Probabilities - questions

[Ibis]

So I'm taking a probability class and there are some homework that I can't figure out how I should figure them out, or if I've figured them out correctly (especially the third one). (And like it says in the rules, I'm not asking you to do my homework for me, I've already spent HOURS (yes, I'm that dumb) on these problems, so just some hints on if I got them right or not...)

First one:
In a restaurant, of the music that is played, F=38%, B=20%, H=14%, J=8%, P=4% and W=16%. Of the men that are present, 90% can dance to F, 60% to B, 70% to H, 50% to J, 5% to P and 80% to W. (let's say capital letters are for music style and small letters are for the dance style)
a) What's the probability that a randomly picked man can't dance to a randomly selected music style?
b) Let's assume that this man can dance to W. What is the probability that he can also dance P (assuming that "dance skills" are independent of each other)?

So do I understand this correctly if a) = P("a music style" AND ("what he can't dance")=1-P("a music style" AND "what he can dance")=1-(F*f+B*b+H*h+J*j+P*p+W*w)=1-(0.38*0.9+0.2*0.6+0.14*0.7+0.08*0.5+0.04*0.05+0.16*0.8)=1-0.73=0.27
And b) =P(P|W)=P(P AND W)=0.05*0.8=0.04

Second one:
A boy has a 20% chance of leaving his wallet in a shop (if it hasn't gone missing already). The boy one day visits shops 1,2,3 and 4 in this particular order.
a) What is the probability that he leaves his wallet in shop s_i, i=1,2,3,4?
b) What is the probability that he leaves his wallet in one of these shops?
c) After the boy gets home he notices that his wallet is missing. What is the probability that he has left it in shop s_i, i=1,2,3,4?

I'm not sure of the difference between a) and b). Do I think correctly if I think that
a) P(s_i)=0.2 for all i=1,2,3,4 as stated in the assignment? and then
c) would be P(s₁)=0.2; P(s₂)=0.8*0.2 (doesn't leave in shop 1, leaves in shop 2)=0.16; P(s₃)=0.8²*0.2=0.128; P(s₄)=0.8⁴*0.2=0.1024
b) I think b should be P(s1 OR s2 OR s3 OR s4)=1-P(didn't leave in any shops)=1-0.8⁴=0.5904

Third:
Power is obtained from two independent sources. One works with the probability of P(A)=0.9 and the other one with P(B)=0.95. If both sources work, power obtained is sufficient. If only one power source works, power is gained sufficiently with P(W₁)=0.8. If both power sources are broken, enough power isn't obtained.
a) What is the probability that i number of power sources work, i=0,1,2?
b) What is the probability of getting enough power?
c) If supplied power is sufficient, what is the probability that i number of power sources are functional, i=0,1,2?

So I marked down P(A)=0.90 -> P(A^c)=1-0.90=0.10, P(B)=0.95->P(B^c)=0.05.
a) P(0 power sources work)=P(A^c AND B^c)=0.10*0.05=0.005
not sure about this one P(1 power source works)=P(A works or B works)=P(A OR B)=0.9+0.95-0.9*0.95=0.995.
P(2 sources work)=P(both work)=P(A AND B)=0.9*0.95=0.855
(shouldn't P(0)+P(1)+P(2)=1?)
b) P(W)=P[(2 sources) OR (1 source)]=P[(A OR B) or (A AND B)|W]=0.855*1+0.14*0.8=0.967
c)P(0|W)=impossible, P(1|W)=P(1 AND W)/P(W)=0.995*0.80/0.967=0.823, P(2|W)=P(2 AND W)/P(W)=P(2)/P(W)=0.855/0.967=0.884
[shouldn't P(1|W)+P(2|W)=1?]
In the third problem I do get the alternative answers of b) P(1 works)=0.14 and c) P(1|W)=0.116 but those probabilities seem low even if they do satisfy the sum=1 rule. And I don't understand why P(1)=P(A OR B)=P(A)+P(B)-P(A AND B) isn't the same as P((A^c and B) OR (A and B^c))?

In some assignments I feel like I can think of 6 different ways to calculate each result and each result gives a different value. :/ So I'd appreciate if anyone can tell me if these are correct and if they aren't, if you could hint where my logic is flawed. Thank you.

[Lilcheeks]

Its been 5 years since I finished my math degree so I'm rusty, but...

I read question 2 the way you did and came up with the same answers.

On question 3, I'm wondering if you need to calculate the P(A*not B) or P(Not A * B) for only 1 being on. It should be a low prob I think since both have a high chance to be on. I got .14 but I may be wrong.

Based on that if I'm not off, B) would be .855 + .14 = .995

C) It looks like you subbed in wrong values for the conditional probability problems "P(1 AND W)/P(W)=0.995*0.80/0.967=0.823" the denominator P(W)=.8 as stated in the initial question.

[Ibis]

Thank you for your response!

C) It looks like you subbed in wrong values for the conditional probability problems "P(1 AND W)/P(W)=0.995*0.80/0.967=0.823" the denominator P(W)=.8 as stated in the initial question.

That's my problem with the conditional probability formula P(B|A)=P(B AND A)/P(A)=P(B)*P(A)/P(A). Doesn't it simplify to just P(B) in all cases and make that formula redundant for calculating P(B|A)?

My reasoning for the P(1|W)= 0.995*0.8/0.967, or actually now =0.14*0.8/0.967 was that P(W)=0.8 when only one is working, and P(W_overall)=0.967 as calculated in part b). Is this logic flawed? So actually P(1|W)=P(1 AND P(W))/P(W_overall)(?).

