.
We need to find P(A|B), the probability that the ball is drawn from Bag 1 given that it is black.
Using Bayes' theorem, we have:
P(A|B) = P(B|A) * P(A) / P(B)
P(B|A) is the probability of drawing a black ball given that Bag 1 is selected. This is equal to 6/10 = 3/5.
P(A) is the probability of selecting Bag 1. This is equal to 1/2.
P(B) is the probability of drawing a black ball. This can be calculated using the law of total probability.
P(B) = P(B|A) * P(A) + P(B|not A) * P(not A)
P(B|not A) is the probability of drawing a black ball given that Bag 2 is selected. This is equal to 3/7.
P(not A) is the probability of not selecting Bag 1. This is equal to 1/2.
Therefore,
P(B) = (3/5) * (1/2) + (3/7) * (1/2) = 39/70
Substituting the values in Bayes' theorem, we have:
P(A|B) = (3/5) * (1/2) / (39/70) = 7/12
Therefore, the probability that the black ball was drawn from Bag 1 is 7/12.
Hence, the correct option is (d) 7/12.
We are given that there are two bags, Bag 1 and Bag 2.
Bag 1 contains 4 white and 6 black balls.
Bag 2 contains 4 white and 3 black balls.
One ball is drawn at random from one of the bags and it is found to be black.
We need to find the probability that it was drawn from Bag 1.
Let A be the event that the ball is drawn from Bag 1.
Let B be the event that the ball is black.
We need to find P(A|B), the probability that the ball is drawn from Bag 1 given that it is black.
Using Bayes' theorem, we have:
P(A|B) = P(B|A) * P(A) / P(B)
P(B|A) is the probability of drawing a black ball given that Bag 1 is selected. This is equal to 6/10 = 3/5.
P(A) is the probability of selecting Bag 1. This is equal to 1/2.
P(B) is the probability of drawing a black ball. This can be calculated using the law of total probability.
P(B) = P(B|A) * P(A) + P(B|not A) * P(not A)
P(B|not A) is the probability of drawing a black ball given that Bag 2 is selected. This is equal to 3/7.
P(not A) is the probability of not selecting Bag 1. This is equal to 1/2.
Therefore,
P(B) = (3/5) * (1/2) + (3/7) * (1/2) = 39/70
Substituting the values in Bayes' theorem, we have:
P(A|B) = (3/5) * (1/2) / (39/70) = 7/12
Therefore, the probability that the black ball was drawn from Bag 1 is 7/12.
Hence, the correct option is (d) 7/12.
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Community Answer
Bag 1 contains 4 white and 6 black balls while another Bag 2 contains ...
Let E1 = event of choosing the bag 1, E2 = event of choosing the bag 2.
Let A be event of drawing a black ball.
P(E1) = P(E2) = 1/2.
Also, P(A|E1) = P(drawing a black ball from Bag 1) = 6/10 = 3/5.
P(A|E2) = P(drawing a black ball from Bag 2) = 3/7.
By using Bayes’ theorem, the probability of drawing a black ball from bag 1 out of two bags is-:
P(E1 | A) = P(E1)P(A | E1)/( P(E1)P(A│E1)+P(E2)P(A | E2))
= (1/2 × 3/5) / ((1/2 × 3/7)) + (1/2 × 3/5)) = 7/12.
Let A be event of drawing a black ball.
P(E1) = P(E2) = 1/2.
Also, P(A|E1) = P(drawing a black ball from Bag 1) = 6/10 = 3/5.
P(A|E2) = P(drawing a black ball from Bag 2) = 3/7.
By using Bayes’ theorem, the probability of drawing a black ball from bag 1 out of two bags is-:
P(E1 | A) = P(E1)P(A | E1)/( P(E1)P(A│E1)+P(E2)P(A | E2))
= (1/2 × 3/5) / ((1/2 × 3/7)) + (1/2 × 3/5)) = 7/12.
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Bag 1 contains 4 white and 6 black balls while another Bag 2 contains 4 white and 3 black balls. One ball is drawn at random from one of the bags and it is found to be black. Find the probability that it was drawn from Bag 1.a)12/13b)5/12c)7/11d)7/12Correct answer is option 'D'. Can you explain this answer?
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