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Math 323 Homework: Probability for Job Apps & Auto Orders, Assignments of Probability and Statistics

Solutions to problems p84.2.25 and p84.2.31 from math 323 homework #2. The first problem deals with calculating the number of ways jobs can be filled when the first person receives a higher salary than the second. The second problem involves finding the probability of certain combinations of orders for automobiles of a certain style, including one blue, one white, and one green, two blues, at least one black, and exactly two orders for the same color.

Typology: Assignments

Pre 2010

Uploaded on 08/31/2009

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Math 323 Homework # 2
P84. 2.25.Seven applicants have applied for two jobs. How many ways can
the jobs be filled if
a. the first person chosen receives a higher salary than the second?
b. there are no differences between the jobs?
Solution: a.
P7
2= 7 ×6 = 42.
b. 7
2!=7×6
1×2= 21.
P84. 2.31. For a certain style of new automobile, the colors blue, white
black, and green are in equal demand. Three successive orders are placed for
automobiles of this style. Find the probability that
a. one blue, one white, and one green are ordered.
b. two blues are ordered.
c. at least one black is ordered.
d. exactly two of the orders are for the same color.
Solution: a. Total is 43= 64. In event, 3 ×2×1 = 6.
p=6
64 =3
32.
b. Locations of blues: 3
2= 3, color of the other order 3
1= 3.
p=3×3
64 =9
64.
c.
p= 1 P(no black)
= 1 33
43=64 27
64 =37
64.
d. Use part b, we get
p= 4 ×9
64 =9
16.
1

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Math 323 Homework # 2

P84. 2.25.Seven applicants have applied for two jobs. How many ways can the jobs be filled if a. the first person chosen receives a higher salary than the second? b. there are no differences between the jobs? Solution: a. P 27 = 7 × 6 = 42.

b. (^) ( 7 2

)

7 × 6

1 × 2

P84. 2.31. For a certain style of new automobile, the colors blue, white black, and green are in equal demand. Three successive orders are placed for automobiles of this style. Find the probability that a. one blue, one white, and one green are ordered. b. two blues are ordered. c. at least one black is ordered. d. exactly two of the orders are for the same color. Solution: a. Total is 4^3 = 64. In event, 3 × 2 × 1 = 6.

p =

b. Locations of blues:

( 3 2

) = 3, color of the other order

( 3 1

) = 3.

p =

3 × 3

c.

p = 1 − P (no black) = 1 −

d. Use part b, we get

p = 4 ×