Probability (Math)
See how to find the probability of an event.
15 examples and their solutions.
Probability
Formula
P(A) = n(A)n(S)
P(A): Probability of an event A happening n(A): Number of ways of A happening
n(S): Total number of ways
P(A): 0 ~ 1
P(A) = 0: A does not happen.
P(A) = 1: A always happens.
Example
A fair die is tossed once.
Find the probability of getting a multiple of 3.
Solution Find the probability of getting a multiple of 3.
A: {3, 6}
P(A) = 26 - [1]
= 13
P(A) = 26 - [1]
= 13
[1]
Multiple of 3: {3, 6}
→ n(A) = 2
Total numbers: {1, 2, 3, 4, 5, 6}
→ n(S) = 6
→ n(A) = 2
Total numbers: {1, 2, 3, 4, 5, 6}
→ n(S) = 6
Close
Example
3 blue marbles, 4 green marbles, 5 red marbles are in a jar.
If a marble is randomly picked from the jar,
P(blue marble) = ?
Solution If a marble is randomly picked from the jar,
P(blue marble) = ?
n(A) = 3
n(S) = 3 + 4 + 5
= 7 + 5
= 12
P(A) = 312
= 14
n(S) = 3 + 4 + 5
= 7 + 5
= 12
P(A) = 312
= 14
Close
Example
For the given spinner,
if you spin the arrow once,
P(X ≥ 4) = ?
Solution if you spin the arrow once,
P(X ≥ 4) = ?
P(A) = 25 - [1]
[1]
X ≥ 4
→ X = 4, 5
→ 2 regions
→ n(A) = 2
Total: 5 regions
→ n(S) = 5
→ X = 4, 5
→ 2 regions
→ n(A) = 2
Total: 5 regions
→ n(S) = 5
Close
Probability: not A
Formula
P(not A) = 1 - P(A)
Example
Numbers: 1 ~ 10
If you randomly pick a number,
P(not a multiple of 3) = ?
Solution If you randomly pick a number,
P(not a multiple of 3) = ?
A: {3, 6, 9} - [1]
P(A) = 310 - [2]
P(not A) = 1 - 310
= 1010 - 310
= 710
P(A) = 310 - [2]
P(not A) = 1 - 310
= 1010 - 310
= 710
[1]
A: Picking a multiple of 3
[1]
P(A): Probability of picking a multiple of 3
Close
Probability: A and B
Example
Numbers: 1 ~ 10
If you randomly pick a number,
P(odd and prime) = ?
Solution If you randomly pick a number,
P(odd and prime) = ?
A: {1, 3, 5, 7, 9}
B: {2, 3, 5, 7} - [1]
A and B: {3, 5, 7}
n(A and B) = 3
P(A and B) = 310
B: {2, 3, 5, 7} - [1]
A and B: {3, 5, 7}
n(A and B) = 3
P(A and B) = 310
[1]
A: Picking an odd number
B: Picking a prime number
B: Picking a prime number
Close
Probability: A or B
Example
Numbers: 1 ~ 10
If you randomly pick a number,
P(odd or prime) = ?
Solution If you randomly pick a number,
P(odd or prime) = ?
A: {1, 3, 5, 7, 9}
B: {2, 3, 5, 7} - [1]
A or B: {1, 2, 3, 5, 7, 9}
n(A or B) = 6
P(A and B) = 610
= 35
B: {2, 3, 5, 7} - [1]
A or B: {1, 2, 3, 5, 7, 9}
n(A or B) = 6
P(A and B) = 610
= 35
[1]
A: Picking an odd number
B: Picking a prime number
B: Picking a prime number
Close
Formula
P(A or B) = P(A) + P(B) - P(A and B)
Example
P(A) = 0.6, P(B) = 0.7, P(A and B) = 0.4
P(A or B) = ?
Solution P(A or B) = ?
