few dumb questions about probability [closed]

I've been working on probability questions for a little while now. The simpler questions are okey, but I can' figure out how to think on these questions.

Question 1:

I have a machine consisting of three parts: A, B and C. The probability of the parts working is: 0.9, 0.85 and 0.95.

What is the probability of the machine working if it only works if at least one of the three parts work?

Question 2

A company that manufactures batteries of a type is divided between three factories. Factory A creates 50% of the batteries, factory B 20% and factory C 30%. You know that batteries from factory A has 95% chanse of lasting longer than 100 hours, for B it's 97% and C it's 98%.

All the batteries are stored in the same warehouse. If you take one battery from there and find that it lasts longer than 100 hours, what is the probability of it coming from factory A?

closed as off-topic by GNUSupporter 8964民主女神 地下教會, John Doe, Rohan, eranreches, Claude LeiboviciDec 23 '17 at 6:52

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More general answer how this works. For question one you have three independent properties, A, B and C which are either true or false. Each machine has a specific combination for A, B, or C. For example if "A is true" "B is false" and "C is false" or $A \bar{B} \bar{C}$ for short. You have two options for each property and three properties so the total number of combination is $2^3=8$. You can divide the total population into 8 different buckets. $ABC$, $AB\bar{C}$, ... $\bar{A}\bar{B}\bar{C}$ .

You can calculate the probability of every bucket by multiplying the individual probabilities. For example $p(A \bar{B} \bar{C})=0.9 \cdot(1-0.85)\cdot(1-0.95)$ . The sum of the probabilities over all buckets will always be 1.

The last part is to identify the buckets that match your question and sum up the probabilities over them. In case of question 1 that's pretty easy, since the only bucket that is "nor working" is $\bar{A}\bar{B}\bar{C}$ . So you simply calculate the probability of this bucket and subtract it from one, $p_1 = 1-p:\bar{A}\bar{B}\bar{C})$

Question 2 works differently. Here is a simple way to go about it: Let's say you start with 1000 batteries. 500 A, 200 B, 300 C. After 100 hours, 95% of A will have failed, so you are left with 475 A. Same for B and C, so after 100 hours you have 475 A, 194 B and 294 C and 963 in total.

That gives you directly the probabilities, i.e $p(A) = 475/963$.

More generally speaking, you have two cascaded probabilities. In order for A to be true, a battery must first be from factory A ($p=0.5$) and last longer than one hundred hours ($p=0.95$). The final probability is simply the product of the two probabilities normalized to the total population. So it's

$$p(A)= \frac{0.5\cdot 0.95}{0.5\cdot 0.95 + 0.2\cdot 0.97 + 0.3\cdot 0.98}$$

Atleast 1 part works is the negation of all parts fail

Therefore $P($atleast 1 work$)$ = $1$- $P($all fail$)$

$P($all fail$)$ = $(1-0.9)(1-0.85)(1-0.95)$