Multiplication Rule for Independent Events
Probability is the likelihood of the occurrence of an event. 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.
Multiplication rule for two independent events:
If A and B are two independent event then; P(A and B) = P(A) · P(B)
Multiplication rule for three independent events:
If A, B, and B are three independent events then; P(A and B and C) = P(A) · P(B) · P(C)
Multiplication Rule for Independent Events - Example Problems
Example 1: Two coins are drawn at one by one with replacement of previous coin in a bag of 12 nickels and 9 quarters. Find the probability of first coin is nickels and second coin is quarters.
Solution:
Lest S be the sample space, n(S) = 12 + 9 = 21
A be the event of drawing a nickels, n(A) = 12
B be the event of drawing a quarters, n(B) = 9
P(A) = `(n(A))/(n(S))` = `12/21` = `4/7`
P(B) = `(n(B))/(n(S))` = `9/21 ` = `3/7`
P(A and B) = P(A) · P(B) = `4/7` · `3/7` = `12/49`
P(A and B) = `12/49`
Example 2: A jar contains 6 dark, 8 milk, and 10 white chocolates. Three chocolates are drawn one by one with replacement. What is the probability of drawing each kind is one time?
Solution:
Lest S be the sample space, n(S) = 6 + 8 + 10 = 24
A be the event of drawing a dark chocolate, n(A) = 6
B be the event of drawing a milk chocolate, n(B) = 8
C be the event of drawing a white chocolate, n(B) = 10
P(A) = `(n(A))/(n(S))` = `6/24` = `1/4`
P(B) = `(n(B))/(n(S))` = `8/24` = `1/3`
P(C) = `(n(C))/(n(S))` = `10/24` = `5/12`
P(A and B and C) = P(A) · P(B) · P(C) = `1/4` · `1/3` · `1/12` = `1/144`
P(A and B and C) = `1/144`
I am planning to write more post on solving complex rational expressions, geometric probability formula. Keep checking my blog.
Multiplication Rule for Independent Events - Practice Problems
Problem 1: Two chocolates are drawn one by one with replacement of previous chocolate in a jar of 10 bitter and 5 dark chocolates. Find the probability of first one is bitter and dark chocolate.
Problem 2: A jar contains 11 dark, 12 milk, and 13 white chocolates. Three chocolates are drawn one by one with replacement.What is the probability of drawing each kind exactly one time.
Answer: 1) `2/9` 2) `143/3888`
Probability is the likelihood of the occurrence of an event. 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.
Multiplication rule for two independent events:
If A and B are two independent event then; P(A and B) = P(A) · P(B)
Multiplication rule for three independent events:
If A, B, and B are three independent events then; P(A and B and C) = P(A) · P(B) · P(C)
Multiplication Rule for Independent Events - Example Problems
Example 1: Two coins are drawn at one by one with replacement of previous coin in a bag of 12 nickels and 9 quarters. Find the probability of first coin is nickels and second coin is quarters.
Solution:
Lest S be the sample space, n(S) = 12 + 9 = 21
A be the event of drawing a nickels, n(A) = 12
B be the event of drawing a quarters, n(B) = 9
P(A) = `(n(A))/(n(S))` = `12/21` = `4/7`
P(B) = `(n(B))/(n(S))` = `9/21 ` = `3/7`
P(A and B) = P(A) · P(B) = `4/7` · `3/7` = `12/49`
P(A and B) = `12/49`
Example 2: A jar contains 6 dark, 8 milk, and 10 white chocolates. Three chocolates are drawn one by one with replacement. What is the probability of drawing each kind is one time?
Solution:
Lest S be the sample space, n(S) = 6 + 8 + 10 = 24
A be the event of drawing a dark chocolate, n(A) = 6
B be the event of drawing a milk chocolate, n(B) = 8
C be the event of drawing a white chocolate, n(B) = 10
P(A) = `(n(A))/(n(S))` = `6/24` = `1/4`
P(B) = `(n(B))/(n(S))` = `8/24` = `1/3`
P(C) = `(n(C))/(n(S))` = `10/24` = `5/12`
P(A and B and C) = P(A) · P(B) · P(C) = `1/4` · `1/3` · `1/12` = `1/144`
P(A and B and C) = `1/144`
I am planning to write more post on solving complex rational expressions, geometric probability formula. Keep checking my blog.
Multiplication Rule for Independent Events - Practice Problems
Problem 1: Two chocolates are drawn one by one with replacement of previous chocolate in a jar of 10 bitter and 5 dark chocolates. Find the probability of first one is bitter and dark chocolate.
Problem 2: A jar contains 11 dark, 12 milk, and 13 white chocolates. Three chocolates are drawn one by one with replacement.What is the probability of drawing each kind exactly one time.
Answer: 1) `2/9` 2) `143/3888`
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