Oh and I noticed that your B) was 0.995. But shouldn't the P(W|1)=0.8 in case only one power supply is functional be taken into account, so that P(W_overall)=P(2)+P(1)*P(W)

I think my marks are confusing as well, don't know what I should call what, but isn't P(W|1)=0.8 the fact given in the initial question and I then need to get P(1|W).

[Lilcheeks]

I just did a quick search on cond prob since its been awhile. The confusion is that P(A|B)= P(A OR B)/P(B) not A and B. Hope that helps.

http://en.wikipedia.org/wiki/Conditional_probability


3.B isn't conditional, at least not the way I read it. I'm reading it as the probability of 1 or 2 sources. I may be wrong though. Since you calculated the probability of 1 source, and calculated the prob of 2 sources in A, its just a matter of adding those 2.

[Ibis]

The way I read it, isn't your answer for 3.B P("1 or 2 work") and not for P(W|"1 or 2") (so probability for one or two power sources working vs. getting enough power when 1 or 2 work)? (and "1 or 2" because if neither work there's no power at all, otherwise it would be "0,1 or 2")

let's mark the upside-down U (= joint, don't know its name? upside-down-union) as m. Doesn't wikipedia (and my lecture "handout") say
P(A|B)=P(A m B)/P(B), not P(A U B)/P(B)? :]

[Lilcheeks]

I'm starting to get confused. My brain doesn't work as well as it used to haha. I'll go back and look...

[Ibis]

I think I now understood why P(A U B)=P(A)+P(B)-P(A U B) doesn't give the correct answer in 3.B. It's not correct because it means A or B or both, not just A or B. So that's clear now. (and really... duh! and I've used the correct logic in another question as well but didn't get it until now)

[Deleted]

First is ok.
2.
b) 20%, same situation as coin toss
a)1=20%, 2=0,8*0,2*100% (16%); 3=0,8*0,84*0,2*100% (13,4%) ; 4= 0,8*0,84*0,866*0,2*100% (i may be totally wrong)
c) isint it same as a) ?

@edit
Damn, my head ! i read your post againn and now it seems that my is sooo wrong but not gonna change it

[Ibis]

@edit
Damn, my head ! i read your post againn and now it seems that my is sooo wrong but not gonna change it

Haha, I know the feeling

[Sunshine]

Your answer for 2 is wrong, because you need to look at the conditional probability.

Royalwolf's second answer is right for A (looking at p(1) = 0.2, p(2) = 0.8*0.2, etc). (Edit: that's what you had for C, yeah.)

I'm pretty sure that B is asking for the total chance that he leaves his wallet in any of the shops (vs not losing it at all).

C is similar, except you know that there's a 100% chance that he's lost his wallet, so there's no "ok, he makes it through all of them and doesn't lose it" option.



(And you may want to take this with a grain of salt, because I slept about an hour last night. )

[Ibis]

Yes, but if A is what I had for C then what is C, I don't really get the exact question I guess, because if A needs to take the conditional probability into account then the difference between A and C is lost on me. :/

---------- Post added 2011-09-15 at 02:56 PM ----------

Time for correct answers!

First one:
In a restaurant, of the music that is played, F=38%, B=20%, H=14%, J=8%, P=4% and W=16%. Of the men that are present, 90% can dance to F, 60% to B, 70% to H, 50% to J, 5% to P and 80% to W. (let's say capital letters are for music style and small letters are for the dance style)
a) What's the probability that a randomly picked man can't dance to a randomly selected music style?
b) Let's assume that this man can dance to W. What is the probability that he can also dance P (assuming that "dance skills" are independent of each other)?

My result for a 1.A was correct. In 1.B I marked correctly that it is the same as P(P|W) but my solution for that was missing the /P(A)-part, so in fact P(P|W) was the same as P(P)*P(W)/P(W)=P(P)=0.05.

Second one:
A boy has a 20% chance of leaving his wallet in a shop (if it hasn't gone missing already). The boy one day visits shops 1,2,3 and 4 in this particular order.
a) What is the probability that he leaves his wallet in shop s_i, i=1,2,3,4?
b) What is the probability that he leaves his wallet in one of these shops?
c) After the boy gets home he notices that his wallet is missing. What is the probability that he has left it in shop s_i, i=1,2,3,4?

2.A was as people here knew correctly, what I had for C. So (s_i=0.8^i-1*0.2, i=1,2,3,4=whatever).
2.B. was correct, 0.59
2.C was the conditional probability for P(s_i|lost his wallet in one of the shops (L)) where the P(has lost his wallet) was the result of 2.B, so it was P(s_i|L)=P(Vi AND L)/P(L)=0.2*0.8^i-1/0.59 (i=1,2,3,4)=....

Third:
Power is obtained from two independent sources. One works with the probability of P(A)=0.9 and the other one with P(B)=0.95. If both sources work, power obtained is sufficient. If only one power source works, power is gained sufficiently with P(W₁)=0.8. If both power sources are broken, enough power isn't obtained.
a) What is the probability that i number of power sources work, i=0,1,2?
b) What is the probability of getting enough power?
c) If supplied power is sufficient, what is the probability that i number of power sources are functional, i=0,1,2?

This I had done correctly.

Next week: probabilities that smell of statistics! (lol, joking... unless I'm not!)