Probability of Dependent Events
The probabilityA measure of how likely it is that something will occur. for any kind of eventA collection of possible outcomes, often describable using a common characteristic, such as rolling an even number with a die or picking a card from a specific suit.—simple, compound, independent, dependent—always follows this basic formula:
`"Probability that an event occurs"=("size of the event space")/("size of the sample space")`
The probability is the ratio of the sizes of the event and sample spacesThe set of all outcomes.. For some situations, like independent and dependent eventsTwo or more events for which the occurrence of one affects the probability of the other(s)., there are ways to calculate these numbers without going through the process of finding and counting possible outcomesA result of a trial. one by one, which is sometimes tedious and prone to mistakes.
To find the probability of dependent events, we can use the Fundamental Counting PrincipleA way to find the number of outcomes in a sample space by finding the product of the number of outcomes for each element. or the permutationsGroupings in which the order of members matters. and combinations factorialAn abbreviated way of writing a product of all whole numbers from `1` to a given number, indicated by that number followed by an exclamation point, as in `3! = 3 * 2 * 1`. formulas to find the sizes of the event and sample spaces.
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The Fundamental Counting Principle finds the number of permutations and combinations as follows:
The factorial formulas calculate permutations and combinations this way:
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Finding the event spaceThe set of possible outcomes in an event: for example, the event “rolling an even number” on a die has the event space of `2`, `4`, and `6`. often requires some extra thought and imagination to identify all of the ways an event can happen.
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Example |
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Problem |
You pull a marble from a bag with `20` red, `20` white, and `10` green marbles. You hold onto it and then pull another marble. What is the probability of pulling a red marble and then pulling a green marble? |
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permutation |
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We want to pull a red marble first and then a green marble, so order matters. This is a permutations problem. |
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`(50!)/((50-2)!)` or `50*49`
`"sample space"= 2450` |
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The size of the sample space is the number of all possible permutations of `2` marbles. The factorial formula for this is `(n!)/((n-k)!)`. In this case, we’re choosing `2` of `50` marbles, so `n = 50` and `k = 2`. (You could also use the Fundamental Counting Principle: There are `50` choices for the first marble and `49` for the second one.) |
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Red, Green |
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The size of the event space is the number of all possible combinations for which the first marble is red and the second marble is green. |
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`20 *10`
`"event space"= 200`
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How many ways can that happen? Don’t make the mistake of thinking there’s only `1` way: red then green! There are `20` different red marbles that can be chosen first. Then there are `10` different green marbles that can be chosen second. The Fundamental Counting Principle says to multiply these to find the number of ways to get red then green. |
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`P("Red," "Green")=200/2450` `=4/49` |
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Now the probability that red then green are pulled is the ratio of all the ways to pull those two colors in that order to all the possible permutations. |
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Answer |
The probability that the pulls are red then green is `4/49`. |
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In finding the event space, it sometimes helps to think of the event we want as removing particular outcomes from individual events. Then we have to find the number of permutations or combinations for the remaining outcomes. In the marble example above, removing a red for the first pull didn’t change the number of green marbles that were available. However, if we wanted the probability of pulling a red and then another red, there are `20*19` ways to do that when order matters.
Let’s look at an example involving combinations:
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Example |
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Problem |
A school organization has `30` members. Four members will be chosen at random for an interview with the school newspaper about the group. What is the probability that Tom and Cindy will be chosen? |
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combination
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There’s no reason for any person to be considered different from any other, based on the order chosen, so the event and sample spaces involve combinations rather than permutations. |
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`(30!)/((30-4)!4!)`
`(30*29*28*27*26*...*2*1)/((26*...*2*1)(4*3*2*1))`
`(26*...*2*1)/(26*...*2*1)*(30*29*28*27)/(4*3*2*1)`
`1*(30*29*28*27)/(4*3*2*1)`
`27,405`
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The size of the sample space is the number of all possible combinations of `4` members. The formula for combinations is `(n!)/((n-k)!k!)`. In this case, we’re choosing `4` of `30` members, so `n = 30` and `k=4`. (You could also use the Fundamental Counting Principle to get the numerator and denominator.) |
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The size of the event space is the number of all possible combinations that include Tom and Cindy. |
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Tom, Cindy, and (two other people)
Find `2` more people from the remaining `28`.
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If Tom and Cindy must be chosen, then we already know `2` of our `4` people. Now the problem is, how can we fill those `2` slots from the remaining `28` members? This is a new combinations problem, choosing `2` of `28` members. |
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`(28!)/((28-2)!2!)`
`(28*27)/(2*1)`
`378` |
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To use the formula, set `n = 28` and `k=2`. |
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`P("Tom and Cindy")=378/(27,405)=2/145`
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Now the probability that Tom and Cindy are both chosen is the ratio of the combinations including them to all the possible combinations. |
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Answer |
The probability that both Tom and Cindy are chosen is `2/145`. |
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Here’s a little arithmetic tip: it may be easier to simplify the probability fraction by leaving the sample and event sizes in factored form. For example:
`P("Tom and Cindy")=(28*27)/(2*1)-:(30*29*28*27)/(4*3*2*1)`
We divide fractions by inverting the divisor and multiplying. Then we can remove common factors in the numerator and denominator:
`(28*27)/(2*1)*(4*3*2*1)/(30*29*28*27)`
`(4*3)/(30*29)`
`(2*2*3)/(2*3*5*29)`
`2/145`
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A bag of marbles has `20` red marbles, `20` white marbles, and `10` green marbles. If we pull out three marbles, what is the probability that we pull out exactly two red marbles?
