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A Truck and a VanDate: 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/
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