1. ## Probability help??

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

The ansers I have are

a = 0.2986
b = 0.0096
c = 0.088

.....?

2. 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
A) P(at least one) = 1 - P(at most one)
= 1 - P(0) - P(1)
$=1 - .92(.95).96(.88)(.95) - .08(.95)(.96)(.88)(.95) - .05(.92)(.96)(.88)(.95)$ $- .04(.92)(.95)(.88)(.95) - .12(.92).95(.96).95 - .05(.92).95(.96)(.88)$

B) P(both water and drainage) = P(W)P(D)
= $.12(.08)$

I will leave the last one up to you.

3. ## Hmm

I get the answer B easily its just the product of P(W) and P(D) but Im still confused as to whats going on. I see you have taken the eventually of each of the servies in turn. but I think the brackets have confused me. as to what you do to these numbers Grr. Probability was never my strong point *sob sob*

4. Originally Posted by juangambino
I get the answer B easily its just the product of P(W) and P(D) but Im still confused as to whats going on. I see you have taken the eventually of each of the servies in turn. but I think the brackets have confused me. as to what you do to these numbers Grr. Probability was never my strong point *sob sob*
$P(A \cap B) = P(A)P(B)$ if independence

$P(A or B) = P(A \cup B)$
$= P(A) + P(B) - P(A \cap B)$

5. I get the answer B easily its just the product of P(W) and P(D) but Im still confused as to whats going on. I see you have taken the eventually of each of the servies in turn. but I think the brackets have confused me. as to what you do to these numbers Grr. Probability was never my strong point *sob sob*
If I understand your problem right, you are being confused by the invisible multiplication signs. These sit between any numbers not separated by another operation (an operation is +,- etc). You will see (or not see them :P) these more and more as you progress in maths.

However, Colby is mistaken anyway.

a) the probability of at least one happening = 1 - the probability none happen.

To get the probability of none happening, just multiply together the probability of each one not happening. So

colby got part b) right and pointed you in the right direction for part c, but just in case you haven't $\cap$ before I will explain in words.