P(A or B) = 0.6 + 0.7 - 0.4
= 1.3 - 0.4
= 0.9
= 1.3 - 0.4
= 0.9
Close
Example
P(A) = 0.5, P(A and B) = 0.4, P(A or B) = 0.8
P(B) = ?
Solution P(B) = ?
0.5 + P(B) - 0.1 = 0.8
P(B) + 0.4 = 0.8
P(B) = 0.4
P(B) + 0.4 = 0.8
P(B) = 0.4
Close
Probability: Mutually Exclusive Events
Formula
P(A or B) = P(A) + P(B)
Mutually exclusive events are the eventsthat don't happen together.
If A and B are mutually exclusive events,
P(A and B) = 0.
→ P(A or B) = P(A) + P(B)
Probability: A or B
Example
Marbles are in a jar.
The probability of picking a blue marble is 0.3.
The probability of picking a green marble is 0.4.
If a marble is randomly picked from the jar,
P(blue marble or green marble) = ?
Solution The probability of picking a blue marble is 0.3.
The probability of picking a green marble is 0.4.
If a marble is randomly picked from the jar,
P(blue marble or green marble) = ?
P(A or B) = 0.3 + 0.4 - [1]
= 0.7
= 0.7
[1]
A: Picking a blue marble
B: Picking a green marble
You cannot pick a blue marble and pick a green marble at the same time.
→ A and B cannot happen together
→ A and B are mutually exclusive events.
→ P(A or B) = P(A) + P(B)
B: Picking a green marble
You cannot pick a blue marble and pick a green marble at the same time.
→ A and B cannot happen together
→ A and B are mutually exclusive events.
→ P(A or B) = P(A) + P(B)
Close
Probability: Independent Events
Formula
P(A and B) = P(A)⋅P(B)
Independent events are the eventsthat do not affect each other.
So P(A) and P(B) do not affect each other.
Example
A fair die and a coin is tossed once.
P(3 and head) = ?
Solution P(3 and head) = ?
P(A) = 16
P(B) = 12 - [1]
P(A and B) = 16⋅12 - [2]
= 112
P(B) = 12 - [1]
P(A and B) = 16⋅12 - [2]
= 112
[1]
A: Getting a 3
(1 ~ 6)
B: Getting a head
(head, tail)
(1 ~ 6)
B: Getting a head
(head, tail)
[2]
Getting a 3 (A) and getting a head (B) don't affect each other.
→ Independent events
→ Independent events
Close
Example
5 blue marbles, 4 green marbles, 3 red marbles are in a jar.
A marble is randomly picked from the jar and replaced.
This is repeated twice.
P(2 blue marbles) = ?
Solution A marble is randomly picked from the jar and replaced.
This is repeated twice.
P(2 blue marbles) = ?
1) 1st pick
n(S) = 5 + 4 + 3
= 9 + 3
= 12
P(A) = 512 - [1]
2) 2nd pick
n(S) = 5 + 4 + 3
= 9 + 3
= 12
P(B) = 512 - [2] [3]
P(A and B) = 512⋅512
= 25144
n(S) = 5 + 4 + 3
= 9 + 3
= 12
P(A) = 512 - [1]
2) 2nd pick
n(S) = 5 + 4 + 3
= 9 + 3
= 12
P(B) = 512 - [2] [3]
P(A and B) = 512⋅512
= 25144
[1]
A: Picking a blue marble in the 1st pick
P(A) = (5 blue marbles)/(12 marbles)
P(A) = (5 blue marbles)/(12 marbles)
[2]
A: Picking a blue marble in the 2nd pick
P(B) = (5 blue marbles)/(12 marbles)
P(B) = (5 blue marbles)/(12 marbles)
[3]
After the 1st pick,
the blue marble is replaced.
→ The marbles in the jar are the same.
→ The 1st pick (A) doesn't affect the 2nd pick (B).
→ A and B are independent events.
the blue marble is replaced.
→ The marbles in the jar are the same.
