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Let us now develop a formula for the conditional probability P E F size 12{P left (E \lline F right )} {} .

Suppose an experiment consists of n size 12{n} {} equally likely events. Further suppose that there are m size 12{m} {} elements in F size 12{F} {} , and c size 12{c} {} elements in E F size 12{E intersection F} {} , as shown in the following Venn diagram.

The figure shows that everything within the square is equal to S. In the Venn diagram m-c is equal to F where c is equal to both E and F.

If the event F size 12{F} {} has occurred, the set of all possible outcomes is no longer the entire sample space, but instead, the subset F size 12{F} {} . Therefore, we only look at the set F size 12{F} {} and at nothing outside of F size 12{F} {} . Since F size 12{F} {} has m size 12{m} {} elements, the denominator in the calculation of P E F size 12{P left (E \lline F right )} {} is m. We may think that the numerator for our conditional probability is the number of elements in E size 12{E} {} . But clearly we cannot consider the elements of E size 12{E} {} that are not in F size 12{F} {} . We can only count the elements of E size 12{E} {} that are in F size 12{F} {} , that is, the elements in E F size 12{E intersection F} {} . Therefore,

P E F = c m size 12{P left (E \lline F right )= { {c} over {m} } } {}

Dividing both the numerator and the denominator by n size 12{n} {} , we get

P E F = c / n m / n size 12{P left (E \lline F right )= { {c/n} over {m/n} } } {}

But c / n = P E F size 12{c/n=P left (E intersection F right )} {} , and m / n = P F size 12{m/n=P left (F right )} {} .

Substituting, we derive the following formula for P E F size 12{P left (E \lline F right )} {} .

For Two Events E size 12{E} {} and F size 12{F} {} , the Probability of E size 12{E} {} Given F size 12{F} {} is

P E F = P E F P F size 12{P left (E \lline F right )= { {P left (E intersection F right )} over {P left (F right )} } } {}
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A single die is rolled. Use the above formula to find the conditional probability of obtaining an even number given that a number greater than three has shown.

Let E size 12{E} {} be the event that an even number shows, and F size 12{F} {} be the event that a number greater than three shows. We want P E F size 12{P left (E \lline F right )} {} .

E = 2,4,6 size 12{E= left lbrace 2,4,6 right rbrace } {} and F = 4,5,6 size 12{F= left lbrace 4,5,6 right rbrace } {} . Which implies, E F = 4,6 size 12{E intersection F= left lbrace 4,6 right rbrace } {}

Therefore, P F = 3 / 6 size 12{P left (F right )=3/6} {} , and P E F = 2 / 6 size 12{P left (E intersection F right )=2/6} {}

P E F = P E F P F = 2 / 6 3 / 6 = 2 3 size 12{P left (E \lline F right )= { {P left (E intersection F right )} over {P left (F right )} } = { {2/6} over {3/6} } = { {2} over {3} } } {} .

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The following table shows the distribution by gender of students at a community college who take public transportation and the ones who drive to school.

Male(M) Female(F) Total
Public Transportation(P) 8 13 21
Drive(D) 39 40 79
Total 47 53 100

The events M size 12{M} {} , F size 12{F} {} , P size 12{P} {} , and D size 12{D} {} are self explanatory. Find the following probabilities.

  1. P D M size 12{P left (D \lline M right )} {}
  2. P F D size 12{P left (F \lline D right )} {}
  3. P M P size 12{P left (M \lline P right )} {}

We use the conditional probability formula P E F = P E F P F size 12{P left (E \lline F right )= { {P left (E intersection F right )} over {P left (F right )} } } {} .

  1. P D M = P D M P M = 39 / 100 47 / 100 = 39 47 size 12{P left (D \lline M right )= { {P left (D intersection M right )} over {P left (M right )} } = { {"39"/"100"} over {"47"/"100"} } = { {"39"} over {"47"} } } {} .
  2. P F D = P F D P D = 40 / 100 79 / 100 = 40 79 size 12{P left (F \lline D right )= { {P left (F intersection D right )} over {P left (D right )} } = { {"40"/"100"} over {"79"/"100"} } = { {"40"} over {"79"} } } {} .
  3. P M P = P M P P P = 8 / 100 21 / 100 = 8 21 size 12{P left (M \lline P right )= { {P left (M intersection P right )} over {P left (P right )} } = { {8/"100"} over {"21"/"100"} } = { {8} over {"21"} } } {} .
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Given P E = . 5 size 12{P left (E right )= "." 5} {} , P F = .7 size 12{P left (F right )=/7} {} , and P E F = .3 size 12{P left (E intersection F right )} {} . Find the following.

