Compound Dependent Events

This page is based on compound dependent events which is a study under dependent events in statistics. In probability, an event is a one or more possible outcomes from an experiment. An event consisting of one or more simple events is called compound event. An event is called dependent event, if an event does affect the other event.

Compound Events Dependent - Examples

Ex 1: A bag contains 20 bulbs, in which 5 bulbs are defective bulbs. If 2 bulbs are taken at one by one without replacement, what is the probability of getting good and defective bulbs?

Sol:

Let S be the sample space, n(S) = 20

Number of good bulbs = 20 - 5 = 15

A be the event of getting a good bulb, n(A) = 15

B be the event of getting a defective bulb, n(B) = 5

P(A) = `(n(A))/(n(S))` = `15/20`= `3/4`

Now 19 balls will be remaining in the box, so

P(B) = `(n(B))/(n(S))` = `5/19`

P(A and B) = `3/4` · `5/19` = `15/76`.

P(Good and Defective bulb) = `15/76` .

Ex 2: A jar contains 7 yellow candies, 5 green candies, and 8 blue candies. If 2 candies are taken one by one without replacement, what is the probability of getting yellow and green candy?

Sol:

Let S be the sample space, n(S) = 7 + 5 + 8 = 20

A be the event of getting a yellow candy, n(A) = 7

B be the event of getting a green candy, n(B) = 5

P(A) = `(n(A))/(n(S))` = `7/20`

Now 19 candies will be remaining in a jar.

P(B) = `(n(B))/(n(S))` = `5/19`

P(A and B) =P(A) · P(B) = `7/20` · `5/19` = `7/76`

P(Yellow and Green) = `7/76`

Practice Problems

Pr 1: A bag contains 15 bulbs, in which 7 bulbs are defective bulbs. If 2 bulbs are taken at one by one without replacement, what is the probability of getting good and defective bulbs?

Pr 2: A jar contains 10 yellow candies, 6 green candies, and 4 blue candies. If 2 candies are taken one by one without replacement, what is the probability of getting yellow and green candy?

Ans: 1) ` 4/15` 2) `3/19`