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sleung
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?
How'd you get that elvis?
conditional 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
My exam says that the answer is 1/21. How do u get that?
oh sorry my P(AB) is wrong, it should be 1.2/30 \[\frac{\frac{1.2}{30}}{\frac{25.2}{30}} = \frac{1.2}{25.2} = \frac{1}{21}\]
How'd you get 1.2 and 25.2? Still not quite clear on that.
1.2 is the avg number of questions guessed correctly ---> 6*(1/5) 25.2 is avg total num of correct questions ---> 24 + 6/5
For 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.
\[48 \int\limits_.5^1 \int\limits_y^1 \int\limits_0^y (xyz) dx dz dy\] \[= 48 \int\limits_.5^1 \int\limits_y^1 \frac{y^3 z}{2} dz dy\] \[= 48 \int\limits_.5^1 (\frac{y^3}{4} - \frac{y^5}{4}) dy\] \[= 12(\frac{y^4}{4} - \frac{y^6}{6}) |_.5^1\] \[= 1 - \frac{10}{64} = \frac{27}{32}\] = 0.84375
how did you go from pre-calc probability to multi-variable calculus?? lol
Haha, don't know - guess i should've asked that in the calc section. I was asked to find a probability though.
Insurance 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?
A fair 6-sided 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.