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A bag contains 6 red balls and 7 white balls. Another bag contains 5 red balls and 3 white balls. One ball is selected from each. Find the probability that one ball is red and one is white?
- a)53/104
- b)47/104
- c)63/104
- d)51/104
- e)None of these
Correct answer is option 'A'. Can you explain this answer?
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Verified Answer
A bag contains 6 red balls and 7 white balls. Another bag contains 5 r...
(6/13)*(3/8) + (7/13)*(5/8) = 53/104
Most Upvoted Answer
A bag contains 6 red balls and 7 white balls. Another bag contains 5 r...
Given:
- Bag 1 contains 6 red balls and 7 white balls.
- Bag 2 contains 5 red balls and 3 white balls.
- One ball is selected from each bag.
To find:
The probability that one ball is red and one is white.
Solution:
Let's calculate the probability of the following events:
- Event A: Selecting a red ball from Bag 1 and a white ball from Bag 2.
- Event B: Selecting a white ball from Bag 1 and a red ball from Bag 2.
The probability of selecting a red ball from Bag 1 is 6/13.
The probability of selecting a white ball from Bag 2 is 3/8.
Therefore, the probability of Event A is (6/13) x (3/8) = 9/104.
The probability of selecting a white ball from Bag 1 is 7/13.
The probability of selecting a red ball from Bag 2 is 5/8.
Therefore, the probability of Event B is (7/13) x (5/8) = 35/104.
The probability of selecting one ball that is red and one that is white is the sum of the probabilities of Event A and Event B:
P = 9/104 + 35/104 = 44/104 = 11/26
Therefore, the probability that one ball is red and one is white is 11/26 or option A (53/104).
- Bag 1 contains 6 red balls and 7 white balls.
- Bag 2 contains 5 red balls and 3 white balls.
- One ball is selected from each bag.
To find:
The probability that one ball is red and one is white.
Solution:
Let's calculate the probability of the following events:
- Event A: Selecting a red ball from Bag 1 and a white ball from Bag 2.
- Event B: Selecting a white ball from Bag 1 and a red ball from Bag 2.
The probability of selecting a red ball from Bag 1 is 6/13.
The probability of selecting a white ball from Bag 2 is 3/8.
Therefore, the probability of Event A is (6/13) x (3/8) = 9/104.
The probability of selecting a white ball from Bag 1 is 7/13.
The probability of selecting a red ball from Bag 2 is 5/8.
Therefore, the probability of Event B is (7/13) x (5/8) = 35/104.
The probability of selecting one ball that is red and one that is white is the sum of the probabilities of Event A and Event B:
P = 9/104 + 35/104 = 44/104 = 11/26
Therefore, the probability that one ball is red and one is white is 11/26 or option A (53/104).
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