You choose one of two identical looking bags at random. One bag has three black marbles and one white marble. The other has three white marbles and one black marble.
After choosing a bag you draw one marble out at random. You notice it is black. You then put it back and draw another marble out of the same bag at random.
What is the probability that the second marble drawn is black?
5/8
To me, the simplest approach would be to contruct a probability tree diagram. Let A be the event that you chose the bag with three black marbles, and B is the event that you chose the bag with three white marbles.
/\
/ \
/ \
/ \
/ \
/ \
½/ \½
/ \
/ \
/ \
/ \
A B
/\ /\
/ \ / \
¼/ \¾ ¾/ \¼
/ \ / \
/ \ / \
W B W B
/\ /\ /\ /\
¼/ \¾ ¼/ \¾ ¾/ \¼ ¾/ \¼
W B W B W B W B
1/32 3/32 3/32 9/32 9/32 3/32 3/32 1/32
The tree is deterministic.
Each fraction along a segment represents the conditional probability of that branch given the previous step.
The last numbers are the probability of each [mutally exclusive] chain of events.
Thus, the probability that the first marble was black is the sum of those events, namely:
3/32 + 9/32 + 3/32 + 1/32 = 16/32 = 1/2.
The probability that two black marbles were selected is the sum of those two events, or:
9/32 + 1/32 = 10/32 = 5/16.
The probability of getting two black marbles, given that the first marble is black, is just the ratio of those probabilities:
(10/32)/(16/32) = 10/16 = 5/8.
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Posted by DJ
on 2003-10-13 15:55:18 |
Solution
