Learn Independent Events
An event is a one or more possible outcomes of a certain experiment. An event is called independent event if one event does not affect the other event. For example, choosing a 7 and 8 in the deck of card with replacement is two independent events. An event consisting of more than one simple event is called compound event. In this lesson we will learn about probability of independent events.
Learn Independent Events – Learn Example Problems
Example 1: A box contains 10 bulbs, in which 5 are red color and 5 are orange color bulbs. Two bulbs are drawn one by one with replacement of first bulb. What is the probability of drawing a red and orange bulb successively?
Solution:
Let S = Sample space, n(S) = 5 + 5 = 10
A be the event of drawing a red ball first, n(A) = 5
B be the event of drawing a orange bulb second, n(B) = 5
P(A) = `(n(A))/(n(S))` =` 5/10` = `1/2`
P(B) = `(n(B))/(n(S))` = `5/10` = `1/2`
P(A and B) = P(A) · P(B) = `1/2` · `1/2` = `1/4`
P(A and B) = `1/4` .
Example 2: A box contains 15 candies, in which 5 are lemon candies, 5 are pineapple candies, and 5 are orange candies. Three candies are drawn one by one with replacement of previous candy. What is the probability of drawing a lemon, pineapple, and orange candy successively?
Solution:
Let S = Sample space, n(S) = 5 + 5 + 5 = 15
A be the event of drawing a lemon candy first, n(A) = 5
B be the event of drawing a pineapple candy second, n(B) = 5
C be the event of drawing a orange candy third, n(C) = 5
A, B, and C are independent event, so one event does not affect the other events.
P(A) = `(n(A))/(n(S))` =` 5/15` = `1/3`
P(B) = `(n(B))/(n(S))` = `5/15` = `1/3`
P(C) = `(n(C))/(n(S))` = `5/15` = `1/3`
P(A and B and C) = P(A) · P(B) · P(C) = `1/3` · `1/3` · `1/3` = `1/27`
P(A and B and C) = `1/27` .
I have recently faced lot of problem while learning cbse class 9 syllabus, But thank to online resources of math which helped me to learn myself easily on net.
Learn Independent Events – Practice Problems
Problem 1: A box contains 25 bulbs, in which 15 are red color and 10 are orange color bulbs. Two bulbs are drawn one by one with replacement of first bulb. What is the probability of drawing a red and orange bulb successively?
Problem 2: A box contains 25 candies, in which 10 are lemon candies, 10 are pineapple candies, and 5 are orange candies. Three candies are drawn one by one with replacement of previous candy. What is the probability of drawing a lemon, pineapple, and orange candy successively?
Answer: 1) ` 6/25` 2) `4/125`
An event is a one or more possible outcomes of a certain experiment. An event is called independent event if one event does not affect the other event. For example, choosing a 7 and 8 in the deck of card with replacement is two independent events. An event consisting of more than one simple event is called compound event. In this lesson we will learn about probability of independent events.
Learn Independent Events – Learn Example Problems
Example 1: A box contains 10 bulbs, in which 5 are red color and 5 are orange color bulbs. Two bulbs are drawn one by one with replacement of first bulb. What is the probability of drawing a red and orange bulb successively?
Solution:
Let S = Sample space, n(S) = 5 + 5 = 10
A be the event of drawing a red ball first, n(A) = 5
B be the event of drawing a orange bulb second, n(B) = 5
P(A) = `(n(A))/(n(S))` =` 5/10` = `1/2`
P(B) = `(n(B))/(n(S))` = `5/10` = `1/2`
P(A and B) = P(A) · P(B) = `1/2` · `1/2` = `1/4`
P(A and B) = `1/4` .
Example 2: A box contains 15 candies, in which 5 are lemon candies, 5 are pineapple candies, and 5 are orange candies. Three candies are drawn one by one with replacement of previous candy. What is the probability of drawing a lemon, pineapple, and orange candy successively?
Solution:
Let S = Sample space, n(S) = 5 + 5 + 5 = 15
A be the event of drawing a lemon candy first, n(A) = 5
B be the event of drawing a pineapple candy second, n(B) = 5
C be the event of drawing a orange candy third, n(C) = 5
A, B, and C are independent event, so one event does not affect the other events.
P(A) = `(n(A))/(n(S))` =` 5/15` = `1/3`
P(B) = `(n(B))/(n(S))` = `5/15` = `1/3`
P(C) = `(n(C))/(n(S))` = `5/15` = `1/3`
P(A and B and C) = P(A) · P(B) · P(C) = `1/3` · `1/3` · `1/3` = `1/27`
P(A and B and C) = `1/27` .
I have recently faced lot of problem while learning cbse class 9 syllabus, But thank to online resources of math which helped me to learn myself easily on net.
Learn Independent Events – Practice Problems
Problem 1: A box contains 25 bulbs, in which 15 are red color and 10 are orange color bulbs. Two bulbs are drawn one by one with replacement of first bulb. What is the probability of drawing a red and orange bulb successively?
Problem 2: A box contains 25 candies, in which 10 are lemon candies, 10 are pineapple candies, and 5 are orange candies. Three candies are drawn one by one with replacement of previous candy. What is the probability of drawing a lemon, pineapple, and orange candy successively?
Answer: 1) ` 6/25` 2) `4/125`
