1. Class records at Rockwood College indicate that a student selected at random
ID: 3350465 • Letter: 1
Question
1. Class records at Rockwood College indicate that a student selected at random has probability 0.62 of passing French 101. For the student who passes French 101, the probability is 0.81 that he or she will pass French 102. What is the probability that a student selected at random will pass both French 101 and French 102? (Round your answer to three decimal places.)
2. Isabel Briggs Myers was a pioneer in the study of personality types. The personality types are broadly defined according to four main preferences. Do married couples choose similar or different personality types in their mates? The following data give an indication.
Similarities and Differences in a Random Sample of 375 Married Couples
Number of Similar Preferences
Number of Married Couples
All four
Three
Two
One
None
29
126
119
66
35
Suppose that a married couple is selected at random.
(a) Use the data to estimate the probability that they will have 0, 1, 2, 3, or 4 personality preferences in common. (Enter your answers to 2 decimal places.)
0
1
2
3
4
(b) Do the probabilities add up to 1? Why should they?
Yes, because they do not cover the entire sample space.No, because they do not cover the entire sample space. Yes, because they cover the entire sample space.No, because they cover the entire sample space.
(c ) What is the sample space in this problem?
0, 1, 2, 3 personality preferences in common1, 2, 3, 4 personality preferences in common 0, 1, 2, 3, 4, 5 personality preferences in common0, 1, 2, 3, 4 personality preferences in common
Similarities and Differences in a Random Sample of 375 Married Couples
Number of Similar Preferences
Number of Married Couples
All four
Three
Two
One
None
29
126
119
66
35
Explanation / Answer
(1)
Let
A = Passing French 101
B = Passing French 102.
Given:
P(A) = 0.62
P(B/A) = 0.81
By Multiplication Rule:
P(AB) = P(A) P(B/A) = 0.62 X 0.81 = 0.5022
So,
Probability that pass both French 101 & French 102 = 0.5022
(2)
Number of similar preferences Number of Married couples
All four 29
Three 126
Two 119
One 66
None 35
-----------------------------------------------------------------
Total 375
(a)
0 1 2 3 4
35/375=0.0933 66/375=0.176 119/375=0.3173 126/375=0.336 29/375=0.0773
(b)
Yes, because they cover the entire sample space.
(c)
Correct option:
0,1,2,3,4 personality preferences in common
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