Math 102 unit 4
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Pace University *
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102
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Mathematics
Date
Jan 9, 2024
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docx
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Faye Forrer
Math 102 Unit 4
1.
Customer account "numbers" for a certain company consists of 2
lower case letters followed by 2
single-digit numbers. How many different account numbers are possible if repetitions of letters and digits are allowed
?
(26)(10) = 260
2.
The president of the college asked a fraternity to submit a list of 7
students to serve as guides at a school function. They have 10
students who are available to help. In how many ways can the students be chosen? (Hint: Does order matter?
120
3.
Lucy was born on 07/10/1979. How many eight-digit codes could she make using the digits in her birthday?
3360
4.
Suppose a jar contains 12 red marbles and 30 blue marbles. If you reach in the jar and pull out 2 marbles at random, find the probability that both are red. (Hint: Is this with or without replacement?)
(12/42)(11/41) = 0.073(100) = 7.3%
5.
The table below shows the number of survey subjects who have received and not received a speeding ticket in the last year, and the color of their car. Find the probability that a randomly chosen person got a speeding ticket given they do not have a red car.
2.26%
6.
Compute the probability of randomly drawing five cards from a deck and getting exactly two Kings. C(52, 5) = 52! / (5! x (52-5)!) = 2,598,960
C(4, 2) = 4! / (2! x (4-2)!) = 6
C(48, 3) = 48! / (3! x (48-3)!) = 17,296
6 x 17,296 = 103,776
(103,776 / 2,598,960)100 = 3.99%
7.
Simplify
(
2
!
)(
7
!
)
(
8
!
)(
(
5
−
3
)
!
)(
3
!
)
3
13
C
6
n! = 13!
13! = 13 x 12 x 11 x 10 x 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1
13! = 6,227,020,800
(n - r)! = (13 - 6)!
(13 - 6)! = 7!
7! = 7 x 6 x 5 x 4 x 3 x 2 x 1
7! = 5,040
r! = 6!
6! = 6 x 5 x 4 x 3 x 2 x 1
6! = 720
13C6 = 6,227,020,800/720 x 5,040
13C6 = 6,227,020,800/3,628,800
13
P
6
13! / (13 - 6)!
13! / 7!
13! = 6,227,020,800
7! = 5,040
6,227,020,800 / 5,040 = 1,235,520
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