What is Compound Probability?

Compound probability refers to the probability of two or more independent events both happening. Independent events are events in which the outcome of one event has no effect on the outcome of another event. For example, flipping a coin twice. If the probability of getting heads is 50 percent, then the chances of getting heads twice is 25 percent. In this article, we will learn about the compound probability.

We can find compound probability by multiplying the probability of the first event by the probability of the second event.

The two types of compound events are mutually exclusive compound events and mutually inclusive compound events. When two events cannot happen at the same time, they are called mutually exclusive compound events. Mutually inclusive compound events are cases where one event can occur with the other.

Formula

P(A and B) = P(A) × P(B)

If mutually exclusive events: P(A or B) = P (A) + P(B).

If mutually inclusive events: P (A or B) = P(A) + P(B) – P(A and B).

Related articles

Bayes theorem of probability

Dependent events

Probability JEE Main Previous Year Questions

Solved Example

Example 1:

If the probability of A to fail in the examination is 0.3 and that for B is 0.2, then the probability that either A or B failing in the examination is:

(a) 0.06

(b) 0.44

(c) 0.5

(d) 0.1

Solution:

Given P(A) = 0.3

P(B) = 0.2

P(AUB) = P(A) + P(B) – P(A ⋂ B)

Here P(A ⋂ B) = P(A) × P(B) = 0.3 × 0.2 = 0.06

P(A or B) = 0.3 + 0.2 – 0.06 = 0.5 – 0.06 = 0.44

Hence, option (b) is the answer.

Example 2:

A and B are events such that P(AUB) = ¾, P(A⋂B) = ¼, P(A’) = ⅔ then P(A’⋂B) is

(a) 5/12

(b) ⅜

(c) ⅘

(d) 5/4

Solution:

Given P(AUB) = ¾, P(A⋂B) = ¼, P(A’) = ⅔

P(A) = 1 – (⅔) = ⅓

P(AUB) = P(A) + P(B) – P(A⋂B)

¾ = ⅓ + P(B) – ¼

P(B) = ¾ + ¼ – ⅓ = ⅔

P(A’⋂B) = P(B) – P(A⋂B)

= ⅔ – ¼ = 5/12

Hence option (a) is the answer.

Example 3: Two dice are tossed once. The probability of getting an even number at the first die or a total of 8 is

(a) 1/36

(b) 3/36

(c) 11/36

(d) 20/36

Solution: 

Let A be event of getting even number on first dice.

Let B be the event of getting sum 8.

n(A) = 18

n(B) = 5

n(A⋂B) = 3

n(S) = 36

P(A⋃B) = P(A)+P(B)-P(A⋂B)

= (18/36)+(5/36)-(3/36)

= 20/36

Hence option (d) is the answer. 

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