→ The 1st pick (A) doesn't affect the 2nd pick (B).
→ A and B are independent events.
Close
Example
A, B: Independent events
P(A) = 47, P(A and B) = 17
P(A or B) = ?
Solution P(A) = 47, P(A and B) = 17
P(A or B) = ?
47⋅P(B) = 17
4⋅P(B) = 7
P(B) = 14
P(A or B) = 47 + 14 - 17 - [1]
= 1628 + 728 - 428
= 16 + 328
= 1928
4⋅P(B) = 7
P(B) = 14
P(A or B) = 47 + 14 - 17 - [1]
= 1628 + 728 - 428
= 16 + 328
= 1928
[1]
P(A or B) = P(A) + P(B) - P(A and B)
Probability: A or B
Probability: A or B
Close
Probability: Dependent Events
Formula
P(A and B) = P(A)⋅P(B')
Dependent events are the eventsthat affect each other.
So P(A) affects P(B):
P(B) → P(B')
Independent events
Example
5 blue marbles, 4 green marbles, 3 red marbles are in a jar.
A marble is randomly picked from the jar and not replaced.
This is repeated twice.
P(2 blue marbles) = ?
Solution A marble is randomly picked from the jar and not replaced.
This is repeated twice.
P(2 blue marbles) = ?
1) 1st pick
n(S) = 5 + 4 + 3
= 9 + 3
= 12
P(A) = 512 - [1]
2) 2nd pick
n(S) = 4 + 4 + 3
= 8 + 3
= 11
P(B) = 411 - [2] [3]
P(A and B) = 512⋅411
= 53⋅111
= 533
n(S) = 5 + 4 + 3
= 9 + 3
= 12
P(A) = 512 - [1]
2) 2nd pick
n(S) = 4 + 4 + 3
= 8 + 3
= 11
P(B) = 411 - [2] [3]
P(A and B) = 512⋅411
= 53⋅111
= 533
[1]
A: Picking a blue marble in the 1st pick
P(A) = (5 blue marbles)/(12 marbles)
P(A) = (5 blue marbles)/(12 marbles)
[2]
B: Picking a blue marble in the 2nd pick
P(B) = (4 blue marbles)/(11 marbles)
P(B) = (4 blue marbles)/(11 marbles)
[3]
After the 1st pick,
the blue marble is not replaced.
→ The marbles in the jar has changed.
→ The 1st pick (A) affects the 2nd pick (B).
→ A and B are dependent events.
the blue marble is not replaced.
→ The marbles in the jar has changed.
→ The 1st pick (A) affects the 2nd pick (B).
→ A and B are dependent events.
Close
Conditional Probability
Formula
P(B|A) = P(A and B)P(A)
P(B|A) means the probability of A and Bwhen A already happened.
B|A is read as
[B bar A] or [B given A].
Example
80% of students like apple.
50% of students like both apple and banana.
If you choose a student who likes apple,
find the probability that the student also likes orange.
Solution 50% of students like both apple and banana.
If you choose a student who likes apple,
find the probability that the student also likes orange.
P(A) = 0.8
P(A and B) = 0.5 - [1]
P(B|A) = 0.50.8
= 58
P(A and B) = 0.5 - [1]
P(B|A) = 0.50.8
= 58
[1]
A: The student likes apple.
B: The student likes banana.
A and B: The student likes both apple and banana.
B: The student likes banana.
A and B: The student likes both apple and banana.
Close
Example
The probability of a student oversleeping is 4%.
The probability of the student oversleeping and getting late for school is 3%.
If the student woke up and realized that he overslept,
find the probability of the student getting late for school.
Solution The probability of the student oversleeping and getting late for school is 3%.
If the student woke up and realized that he overslept,
find the probability of the student getting late for school.
P(A) = 0.04
P(A and B) = 0.03
P(B|A) = 0.30.4
= 34
P(A and B) = 0.03
P(B|A) = 0.30.4
= 34
Close