Question

Any statistics people on here!? I need help with probability using the multiplication rule!!!

Answer 1

All right, what's your question?

Answer 2

I just cannot figure out how they got this answer.

Answer 3

So you start by getting the NON defective ones in a fraction which is 669-160 = 509.
so 509/669 are Not defective

Answer 4

and then I dont know what to do after that because i've confused on why that isnt the answer.

Answer 5

i'm*

Answer 6

All right, so we have:
\[
\frac{160}{669}
\]CD's are defective. The probability of us choosing one that is not defective:
\[
\frac{669-160}{669}=\frac{509}{669}
\]And, continually, as we choose more CD's, we have less chance of picking all non-defective, in a row, our probability becomes:
\[
\frac{_{509}P_8}{_{669}P_8}=\frac{509\cdot508\cdot507\cdot506\cdot505\cdot504\cdot503\cdot502}{669\cdot668\cdot667\cdot666\cdot665\cdot664\cdot663\cdot662}\approx0.1108
\]And, that should be our answer.

Answer 7

but it says the answer is .112
:/

Answer 8

how about we try another one? something a little bit different

Answer 9

I know. You should have a bound of error, since some calculators use scientific notation to compute these large numbers and lose precision. But, yes, go for it.

Answer 10

again the answer is given, but i cant figure out why

Answer 11

I'm not sure what BOTH is referring to, so I dont know which two to multiply against eachother.

Answer 12

So, we start with 48 possible cases involving an intoxicated driver, out of 166:
\[
\frac{48}{166}
\]We have a mutually exclusive even, since, if we pick our initial case to be intoxicated, both numbers are lowered by unity, so we have, the probability of picking a second intoxicated driver is:
\[
\frac{47}{165}
\]These are mutually exclusive events, so we have:
\[
P(A\wedge B)=\frac{48}{166}\cdot\frac{47}{165}\approx0.082365
\]

Answer 13

So if the question asked if 3 different deaths were selected without replacement it would be
(48/166)*(47/165)*(46/164) =.023?

Answer 14

Yep.

Answer 15

I get that one! thank you. how about another one, similar situation

Answer 16

Sure

Answer 17

so this one has intercepting numbers, and it says WITH replacement, so the total number doesnt change. but when i tried to number it out, it's not the right answer.

Answer 18

So, what is the probability of picking a Group B?
\[
\frac{39}{249}
\]And, now that we have our Group B, what is the probability of picking \(\text{Rh}^+\) from within that group?
\[
\frac{37}{39}
\]Which means, for one person:
\[
\Pr(B \wedge \text{Rh}^+)=\frac{37}{39}\cdot\frac{39}{249}=\frac{37}{249}\approx 0.1485944
\]This means, since we are drawing with replacement, that drawing two people is:
\[
\left(\frac{37}{249}\right)^2\approx0.02208
\]Which is what we wished to find.

Answer 19

wait i see you have 37/39 * 39/249 = 37/249

Answer 20

why dont you use the 37/39?

Answer 21

also, how do you configure the second portion of the question?

Answer 22

For the latter case, drawing without replacement, we go by the same process, but compute each probability individually, and, instead of squaring the probability, we multiply the last two in the end.
As for the \(\frac{37}{39}\)... it is used. It's just when multiplied out it comes out to such.

Answer 23

The second portion? Well, we know the probability of picking two people, with replacement, is the same, so we simply multiply the result with itself.

Answer 24

hmm okay so i will attempt to see if this is right! thank you LolWolf!

Answer 25

have time for another? lol

Answer 26

I have one more

Answer 27

Sure, go for it.

Answer 28

need help with B and C

Answer 29

For B, both events are mutually exclusive, so:
\[
\Pr(A\wedge B)=\Pr(A)\cdot\Pr(B)=0.063\cdot0.063=0.003969
\]

Answer 30

okay that makes sense

Answer 31

For C, we have to compute the inverse (because the probability of multiplying both is actually interpreted as "what is the probability of both alarm clocks going off?)
We know:
\[
\Pr(\neg A)=1-\Pr(A)
\]So, we have that, the probability of both *not* sounding is:
\[
\Pr(A\wedge B)=0.003969
\]Therefore, the probability of at least ONE sounding is:
\[
\Pr(\neg(A\wedge B))=1-\Pr(A\wedge B)=1-0.003969=0.996031
\]For these, I recommend reading De Morgan's laws.

Answer 32

http://en.wikipedia.org/wiki/De_Morgan%27s_laws

Answer 33

thank you!! i'll check that out right now! thank you for all your help. you have been great!!!!

Answer 34

Sure thing