Associated Topics || Dr. Math Home || Search Dr. Math

A Truck and a Van

```
Date: 07/11/99 at 04:02:44
From: Pekkle
Subject: Probability

A small cartage company has two vehicles, a truck and a van. The
truck is used 75% of the time; both vehicles are used together 40%
of the time; and neither of the vehicles is used for 10% of the time.

Let T be the event "the truck is used" and V be the event "the van is
used."

a) complete the statement P(T U V)= P(T) + _____?
Hence find the proportion of the time that the van is used.

b) Show that the two vehicles are not statistically independent in
their use.

c) Given that the van is being used, find the probability that the
truck is also being used. (Use a probability tree.)

Thanks.
```

```
Date: 07/11/99 at 07:24:50
From: Doctor Anthony
Subject: Re: Probability

>Let T be the event "the truck is used" and V be the event "the van
>is used."

P(T or V) = P(T) + P(V) - P(T and V)

0.9   = 0.75 + P(V) - 0.4

P(V) = 0.9 - 0.75 + 0.4

= 0.55

So the van is used 55% of the time.

>b) Show that the two vehicles are not statistically independent in
>their use.

To be statistically independent we require

P(T and V) = P(T) x P(V)

0.4     = 0.75 x 0.55

= 0.4125  NOT TRUE

so the two are NOT statistically independent.

>c) Given that the van is being used, find the probability that the
>truck is also being used. (Use a probability tree.)

| T      not T   |   Total
------|----------------|-----------
V  |0.4             |0.55
not V  |         0.10   |0.45
------|----------------|------------
Total  |0.75     0.25   |1.00

The above table can be filled in as shown. It is now possible to
complete it to make the row and column totals agree.  We complete the
table as follows:

| T      not T   |   Total
------|----------------|-----------
V  |0.4      0.15   |0.55
not V  |0.35     0.10   |0.45
------|----------------|------------
Total  |0.75     0.25   |1.00

And now from the table we can answer the last question. "Given that
the van is being used" means we are in row V with a total sample space
of 0.55.

Then the probability that the truck is also being used is 0.4/0.55

=  0.7272..

=  8/11

- Doctor Anthony, The Math Forum
http://mathforum.org/dr.math/
```

```
Date: 07/13/99 at 04:39:29
From: Pekkle
Subject: Probability

Thank you for answering my question. I have another:

The following table shows how many of the 1992 graduates from a
university had jobs by May. (The following tables are together.)

Occupation Agriculture Business Education Arts Science Engineering
----------------------------------------------------------------------
Permanent     209        110      158      17     48       85
employment
----------------------------------------------------------------------
Seeking        47         17       21      37     62       11
employment
----------------------------------------------------------------------
Not seeking    22         15        8      15     14        5
employment
----------------------------------------------------------------------
Totals        278        142      187      69    124      101
----------------------------------------------------------------------

Occupation  Social Sciences  Medicine  Totals
-------------------------------------------------------
Permanent        96            39         762
employment
-------------------------------------------------------
Seeking          68            31         267
employment
-------------------------------------------------------
Not seeking      31             7         117
employment
-------------------------------------------------------
Totals          195            50        1146

Consider the following events:

A: that a randomly selected graduate is either from the education or
social sciences faculties
B: that a randomly selected graduate is seeking employment

(required to show your reasoning and working clearly for parts (a)
and (b).

a)  The events A and B are:
A. Complementary but not independent
B. Mutually exclusive but not independent
C. Mutually exclusive and independent
D. Neither mutually exclusive nor independent
E. Not mutually exclusive, but independent

b)  The conditional probability of A happening, given that B has
happened, and the probability of either A or B or both happening are:
A. P(A/B)=0.333     P(A or B)=0.566
B. P(A/B)=0.333     P(A or B)=0.489
C. P(A/B)=0.078     P(A or B)=0.566
D. P(A/B)=0.078     P(A or B)=0.489
E. P(A/B)=0.333     P(A or B)=0.078

Thanks a lot.
```

```
Date: 07/13/99 at 16:52:21
From: Doctor Anthony
Subject: Re: Probability

>Consider the following events:
>A: that a randomly selected graduate is either from the education or
>social sciences faculties

= 187 + 195 = 382    out of total of 1146

>B: that a randomly selected graduate is seeking employment

= 267  out of 1146

>(required to show your reasoning and working clearly for parts (a)
>and (b).
>
>a)  The events A and B are:
>    A. Complementary but not independent

To be complementary you require P(B) = 1 - P(A)

That is  267/1146 = 1 - 382/1146

This is clearly NOT TRUE

To be independent you require

P(A and B) = P(A) x P(B)

Then P(A and B) = (21 + 68)/1146  =  89/1146

and P(A) x P(B) = 382/1146 x 267/1146 = 1/3 x 89/382 = 89/1146

since these two probabilities are equal, the two events are
independent.

We conclude that the events are NOT complementary but ARE independent.

>    B. Mutually exclusive but not independent

Since there are numbers that are in BOTH A and B the events are NOT
mutually exclusive.

We have already seen that the events are independent, so wrong again
on both counts.

>    C. Mutually exclusive and independent

Not Mutually exclusive - see previous answer. Yes, they are
independent.

>    D. Neither mutually exclusive nor independent

Not mutually exclusive but they ARE independent.

>    E. Not mutually exclusive, but independent

Correct.  Not mutually exclusive, and they are independent.
>
>b)  The conditional probability of A happening, given that B has
>happened, and the probability of either A or B or both happening are:

If B has happened the sample space is limited to the 'Seeking
Employment' row.

The numbers in this row for Education or Soc.Science  = 21 + 68 = 89

So  P(A|B) = 89/267  = 1/3

P(A or B) = P(A) + P(B) - P(A and B)
= 382/1146 + 267/1146 - 89/1146
= 560/1146
= 0.488656

So the two answers are  P(A|B) = 1/3  and P(A or B) = 0.488656

These agree with the answers given under (B) below.

>    B. P(A/B)=0.333     P(A or B)=0.489

- Doctor Anthony, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
High School Probability

Search the Dr. Math Library:

 Find items containing (put spaces between keywords):   Click only once for faster results: [ Choose "whole words" when searching for a word like age.] all keywords, in any order at least one, that exact phrase parts of words whole words

Submit your own question to Dr. Math
Math Forum Home || Math Library || Quick Reference || Math Forum Search