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 11 months ago
A student who is taking a 30 question multiple choice test knows 24 of the answers. If the student doesn't know the answer, he chooses uniformly from 1 of 5 choices. Given that he gets a randomly chosen question right, what is the probability he guessed on the question?
 11 months ago
A student who is taking a 30 question multiple choice test knows 24 of the answers. If the student doesn't know the answer, he chooses uniformly from 1 of 5 choices. Given that he gets a randomly chosen question right, what is the probability he guessed on the question?

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sleung
 11 months ago
Best ResponseYou've already chosen the best response.0How'd you get that elvis?

dumbcow
 11 months ago
Best ResponseYou've already chosen the best response.1conditional probability \[P(A  B) = \frac{P(A and B)}{P(B)}\] probability he guesses and gets it right = 1/5 probability the question is right = (24+6/5)/30

sleung
 11 months ago
Best ResponseYou've already chosen the best response.0My exam says that the answer is 1/21. How do u get that?

dumbcow
 11 months ago
Best ResponseYou've already chosen the best response.1oh sorry my P(AB) is wrong, it should be 1.2/30 \[\frac{\frac{1.2}{30}}{\frac{25.2}{30}} = \frac{1.2}{25.2} = \frac{1}{21}\]

sleung
 11 months ago
Best ResponseYou've already chosen the best response.0How'd you get 1.2 and 25.2? Still not quite clear on that.

dumbcow
 11 months ago
Best ResponseYou've already chosen the best response.11.2 is the avg number of questions guessed correctly > 6*(1/5) 25.2 is avg total num of correct questions > 24 + 6/5

sleung
 11 months ago
Best ResponseYou've already chosen the best response.0For 0<x<y<z<1, the joint density of (X,Y,Z) is given by f(x,y,z)=48xyz. Find P(Y>1/2). I got .78, but my exam says it's .84.

dumbcow
 11 months ago
Best ResponseYou've already chosen the best response.1\[48 \int\limits_.5^1 \int\limits_y^1 \int\limits_0^y (xyz) dx dz dy\] \[= 48 \int\limits_.5^1 \int\limits_y^1 \frac{y^3 z}{2} dz dy\] \[= 48 \int\limits_.5^1 (\frac{y^3}{4}  \frac{y^5}{4}) dy\] \[= 12(\frac{y^4}{4}  \frac{y^6}{6}) _.5^1\] \[= 1  \frac{10}{64} = \frac{27}{32}\] = 0.84375

dumbcow
 11 months ago
Best ResponseYou've already chosen the best response.1how did you go from precalc probability to multivariable calculus?? lol

sleung
 11 months ago
Best ResponseYou've already chosen the best response.0Haha, don't know  guess i should've asked that in the calc section. I was asked to find a probability though.

sleung
 11 months ago
Best ResponseYou've already chosen the best response.0Insurance losses L in a given year have a lognormal distribution with L=e^X, where X is a normal random variable with mean 3.9 and standard deviation 0.8. If a $100 deductible and a $50 benefit are imposed, what is the probability the insurance company will pay the benefit limit given that a loss exceeds the deductible?

sleung
 11 months ago
Best ResponseYou've already chosen the best response.0A fair 6sided die is rolled 1,000 times. Using a normal approximation with a continuity correction, what is the probability the number of 3's rolled is greater than 150 and less than 180? I'm supposed to get .78, but instead I'm getting .81.
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