intersection - the set containing all elements of A that also belong to B or equivalently, all elements of B that also belong to A.

Formula: A ∩ B

(A intersection B) U (A intersection B')

(A intersection B) U (A intersection B')
This is the [B]Universal Set U[/B].
Everything that isn't A and isn't B is everything else.

100 students were interviewed. 72 ate at A, 52 ate at B. How many ate at A and B?

P(A U B) = P(A) + P(B) - P(A intersection B)
100 = 72 + 52 - P(A intersection B)
P(A intersection B) = 72 + 52 - 100
P(A intersection B) = P(who ate at A and B) = 24

2x^2+4x < 3x+6

2x^2+4x < 3x+6
Subtract 3x from both sides:
2x^2 + x < 6
Subtract 6 from both sides
2x^2 + x - 6 < 0
Using our [URL='http://www.mathcelebrity.com/quadratic.php?num=2x%5E2%2Bx-6&pl=Solve+Quadratic+Equation&hintnum=+0']quadratic calculator[/URL], we get:
x < 1.5 and x < -2
When we take the intersection of these, it's [B]x < 1.5[/B]

45 students, 12 taking spanish, 15 taking chemistry, 5 taking both spanish and chemistry. how many s

45 students, 12 taking spanish, 15 taking chemistry, 5 taking both spanish and chemistry. how many students are not taking either?
Let S be the number of students taking spanish and C be the number of students taking chemistry:
We have the following equation relating unions and intersections:
P(C U S) = P(C) + P(S) - P(C and S)
P(C U S) = 15 + 12 - 5
P(C U S) = 22
To get people that aren't taking either are, we have:
45 - P(C U S)
45 - 22
[B]23[/B]

If A and B are independent events with P(A) = 0.2 and P(B) = 0.6, then P(A U B)=

If A and B are independent events with P(A) = 0.2 and P(B) = 0.6, then P(A U B)=?
We know the following formula for the probability of 2 events:
P(A U B) = P(A) + P(B) - P(A intersection B)
We're told A and B are independent, which makes P(A intersection B) = 0. So we're left with:
P(A U B) = P(A) + P(B) - P(A intersection B)
P(A U B) = 0.2 + 0.6 - 0
P(A U B) = [B]0.8[/B]

If n(A)=1200, n(B)=1250 and n(AintersectionB)=320, then n(AUB) is

If n(A)=1200, n(B)=1250 and n(AintersectionB)=320, then n(AUB) is
We know that:
n(AUB) = n(A) + n(B) - n(AintersectionB)
Plugging in our given numbers, we get:
n(AUB) = 1200 + 1250 - 320
n(AUB) = [B]2130[/B]

Let A and B be independent events with P(A) = 0.52 and P(B) = 0.62. a. Calculate P(A ? B).

Let A and B be independent events with P(A) = 0.52 and P(B) = 0.62. a. Calculate P(A ? B).
With independent events, the intersection probability is found by:
P(A ? B) = P(A) * P(B)
P(A ? B) = 0.52 * 0.62
P(A ? B) = [B]0.3224[/B]

Refer to a bag containing 13 red balls numbered 1-13 and 5 green balls numbered 14-18. You choose a

Refer to a bag containing 13 red balls numbered 1-13 and 5 green balls numbered 14-18. You choose a ball at random.
a. What is the probability that you choose a red or even numbered ball?
b. What is the probability you choose a green ball or a ball numbered less than 5?
a. The phrase [I]or[/I] in probability means add. But we need to subtract even reds so we don't double count:
We have 18 total balls, so this is our denonminator for our fractions.
Red and Even balls are {2, 4, 6, 8, 10, 12}
Our probability is:
P(Red or Even) = P(Red) + P(Even) - P(Red and Even)
P(Red or Even) = 13/18 + 9/18 - 6/18
P(Red or Even) = 16/18
Using our [URL='https://www.mathcelebrity.com/fraction.php?frac1=16%2F18&frac2=3%2F8&pl=Simplify']Fraction Simplify Calculator[/URL], we have:
P(Red or Even) = [B]16/18[/B]
[B][/B]
b. The phrase [I]or[/I] in probability means add. But we need to subtract greens less than 5 so we don't double count:
We have 18 total balls, so this is our denonminator for our fractions.
Green and less than 5 does not exist, so we have no intersection
Our probability is:
P(Green or Less Than 5) = P(Green) + P(Less Than 5) - P(Green And Less Than 5)
P(Green or Less Than 5) = 5/18 + 4/18 - 0
P(Green or Less Than 5) = 9/18
Using our [URL='https://www.mathcelebrity.com/fraction.php?frac1=9%2F18&frac2=3%2F8&pl=Simplify']Fraction Simplify Calculator[/URL], we have:
P(Red or Even) = [B]1/2[/B]

Set Notation

Given two number sets A and B, this determines the following:

* Union of A and B, denoted A U B

* Intersection of A and B, denoted A ∩ B

* Elements in A not in B, denoted A - B

* Elements in B not in A, denoted B - A

* Symmetric Difference A Δ B

* The Concatenation A · B

* The Cartesian Product A x B

* Cardinality of A = |A|

* Cardinality of B = |B|

* Jaccard Index J(A,B)

* Jaccard Distance J_{σ}(A,B)

* Dice's Coefficient

* If A is a subset of B

* If B is a subset of A

* Union of A and B, denoted A U B

* Intersection of A and B, denoted A ∩ B

* Elements in A not in B, denoted A - B

* Elements in B not in A, denoted B - A

* Symmetric Difference A Δ B

* The Concatenation A · B

* The Cartesian Product A x B

* Cardinality of A = |A|

* Cardinality of B = |B|

* Jaccard Index J(A,B)

* Jaccard Distance J

* Dice's Coefficient

* If A is a subset of B

* If B is a subset of A

There are 100 people in a sport centre. 67 people use the gym. 62 people use the swimming pool. 5

There are 100 people in a sport centre. 67 people use the gym. 62 people use the swimming pool. 56 people use the track. 38 people use the gym and the pool. 31 people use the pool and the track. 33 people use the gym and the track. 16 people use all three facilities. A person is selected at random. What is the probability that this person doesn't use any facility?
WE use the compound probability formula for 3 events:
[LIST=1]
[*]Gym use (G)
[*]Swimming pool use (S)
[*]Track (T)
[/LIST]
P(G U S U T) = P(G) + P(S) + P(T) - P(G Intersection S) - P(G Intersection T) - P(S Intersection T) + P(G Intersection S Intersection T)
[LIST]
[*]Note: U means Union (Or) and Intersection means (And)
[/LIST]
Plugging our numbers in:
P(G U S U T) = 67/100 + 62/100 + 56/100 - 38/100 - 31/100 - 33/100 + 16/100
P(G U S U T) = (67 + 62 + 56 - 38 - 31 - 33 + 16)/100
P(G U S U T) = 99/100 or 0.99
What this says is, the probability that somebody uses at any of the 3 facilities is 99/100.
The problem asks for none of the 3 facilities, or P(G U S U T)'
P(G U S U T)' = 1 - P(G U S U T)
P(G U S U T)' = 1 - 99/100
P(G U S U T)' = 100/100 - 99/100
P(G U S U T)' = [B]1/100 or 0.1[/B]

Venn Diagram

This lesson walks you through what a Venn diagram is, the Venn diagram for A union B, A intersection B, and A Complement.

Venn Diagram (2 circles)

Given two circles A and B with an intersection piece of C, this will calculate all relevant probabilities of the Venn Diagram.