# Help me understand events/sample space

## Homework Statement

1.

Suppose that A, B, and C are 3 independent events such that Pr(A)=1/4, Pr(B)=1/3 and Pr(C)=1/2.

a. Determine the probability that none of these events will occur.

Is it just:

(1-P(a))(1-P(b))(1-P(c)) = 3/4 * 2/3 * 1/2 = 1/4

## The Attempt at a Solution

I tried to do 1. another way:

The probability that all theses events will occur: 1/4 * 1/3 * 1/2 = 1/24

1-(1/24) = 23/24

Obviously this is wrong. Is the reason it is wrong, because: the complement of "all of these events will occur" is that "not all of these events will occur," meaning, it is not "none of these events will occur."

None of these events will occur is included in the compliment 1-(1/24), but so is that 1 of the events occur, and that 2 of the events occur, etc.

Am I right in my reasoning?

• PeroK

the (1) is correct.
for reference, in general the answer can be found by calculating multinomial distribution.
in (3), the 23/24 probability is sum of "no events", "A only", "B only", "C only", "A&B", "A&C", "B&C".

PeroK
Homework Helper
Gold Member
2020 Award

## Homework Statement

1.

Suppose that A, B, and C are 3 independent events such that Pr(A)=1/4, Pr(B)=1/3 and Pr(C)=1/2.

a. Determine the probability that none of these events will occur.

Is it just:

(1-P(a))(1-P(b))(1-P(c)) = 3/4 * 2/3 * 1/2 = 1/4

## The Attempt at a Solution

I tried to do 1. another way:

The probability that all theses events will occur: 1/4 * 1/3 * 1/2 = 1/24

1-(1/24) = 23/24

Obviously this is wrong. Is the reason it is wrong, because: the complement of "all of these events will occur" is that "not all of these events will occur," meaning, it is not "none of these events will occur."

None of these events will occur is included in the compliment 1-(1/24), but so is that 1 of the events occur, and that 2 of the events occur, etc.

Am I right in my reasoning?

Yes, that's it exactly.

Ray Vickson
Homework Helper
Dearly Missed

## Homework Statement

1.

Suppose that A, B, and C are 3 independent events such that Pr(A)=1/4, Pr(B)=1/3 and Pr(C)=1/2.

a. Determine the probability that none of these events will occur.

Is it just:

(1-P(a))(1-P(b))(1-P(c)) = 3/4 * 2/3 * 1/2 = 1/4

## The Attempt at a Solution

I tried to do 1. another way:

The probability that all theses events will occur: 1/4 * 1/3 * 1/2 = 1/24

1-(1/24) = 23/24

Obviously this is wrong. Is the reason it is wrong, because: the complement of "all of these events will occur" is that "not all of these events will occur," meaning, it is not "none of these events will occur."

None of these events will occur is included in the compliment 1-(1/24), but so is that 1 of the events occur, and that 2 of the events occur, etc.

Am I right in my reasoning?

You are correct, and there is sound reasoning to justify that fact, as follows. If we denote the complement of any event ##E## as ##\bar{E}##, then
$$\{ \text{none occur} \} = \overline{A \cup B \cup C},$$
because the event that at least one occurs is ##A \cup B \cup C,## so the complement of that is the event that none occurs.

However, there is a general set-theoretic result:
$$\overline{ \bigcup_{i=1}^n A_i } = \bigcap_{i=1}^n \overline{A_i}$$ That is, the complement of a union is the intersection of the complements.