Question 1154917: There are three boxes containing black and white balls. The number of each color is Box 1, 4 white and 2 black; Box 2, 3 white and 2 black; Box 3, 1 white and 2 black.
A) What is the probability that the last ball, drawn from box 3 is white?
B) What is the probability that the last ball drawn is black?
Found 2 solutions by ikleyn, greenestamps:
Answer by ikleyn(52781)   (Show Source):
You can put this solution on YOUR website! .
The description of this probability experiment is incomplete.
Therefore, the problem CAN NOT be solved and the answer can not be given.
To be complete, it must say, how many balls are drawn; whether they were drawn with the replacement or not.
May I ask you: for what reason do you post problems to this forum, when you don't know how to formulate them correctly ?
May you clarify me, please ?
For what reason do you disturb the tutors ?
Answer by greenestamps(13200)  (Show Source):
You can put this solution on YOUR website!
A) There are 3 balls in box 3; 1 of them is white:
P(last ball drawn from box 3 is white) = 1/3.
B) In the 3 boxes together, there are a total of 14 balls, of which 8 are white:
P(last ball drawn (from any box) is white) = 8/14 = 4/7.
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Note that I ignored the comma (",") in part A. That makes the probability to be found simply the probability that the white ball is the last one drawn from box 3.
If the question in part A had been written with two commas --
"What is the probability that the last ball, drawn from box 3, is white?"
then that would be a different probability: the probability that the last ball drawn is white, AND THAT IT COMES FROM BOX 3. That would of course be a different problem.
The question as written, with the single comma, is grammatically incorrect and therefore open to different interpretations.
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