# Thread: Need formulas for probability

1. Good post? |

## Need formulas for probability

There are ceratin formulas for probabilities

For example

P(Not A and Not B) = 1- P (A OR B)

Please share some more formulas if you have in mind.

CB

2. Good post? |
1. If two events are mutually exclusive (i.e. they cannot occur at the same time), then the probability of them both occurring at the same time is 0. then: P(A and B) = 0 and P(A or B) = P(A) + P(B)

2. if two events are not-mutually exclusive (i.e. there is some overlap)
then: P(A or B) = P(A) + P(B) - P(A and B)

3. If events are independent (i.e. the occurrence of one does not change the probability of the other occurring), then the probability of them both occurring is the product of the probabilities of each occurring. Then: P(A and B) = P(A) * P(B)

4. If A, B and C are not mutually exclusive events, then P(A or B or C) = P(A) + P(B) + P(C) - P(A and B) - P(B and C) - P(C and A) + P(A and B and C)

Either:
P(A or B)= P(A) + P(B) - P(A and B)

Where, P(A and B) = P(A) * P(B)
Neither:
P(Not A and Not B)= P(A') * P(B')
Where, P(A') = 1 - P(A), P(B') = 1- P(B)

Thanks.

3. Good post? |
Thank you skychild for sharing...

One thing i need to clarify. Is P(at least one) = P(A U B) true

Also another formulas that i found are

1. P(at least one) = 1 - P(None)

2. P(Not A and Not B) = 1 - P(A U B)
Also P(Not A Or Not B) = 1 - P(A AND B)

3. P(A ' ) * P(B ' ) = 1 - P(A U B)

Please someone confirm whether they are correct or not.

Regards,

CB

4. Good post? |
one interesting fact....

Suppose we have 2 red, 2 green and 2 blue balls.

When question says neither is blue ---- P(Neither Blue) = P(B' ) * P(B') ------ This is right

When question says both are not blue ----- P(Both Not Blue) = P(B' ) * P(B') ------ This is wrong

The correct formula is P(Both Not Blue) = P(B' ) * P(B') + P(B')*P(B)+P(B)*(B')

This means that both not blue means only those combinations where both the balls together are blue but combinations where 1st one alone or 2nd one alone is blue is not possible.

CB

5. Good post? |
One more thing to clarify.

Is it right that P(Both not Blue) = 1 - P(Both Blue)