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?

Question

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 · Answer

30!/(24!)  * 1/5

anonymous · Answer

How'd you get that elvis?

dumbcow · Answer

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

anonymous · Answer

My exam says that the answer is 1/21.  How do u get that?

dumbcow · Answer

oh sorry my P(AB) is wrong, it should be 1.2/30
$$\frac{\frac{1.2}{30}}{\frac{25.2}{30}} = \frac{1.2}{25.2} = \frac{1}{21}$$

anonymous · Answer

Thanks a lot.

dumbcow · Answer

yw

anonymous · Answer

How'd you get 1.2 and 25.2?  Still not quite clear on that.

dumbcow · Answer

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

anonymous · Answer

Thanks so much

anonymous · Answer

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.

dumbcow · Answer

$$48 \int\limits_.5^1 \int\limits_y^1 \int\limits_0^y (xyz) dx dz dy$$
$$= 48 \int\limits_.5^1 \int\limits_y^1 \frac{y^3 z}{2} dz dy$$
$$= 48 \int\limits_.5^1 (\frac{y^3}{4} - \frac{y^5}{4}) dy$$
$$= 12(\frac{y^4}{4} - \frac{y^6}{6}) |_.5^1$$
$$= 1 - \frac{10}{64} = \frac{27}{32}$$
= 0.84375

dumbcow · Answer

how did you go from pre-calc probability to multi-variable calculus?? lol

anonymous · Answer

Haha, don't know - guess i should've asked that in the calc section.  I was asked to find a probability though.

anonymous · Answer

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?

anonymous · Answer

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.