To find the probability of the second of two dependent events is a pretty easy task.
A good example is pulling a card out of a standard deck of 52 cards. Lets say you want to pull a red jack, then a black ace.
First, the probability of the red jack. there are 2 in a deck so the probability is 2/52 or,reduced, 1/26.
Then find for the black aces. There are 2 in a deck, but now since there are only 51 cards in the deck now that you already took one out, the probability is 2/51.
If you wanted to find the combined probability, you would multiply the two numbers.
1/26x2/51= 2/1326 or, reduced, 1/663
2006-09-23 16:38:22
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answer #1
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answered by american_girl_24_7 1
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Hi. Assuming the first event occurs favorably and the second event's probablility is known then it is simply the second events probability. The total would be #1 times #2 as a percent. (30% times 60% equals 18%.)
2006-09-23 16:26:12
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answer #2
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answered by Cirric 7
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Use Bayes' theorem.
(let A intersection B be denoted by AB).
P (AB) = P (A/B) * P(B).= P(B/A)*P(A)
So, P(B) = P(AB)/P(A/B).
2006-09-23 17:47:37
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answer #3
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answered by jinxy 2
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