There are 6 black, 2 brown, 4 green, and 3 white balls in a bag. If 5 balls are picked up randomly, what is the probability that at least one ball is black?
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Answer:
There are two ways to interpret this question: either no green balls are selected or one black ball and one which ball is selected.
There are 12 balls in total.
Assuming no green balls are to be selected, then the blacks balls and white balls can be lumped into the same group of 7. On the first draw, the chances of drawing one of the group would be (7 of 12) and on the second draw the chances would be (6 of 11). So the chances of selecting twice in a row from the combined group would be:
(712×611)=722=0.318¯¯¯¯¯ .
Assuming one black ball and one white ball is to be selected, and noting that there are two orders (WB and BW), the chances would be:
Answer:
There are two ways to interpret this question: either no green balls are selected or one black ball and one which ball is selected.
There are 12 balls in total.
Assuming no green balls are to be selected, then the blacks balls and white balls can be lumped into the same group of 7. On the first draw, the chances of drawing one of the group would be (7 of 12) and on the second draw the chances would be (6 of 11). So the chances of selecting twice in a row from the combined group would be:
(712×611)=722=0.318¯¯¯¯¯ .
Assuming one black ball and one white ball is to be selected, and noting that there are two orders (WB and BW), the chances would be:
2(412×311)=211=0.18¯¯¯¯¯ .
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