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(s_{1})=0.2; P(s_{2})=0.8*0.2 (doesn't leave in shop 1, leaves in shop 2)=0.16; P(s_{3})=0.8^{2}*0.2=0.128; P(s_{4})=0.8^{4}*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)=1-0.8^{4}=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(W_{1})=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 oneP(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.