SOLUTION: 4. One bag contains 3 white balls and 4 black balls, and a second bag contains 4 white balls and 5 black balls. One ball is drawn from the first bag and placed unseen in the second
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-> SOLUTION: 4. One bag contains 3 white balls and 4 black balls, and a second bag contains 4 white balls and 5 black balls. One ball is drawn from the first bag and placed unseen in the second
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Question 1132098: 4. One bag contains 3 white balls and 4 black balls, and a second bag contains 4 white balls and 5 black balls. One ball is drawn from the first bag and placed unseen in the second bag. What is the probability that the ball drawn from the second bag is white given that the ball drawn from the first bag is black?
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could you please tell me how to solve this question i really dont understand the question neither its solution Answer by ikleyn(52800) (Show Source):
We are given that there were 4 white balls and 5 black balls in the second bag initially.
We are also given that the ball drawn from the first bag and placed in the second bag was black.
So, after this transfer, there are 4 white balls and 5+1 = 6 black balls in the second bag.
Hence, the probability that the ball drawn from the second bag is white is equal to
= = = 0.4 = 40%. ANSWER
Solved.
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The key step to solve this problem is TO READ its condition attentively.
The rest is TRIVIAL.
This problem is for the 3rd grade students of a primary school.
It does not require any special knowledge or skills - except the skill to think logically (I mean - the skill
to make one simple evident logical conclusion after reading the text).
Then the notion/conception "probability" is clear at the intuitive level.
