Joint Probability |
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Joint probability is used in multi-stage
experiments when we want to find out how likely it is that two
(or more) events happen at the same time. An example is drawing three
cards and getting all jacks P(JJJ). We can use a
tree diagram to figure out this type of probability. This type
of probability depends on whether the experiment is done with
or without replacement.
With replacement means
that the object chosen on one stage is returned to the sample space before
the next choice is made. Independent events are considered with replacement
since they do not affect each other, interfere with each other, or cause
each other. For example, tossing a head on the first toss does not affect
the outcome of flipping the coin a second time.
The probability that independent events A and B happen simultaneously is found by using the multiplication rule, or the product of the individual probabilities.
Without replacement uses the same idea, only we consider the change in the sample space if the first choice is not replaced. These can be considered dependent events, since changing the sample space changes the probability. We still use the multiplication rule, but the numerator and/or denominator decrease(s) for each of the stages. |
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WITHOUT REPLACEMENT: If we draw two socks from this drawer, what are the chances we will get a pair of black socks? Well, on our first pick there are two black socks out of five, and on our second pick there is one black sock left out of four. So P(black,black) = (2/5)(1/4) = 2/20. |
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WITHOUT
REPLACEMENT: If you draw two cards from the deck
without replacement, what is the probability that they will both be aces?
P(AA) = (4/52)(3/51) = 1/221.
What is the probability that the second card will be an ace if the first card is a king? P(A|K) = 4/51 since there are four aces in the deck but only 51 cards left after the king has been removed. |
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WITH REPLACEMENT: Find the probability of drawing three queens in a row, with replacement. We pick a card, write down what it is, then put it back in the deck and draw again. To find the P(QQQ), we find the probability of drawing the first queen which is 4/52. The probability of drawing the second queen is also 4/52 and the third is 4/52. We multiply these three individual probabilities together to get P(QQQ) = P(Q)P(Q)P(Q) = (4/52)(4/52)(4/52) = .00004 which is very small but not impossible. |
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WITH REPLACEMENT: What is the probability of tossing a fair coin twice in a row and getting heads both times? Since the probability of tossing a head (independent events) is 1/2 each time P(HH) = (1/2)(1/2) = 1/4. |
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If you are unfamiliar
with a standard deck of cards, click
here. |
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