statistics help???

Question

I wondered if some one can help me and show working out the calks to this question...

A contractor has to excavate for foundations in an area for which there are no plans of existing services. It is anticipated that he may cut across one or more of the service lines whilst excavating. The following probabilities have been calculated for each of the following services:

Water supply - 0.08
Gas - 0.05
Telephone Cable – 0.04
Drainage – 0.12
Electricity – 0.05

Calculate the probability of cutting across

a) at least one of the services
b) both water and drainage
c) electricity or telephone cable

 I have the answers but I don’t know how to get them no putting them on here so I can check your answers too.

Hope someone can help

Regards

J

Answer 1

I don't think adding the probabilities is the correct way for calculating problem A. Think of it this way. You have three pennies. Each one has a probability of 0.5 of coming up heads. If you add up 0.5 + 0.5 + 0.5 you get 1.5, which is greater than 1 (1 being 100% certainty). That clearly doesn't make any sense. The true probability that at least one penny will come up heads is 0.875.

The way to figure out a problem like this is to find the product of all the probabilities of each item NOT happening and then subtract that from 1. The probability of cutting the water supply is 0.08 so the probability of not cutting it is 0.92 (1 - 0.08). If you multiply the probabilities of each one not happening it looks like this:

0.92 * 0.95 * 0.96 * 0.88 * 0.95 = 0.70143744

That's the probability of not cutting any lines at all so the probability of cutting at least one is:

1 - 0.70143744 = 0.29856256

Problem B can be solved simply by multiplying the two probabilities together: 0.08 * 0.12 = 0.0096

Problem C is similar to problem A and can be solved in the same way. The probability of not cutting the electricity or telephone lines is 0.95 * 0.96 = 0.912. So the probability of cutting at least one of them is 1 - 0.912 = 0.088.

Hope this helps.

Answer 2

The probability of cutting at least one of the services is the sum of all of the probabilities - you don't care which service is cut, only that at least one gets cut.

Now the probability of cutting both water and drainage is the product of the those two probabilities. Think of this in terms of sets. Each probability is a set. To cut the two services mentioned requires their sets to intersect which is the equivalent of multiplying the probabilities.

The final one is the sum of the electrical and phone cable probabilities. Again, from a set stand point, this is the union of the two sets which is equivalent to addition.

Answer 3

A) Water supply - 0.08
B) Gas - 0.05
C) Telephone Cable – 0.04
D) Drainage – 0.12
E) Electricity – 0.05

Calculate the probability of cutting across

a) at least one of the services

P[A or B or C or D or E] = .08 + .05 + .04 + .12 + .05 = .34

b) both water and drainage

P[A and D] = .08 * .12 = .0096

c) electricity or telephone cable

P[C or E] = .04 + .05 = .09

Answer 4

Its simple:

If its a, b OR c, you add 'em up.

If its a, b AND c, you multiply 'em.