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anonymous
 2 years 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?
anonymous
 2 years 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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anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0How'd you get that elvis?

dumbcow
 2 years 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

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

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

anonymous
 2 years 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
 2 years 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

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

anonymous
 2 years 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.

anonymous
 2 years 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?

anonymous
 2 years 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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