  1. P E F size 12{P left (E \lline F right )} {}
  2. P F E size 12{P left (F \lline E right )} {} .

We use the conditional probability formula P E F = P E F P F size 12{P left (E \lline F right )= { {P left (E intersection F right )} over {P left (F right )} } } {} .

  1. P E F = . 3 . 7 = 3 7 size 12{P left (E \lline F right )= { { "." 3} over { "." 7} } = { {3} over {7} } } {} .
  2. P F E = . 3 / . 5 = 3 / 5 size 12{P left (F \lline E right )= "." 3/ "." 5=3/5} {} .
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Given two mutually exclusive events E size 12{E} {} and F size 12{F} {} such that P E = . 4 size 12{P left (E right )= "." 4} {} , P F = . 9 size 12{P left (F right )= "." 9} {} . Find P E F size 12{P left (E \lline F right )} {} .

Since E size 12{E} {} and F size 12{F} {} are mutually exclusive, P E F = 0 size 12{P left (E intersection F right )=0} {} . Therefore,

P E | F = 0 .9 = 0 size 12{P left (E intersection F right )=0} {} .

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Given P F E = . 5 size 12{P left (F \lline E right )= "." 5} {} , and P E F = . 3 size 12{P left (E intersection F right )= "." 3} {} . Find P E size 12{P left (E right )} {} .

Using the conditional probability formula P E F = P E F P F size 12{P left (E \lline F right )= { {P left (E intersection F right )} over {P left (F right )} } } {} , we get

P F E = P E F P E size 12{P left (F \lline E right )= { {P left (E intersection F right )} over {P left (E right )} } } {}

Substituting,

. 5 = . 3 P E size 12{ "." 5= { { "." 3} over {P left (E right )} } } {} or P E = 3 / 5 size 12{P left (E right )=3/5} {}

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In a family of three children, find the conditional probability of having two boys and a girl, given that the family has at least two boys.

Let event E size 12{E} {} be that the family has two boys and a girl, and let F size 12{F} {} be the probability that the family has at least two boys. We want P E F size 12{P left (E \lline F right )} {} .

We list the sample space along with the events E size 12{E} {} and F size 12{F} {} .

S = BBB , BBG , BGB , BGG , GBB , GGB , GGG size 12{S= left lbrace ital "BBB", ital "BBG", ital "BGB", ital "BGG", ital "GBB", ital "GGB", ital "GGG" right rbrace } {}

E = BBG , BGB , GBB size 12{E= left lbrace ital "BBG", ital "BGB", ital "GBB" right rbrace } {} and F = BBB , BBG , BGB , GBB size 12{F= left lbrace ital "BBB", ital "BBG", ital "BGB", ital "GBB" right rbrace } {}

E F = BBG , BGB , GBB size 12{E intersection F= left lbrace ital "BBG", ital "BGB", ital "GBB" right rbrace } {}

Therefore, P F = 4 / 8 size 12{P left (F right )=4/8} {} , and P E F = 3 / 8 size 12{P left (E intersection F right )=3/8} {} .

And

P E F 3 / 8 4 / 8 = 3 4 size 12{P left (E \lline F right ) - { {3/8} over {4/8} } = { {3} over {4} } } {} .

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At a community college 65% of the students use IBM computers, 50% use Macintosh computers, and 20% use both. If a student is chosen at random, find the following probabilities.

  1. The student uses an IBM given that he uses a Macintosh.
  2. The student uses a Macintosh knowing that he uses an IBM.

Let event I size 12{I} {} be that the student uses an IBM computer, and M size 12{M} {} the probability that he uses a Macintosh.

  1. P I M = . 20 . 50 = 2 5 size 12{P left (I \lline M right )= { { "." "20"} over { "." "50"} } = { {2} over {5} } } {}
  2. P M I = . 20 . 65 = 4 13 size 12{P left (M \lline I right )= { { "." "20"} over { "." "65"} } = { {4} over {"13"} } } {} .
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Questions & Answers

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the amount of a good that buyers are willing and able to purchase
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The people living within a political or geographical boundary.
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Source:  OpenStax, Applied finite mathematics. OpenStax CNX. Jul 16, 2011 Download for free at http://cnx.org/content/col10613/1.5
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