A) `2/27`
B) `38/245`
C) `1/(117,600)`
D) `19/196`
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We've been using the Fundamental Counting Principle and the permutation and combination formulas to calculate the probability of a series of events as a whole. We can also calculate probability one event at a time. We can use the following rule to calculate the probability of independent eventsTwo or more events for which the occurrence of one does not affect the probability of the other(s).:
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If `A` and `B` are independent events, `P(A" and " B)=P(A)*P(B)`.
In general, for any number of independent events, the probability that all the events happen is the product of the probabilities that the individual events happen. |
So we can find the probability of many independent events by finding the probability of each individual event on its own, and then multiplying them all together. Each individual event would have the same probability even if none of the other events occurred.
The probability of dependent events can be found in almost the same way. Consider some of the dependent events we have worked with, like the marbles example above. The probability for that example also involved products:
`P("exactly " 2 " red marbles")=(20*19*30)/(3*2*1)-:(50*49*48)/(3*2*1)=(20*19*30)/(50*49*48)`
If we look at this as three separate events, what are the individual probabilities?
First pull: `P("red")=20/50`
Second pull: `P("red")=19/49`
Third pull: `P("not red")=30/48`
Notice the product of these individual probabilities:
`(20*19*30)/(50*49*48)`
This is the same result! So we are able to calculate the probability of a series of dependent events by finding the probability of each individual event and then multiplying them all together. But, the individual probabilities are not the same as if the individual events were occurring alone, as was the case for independent events. With dependent events, the probabilities for later events are different than they would have been if they occurred alone.
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If `A` and `B` are dependent events, `P(A" and "B) = P(A)*P(B" after "A)` where `P(B" after "A)` is the probability that `B` occurs after `A` has occurred.
In general, for any number of independent events, the probability that all the events happen is the product of the probabilities that the individual events happen, provided the occurrence of an earlier event is included when finding the probabilities of later events. |
Let's try this method on a problem:
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Example |
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Problem |
You accidentally sit on your cellphone, and every time you shift your weight, it randomly dials a different number from your phone book. You have `16` numbers stored, and `4` of those are called before you stand up. What’s the probability that Mary, Lulu, Bo, and Dan all got a phone call? |
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Mary is called Lulu is called Bo is called Dan is called |
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First let’s define the individual events. Order does not matter. |
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`P("first")=4/16`
`P("second")=3/15`
`P("third")=2/14`
`P("fourth")=1/13`
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Now let's figure out the probability of each event. For the first call, there are `4` numbers in the event space and `16` numbers to choose from. If the first call does go to Mary, Lulu, Bo, or Dan, for the second call there will be `3` chances to call someone in the event and `15` remaining numbers that could possibly be called. For the `3"rd"` call, there are `2` choices from the `14` remaining numbers, and then only `1` out of `13` numbers for the last call. Use these numbers to find the probabilities of each call. |
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`P("all "4" called")=(4/16)(3/15)(2/14)(1/13)`
`(1/4)(1/5)(1/7)(1/13)`
`1/(1,820)` |
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The probability that all four individual events occur is the product of the individual probabilities. |
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Let’s find the probability again using the combinations formula. We want a combination because the order doesn’t matter. |
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`(16!)/((16-4)!4!)=(16*15*14*13)/(4*3*2*1)`
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The sample space is the number of combinations of `4` phone numbers from `16`. Here, `k = 4` and `n=16`. |
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`(4!)/((4-4)!4!)=(4*3*2*1)/(1*4*3*2*1)=1`
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The event space is the number of combinations for choosing `4` out of `4` phone numbers. There is only `1` way to do that—all `4` are called! We can check this with the formula (remember that `0! =1`). |
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`1/(((16*15*14*13)/(4*3*2*1)))=(4*3*2*1)/(16*15*14*13)=1/(1,820)`
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The probability is the ratio of the event and sample sizes. Notice that the result is the same we got for calculating individually. |
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Answer |
`P("Mary, Lulu, Bo, Dan called")=1/(1,820)` |
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A group of eight friends are playing a board game in which the players race to the last spot on the board. The friends agree to play for first, second, and third place.
Assuming all the players are equally matched (so who wins is completely random), what is the probability that Juanita will place first, and then either Bill or Susan places second and either Bill or Susan places third?
A) `1/168`
B) `1/336`
C) `1/8`
D) `1/128`
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There are several ways to find the probabilities of dependent events. The sequence of events can be treated as a whole, in which case the Fundamental Counting Principle or the permutations and combinations formulas are used to find the sizes of the event and sample spaces.
Or each event can be treated separately, in which case the probabilities of each event are multiplied to find the probability of the entire chain of events:
If `A` and `B` are dependent events, `P(A" and "B)=P(A)*P(B" after "A)` where `P(B" after "A)` is the probability that `B` occurs after `A` has occurred.
Regardless of method, it's important to note that when events are dependent, the size of both the event and sample spaces can change over time, because the occurrence of one event affects the outcomes of later events.