How to Solve Compound Probability Problems

An event is an occurrence that can be calculated by a given level of certainty. For example, when we say that the probability that it will rain today is high or low, we are stating that the event may or may not happen. The ways in which an event can happen are the outcomes of an event. Many events that you see around you can have different outcomes. Compound probability is a term relating to the likeliness of two independent events occurring. In this article, we will learn how to solve compound probability examples.

What is Compound Probability?

Compound probability refers to the probability of two or more independent events both happening. We can find the compound probability by multiplying the probability of the first event by the probability of the second event.

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).

Steps to Find Compound Probability

1. If events are independent, find the product of the probabilities.

2. If events are mutually exclusive, find the sum of the probabilities.

3. If events are mutually inclusive, find the sum of probabilities and subtract the probability of both events occurring.

Compound Probability Word Problems

Problem 1.

A store contains 35 light green parakeets (14 females and 21 males) and 44 sky blue parakeets (28 females and 16 males). You randomly choose 1 of the parakeets. What is the probability that it is a female or sky blue parakeet?

Solution:

Let P(F) denotes the probability of females and P(SB) denotes the probability of sky blue parakeets

P(Female) = 42/79

P(Sky blue) = 44/79

P(F or SB) = P(F) + P(SB) – P(F and SB)

= (42/79) + (44/79) – (28/79)

= 58/ 79

Hence, the required probability is 58/79.

Problem 2.

A card is drawn from a well-shuffled deck of 52 cards. What is the probability of getting a king or an ace?

Solution:

Let A be the event of getting a king and B be the event of getting an ace.

P(A) = 4/52 = 1/13

P(B) = 4/52 = 1/13

We know that, drawing a well-shuffled deck of 52 cards ‘getting a king’ and ‘getting an ace’ are mutually exclusive events.

So P(A or B) = P(A) + P(B)

= (1/13) + (1/13)

= 2/13

Hence the required probability is 2/13.

Problem 3.

Let X and Y be two events such that P(X|Y) = 1/2. P(Y|X) = 1/3 and P(X⋂Y) = 1/6. Which of the following is (are) correct.

(a) P(X U Y) = 2/3

(b) X and Y are independent

(c) P(E) = 2/5, P(F) = 1/5

(d) P(E) = 3/5, P(F) = 4/5

Solution:

Given P(X|Y) = 1/2. P(Y|X) = 1/3 and P(X ⋂ Y) = 1/6

P(X ⋂ Y) = P(X) . P(Y|X)

1/6 = P(X) (1/3)

So P(X) = 1/2

P(X⋂Y) = P(Y) . P(X|Y)

⇒ P(Y) = 1/3

P(X⋂Y) = P(X).P(Y)

⇒ X and Y are independent.

P(XUY) = P(X) + P(Y) – P(X⋂Y)

= ½ + 1/3 – 1/6

= ⅔

P(X’⋂Y) = P(Y) – P(X⋂Y) = ⅓ – ⅙ = ⅙

Hence option (a) and (b) are correct.

Problem 4.

A box contains 50 tickets numbered 1 to 50. If a ticket is drawn at random, what is the probability that the number drawn is a multiple of 3 or 5?

Solution:

Let A be the event of getting a multiple of 3.

Y be the event of getting a multiple of 5.

The events of getting a multiple of 3 = {3,6,9,12,15,18,21,24,27,30,33,36,39,42,45,48}

Total number of multiple of 3 = 16

So P(A) = 16/50 = 8/25

The events of getting a multiple of 5 = {5, 10, 15, 20, 25, 30, 35, 40, 45, 50}

Total number of multiple of 5 = 10

P(B) = 10/50 = 1/5

Common outcomes are 15, 30, and 45.

So the number of common multiple of both 3 and 5 = 3

P(A and B) = P(A ∩ B) = 3/50

P(A ∪ B) = P(A) + P(B) – P(A ∩ B)

= (8/25) + (1/5) – (3/50)

= 23/50

Hence, the required probability is 23/50.

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