help!!!!Basic Probability(2)

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2013-10-12 7:13 am

1.A survey was conducted recently to study the problems about shopping on the
Internet. It was found that 75% of the orders were received on time. Of those
orders that were on time, 80% fulfilled the order requirement completely; of
those orders that were late, only 45% fulfilled the order requirement completely.

a.Find the probability that an order is on time and fulfils the order
requirements completely.
b.Find the probability that an order fulfils the requirements completely.
c.Suppose an order fulfils the requirements completely. Find the probability that this order is late.

2.A new-car buyer has to make an order of a new car by indicating his choice from 4 body styles, 3 different engines, and 10 colours. How many different cars can be resulted?

3.The following letters, A B C D E F, are arranged from the left to the right to
form “words” (A “word” needs not make sense).
a.How many ways to arrange them?
b.How many ways to arrange them if A cannot be the first letter?

4.A set of five different textbooks to be placed on a shelf from the left to the right.
a.How many arrangements will be resulted?
There are five different textbooks but only three books will be placed on theshelf.
b.How many arrangements will be resulted?
c.Without considering the order of the books on the shelf, how many combinations will be resulted?

回答 (1)

fooks

2013-10-12 8:07 am

✔ 最佳答案

1.
(a)
O: the order is received on time
O': the order is received late (NOT on time)
F: the order fulfills the requirement completely
F': the order does NOT fulfill the requirement completely

P(O) = 75% = 0.75
P(O') = 1 - 25% = 0.25

P(F|O) = 80%
P(F∩O) / P(O) = 0.8
P(F∩O) / 0.75 = 0.8

The required probability
= P(F∩O)
= 0.6

(b)
P(F|O') = 45%
P(F∩O') / P(O') = 0.45
P(F∩O') / 0.25 = 0.45
P(F∩O') = 0.1125

The required probability
= P(F)
= P(F∩O) + P(F∩O')
= 0.6 + 0.1125
= 0.7125

(c)
The required probability
= P(O'|F)
= P(F∩O') / P(F)
= 0.1125 / 0.7125
= 3/19
» 0.1579

2.
Number of cars can be resulted
= 4C1 x 3C1 x 10C1
= 4 x 3 x 10
= 120

3.
(a)
Number of ways to arrange them without restriction
= 6P6
= 6!
= 720

(b)
Number of ways to arrange them if A is the first letter
= 1P1 x 5P5
= 1 x 5!
= 120

Number of ways to arrange them if A is NOT the first letter
= 720 - 120
= 600

4.
(a)
Number of arrangements
= 5P5
= 5!
= 120

(b)
Number of arrangements
= 5P3
= 5!/2!
= 60

(c)
Number of combinations
= 5C3
= 5!/3!2!
= 10

參考: fooks

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