[DyNaFeN]

hello.. I have an exam and a homework I couldn't solve all of it..
I know it's a lot of work.. but I couldn't solve 5 question

1)http://img696.imageshack.us/img696/7222/p46exercises24.jpg

2) If the probability that each component will operate is shown, what is the probability that a current will run through the circle shown.

http://img692.imageshack.us/img692/4213/90977388.jpg

3)http://img251.imageshack.us/img251/658/p54exercises26.jpg


4)http://img202.imageshack.us/img202/346/p58exercises27.jpg
if the company has 1% chance of incorrectly classifying the good item as defective.


5) http://img3.imageshack.us/img3/4338/p64exercises28.jpg


it would be greatly appreciated
thanks in advance

[Unknown008]

1 I'll use some notations here:
Let S represent the sample space (total failures)
Let G represent the set of failures due to gas leaks
Let E represent the set of failures due to electrical failure

(a) P(G) = n(G)/n(S)

(b) Use your formula for conditional probability:



(c) The same here, but inversed.



2. For this one, you have to find the probability that there is a complete circuit. There are 10 ways through which the current might flow. You need to find the probability of each of them. I don't see any other way than that. I know, it'll be long.

3. P(all are 1) = P(1st is 1) x P(2nd is 1) x... x P(10th is 1)

Similarly;

P(all are 0) = P(1st is 0) x P(2nd is 0) x... x P(10th is 0)

You can do it the long way, or you can apply what you know of the binomial distribution to solve it. Anyway, to solve the last part, you will have to use the binomial distribution.

Let X be the event there is a 1.
X~B(0.5, 10)

P(X = 5) = 10C5(0.5)^5(0.5)^5

4.
For this one, it's best to draw a probability tree. The first two branches will be Non defective and defective outcomes upon production. Then, from the branch of non defective, there is correctly identified as non defective and incorrectly classified as defective. From the branch of defective, there is correctly identified as defective and incorrectly classified as non defective.

The rest should then be very easy if you fill out the probability tree.

Look for the instances where the items are classified as defective, that is, non defective item incorrectly identified or defective item correctly identified.

For the second question, use the same formula as I said above.

5. Come on, this one is easy, use what you have learned again above to solve it.

P(A) = P(One dominant red gene) = P(Gene 1 is dominant red and gene 2 is recessive red) +P(Gene 2 is dominant red and gene 1 is recessive red)

Having a probability tree here would also help.

Post your answers.

[morgaine300]

Please don't double post. This was being addressed over here:
https://www.askmehelpdesk.com/mathem...on-456239.html

But unfortunately you've already cheated.