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Say 2 indentical bags contain different numbers of and counters.
A bag is chosen AT RANDOM and a single counter removed from it without looking.
bag A : 2 3
Bag B : 3 4
If the counter was what was the probability that the bag was Bag A??
EYE AM FRIENDLY, THAT'S O U NEED 2 NO,
psst,
Don't trust strangers, EYE AM FRIENDLY, THAT'S O U NEED 2 NO ..........psst, Don't trust strangers......
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I got
2 any different anwsers? please HELP
5
EYE AM FRIENDLY, THAT'S O U NEED 2 NO,
psst,
Don't trust strangers, EYE AM FRIENDLY, THAT'S O U NEED 2 NO ..........psst, Don't trust strangers......
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The probability of choosing bag A, then getting a from it is 1/2 * 2/5 = 1/5.
The probability of choosing bag B, then getting a from it is 1/2 * 3/7 = 3/14.
The overall probability of getting a is the sum of these two, which is 29/70.
Therefore, given that you picked a , the probability of you having chosen bag A is
(1/5) / (29/70) = 14/29.
Why did the vector cross the road?
It wanted to be normal.
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Thank you very much
EYE AM FRIENDLY, THAT'S O U NEED 2 NO,
psst,
Don't trust strangers, EYE AM FRIENDLY, THAT'S O U NEED 2 NO ..........psst, Don't trust strangers......
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