Sta107 Online Tutorial

Defintion 1.5-1

Example 1:

At UTM, the probability that a student is taking STA107 and MAT138 is 0.32. The probability that a student is taking only MAT138 is 0.62. What is the probability that a student is taking STA107 given that they are taking MAT138.

Solution:

Let A be the event that a student is taking STA107. Let B be the event that a student is taking MAT138. So P(STA107|MAT138)=P(STA107 ∩ MAT138) / P(MAT138) = 0.32/0.62 = 0.52

Example 2:

A professor gave her class 2 tests. 29% of the class passed both tests, and 37% of the class passed the first test. What percentage of the class passed the second test given that they passed the first test?

Click here to see an interactive venn diagram that will illustrate the concept of conditional probability!

Multiplication Rule for Conditional Probability

Example 3:

Consider drawing two cards, one after the other, without replacement, from a deck of 52 cards. Find the probability that both cards are Hearts.

Solution:

Let A be the event of drawing a Heart on the first draw. Let B be the event of drawing a Heart on the second draw. We want to find P(A ∩ B). So P(A)=13/52, because there are 13 Hearts in a deck of cards. Given that the first card was a Heart, there would be 12 Hearts left in the 51 cards. So the probability of drawing a Hearts on the second draw is P(B|A) =12/51. So P(A ∩ B)=(13/52) x (12/51)

What is the probability of drawing 3 Hearts, without replacement?

1. (13/52)x(12/52)x(11/52)
2. (13/52)x(13/52)x(13/52)
3. (13/52)x(12/51)x(11/50)

Solution

Law of Total Probability

Example 4:

A company manufacturing candy canes has 3 production machines M1, M2 and M3. The percent of daily production of candy canes and the percentage of broken candy canes for each machine is given in the table below:

Table 1: Daily Production of Candy Canes

Machine

% of Daily Production

% of Broken Candy Canes

M1

11%

2%

M2

56%

1%

M3

33%

3%

So 11% of the daily production of candy canes come from Machine 1, and 2% of those candy canes are broken. In the same way, M2 and M3 produce 56% and 33% of the daily production of candy canes respectively, where 1% and 3% of those candy canes are broken respectively.

Let B represent the event of a selected candy can being broken.

A: What is P(M1)?

(Round to 2 decimal places)

B: What is P( B|M3) ?

The probability that a selected candy cane is defective is:

P(B) = P(B ∩ M1)+P(B ∩ M2) +P(B ∩ M3)

=P(M1)P(B|M1) + P(M2)P(B|M2) + (M3)P(B|M3)

So P(M1)P(B|M1)=(0.11)(0.01)= 0.0011.

C: What is P(M2)P(B|M2)?

(Round to 4 decimal places)

D: What is P(M3)P(B|M3)?

(Round to 4 decimal places)

Solution

The probability that a selected candy cane is broken is:

P(B) = P(B ∩ M1)+P(B ∩ M2) +P(B ∩ M3)

= P(M1)P(B|M1) + P(M2)P(B|M2) + P(M3)P(B|M3)

=0.0177

The probability that a selected broken candy cane was produced in M1 is:

P(M1|B)= P( M1 ∩ B) / P(B) = (0.0022) / (0.0177) = 0.12

E: What is the probability that a selected broken candy cane came from M2?

(Round to 2 decimal places)

F: What is the probability that a selected broken candy cane came from M3?

Useful Web Resources

Conditional Probability and Independence

Conditional Probability- Wikipedia, the free encyclopedia

Stats: Conditional Probability

Solutions

Solution to Example 2

Let A be the event that the class passed the first test. The B be the event that the class passed the second test.

P(B|A)=0.29/0.37=0.78

Solution to Example 3

Let A be the event of drawing a Heart on the first draw. Let B be the event of drawing a Heart on the second draw. Let C be the event of drawing a Heart on the third draw.We know that P(A ∩ B ∩ C) = P(A)P(B|A)P(C|A and B). So P(A)=13/52, because there are 13 Hearts in a deck of cards. Given that the first card was a Heart, there would be 12 Hearts left in the 51 cards. So the probability of drawing a Hearts on the second draw is P(B|A) =12/51. Given that the first two cards were Hearts, there would be 11 Hearts in the remaining 50 cards. So P(C| A and B)=(13/52) x (12/51) x (11/50).

Solution to Example 4

A:The daily production of candy canes from M1 is 11%. So P(M1)=0.11

B: The percentage of broken candy canes from M2 is P(B|M3)=0.03

C: P(M2)P(B|M2)=(0.56)(0.01)= 0.0056

D: P(M3)P(B|M3)=(0.33)(0.03)= 0.0099

E: The probability that a selected broken candy cane came from M2 is P(M2|B)= P( M2 and B) / P(B) =(0.56)(0.01)/(0.0177)=0.32

F: The probability that a selected broken candy cane came from M3 is P(M3|B)= P( M3 and B) / P(B) =(0.33)(0.03)/(0.0177)=0.56

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