Solutions for Practice Final
Katherine
* denotes questions that David went over in review so the answer is 99.999% correct Let me know what you got differently! Comment if you want a detailed description on how I got the answer C1: 4*5*3 C2: 8!/ (2!*3!) C3: 12 Choose 10 C4: 11/100 C5: 2/9 C6: [(3/10)*(2/9) + (7/10)*(3/10)] C7: [(20 choose 12)* (7/10)^12 * (3/10)^8] C8*: .1 < 1/[(.1^2) (4) (100,000)] C9*: [(7/8)^10]* [(1/8)^1] C9 is tricky because 10 answered correctly means that there are 11 times that you need to play the game (10 answered correctly + 1 answered incorrectly to stop the game) C10: 1/ (1/8)=8 C11: c=1 C12*: P(x<10) = 1-e^-2 Lambda is found because E[X]=1/Lambda 5=1/Lambda Lambda = 1/5 t=10 days
Katherine: March 19, 2015, 1:21 p.m.
A1: 5^3 This works because this takes into consideration the case where all are repeated, none are repeated, and two are repeated
Katherine: March 19, 2015, 1:30 p.m.
A8*: [1-(5/6)^6]
Katherine: March 19, 2015, 1:34 p.m.
A9: E[X]= E[XI1]p(1)+E[XI2]p(2)+E[XI3]p(3)+E[XI4]p(4)+E[XI5]p(5)+E[XI6]p(6)=(1/6)[(6/1)+(6/2)+(6/3)+(6/4)+(6/5)+(6/6)]
Katherine: March 19, 2015, 1:35 p.m.
A10*: n>6945
Katherine: March 19, 2015, 1:41 p.m.
z=max(X,Y) p(z<x)=p(max(x,y)<x=p(X<x)p(Y<x) f(x)=uniformally distributed=(x/2) when 0<x<2 f(y)=exponentially distributed and lambda is 3 = 1-e^-3t when t>0 Multiply the two functions together THEN differentiate to get to the Cumulative Mass Function
Katherine: March 19, 2015, 1:48 p.m.
A12*: When you see the words EXACTLY one decays in the first hour and EXACTLY two decay in the second hour means that it is binomial P(x<3)=1/3 Lambda = [ln (3/2)]/3 {(8 choose 1)(Probability that it doesn't decay)^7(Probability that it does decay within 1 hour)^1}*{(7 choose 2) (probability that it does decay within the hour)^2 (Probability that it doesn't decay within the hour)^5}
PeterH: March 19, 2015, 2:20 p.m.
I think C10 is actually 7. 8 describes the number of trials including one wrong answer. So if you subtract that one wrong answer from 8, you get 7.
PeterH: March 19, 2015, 2:44 p.m.
I feel like A1 can't be as easy as 3 generations of 5 choices. But I don't know how to do it otherwise.
betht123: March 19, 2015, 2:50 p.m.
Why can't we use the hand method to solve A1?
dewfallen: March 19, 2015, 3:55 p.m.
I feel that for A11 if you differentiated it after you multiplied the two CDF's together you would get a PDF instead of a CDF for Z.
dewfallen: March 19, 2015, 3:55 p.m.
I feel that for A11 if you differentiated it after you multiplied the two CDF's together you would get a PDF instead of a CDF for Z.
Lilyhui: March 19, 2015, 5:24 p.m.
A1 is actually just 5^3 and you dont have to put a 3! because the ice cream is stacked top, middle, and bottom so already distinguished
frokrab: March 19, 2015, 9:57 p.m.
Do you need the (20 choose 12) for C7? Does the ordering matter?
frokrab: March 19, 2015, 10 p.m.
Also for C10 wouldn't 8 be the question that you get wrong, so you're expected to get only 7 right?
Lilyhui: March 19, 2015, 11:57 p.m.
Yes, you need 20Choose12 for c7
Lilyhui: March 19, 2015, 11:59 p.m.
20 Choose 12 means you choose 12 from the 20 for the red ball, not that the order matters (then it would be 20 pick 12)
Lilyhui: March 20, 2015, midnight
yup c10 is 7
weisbart: March 20, 2015, 12:54 a.m.
PeterH is right with C10. While you have 8 trials, you have only 7 successes.
weisbart: March 20, 2015, 12:59 a.m.
A1 is $5^3$. I made up the problem to be difficult, found a difficult solution, and only realized there was an easier solution. There's always something new to learn!
adikantawala: March 20, 2015, 11 a.m.
I got (3/10) + (2/9) for c6. Is this right? I don't see why it would be C6: [(3/10)*(2/9) + (7/10)*(3/10)].