## tomo 2 years ago A multiple-choice test has five possible answers for each questions. On any given question a student has probability .75 to know the right answer and probability .25 to have to guess. Since there are five possible answers, the probability he guesses correctly is .25. If a student answered the question correctly, what is the probability that he guessed?

1. tomo

give me one second to show my work

2. tomo

P(knows correct) = .75 P(guess) = .25 Is the probability he guesses correctly .25 or is it 1/5 = .2? I believe that i am looking for P(guess given that it is correct) ??

3. satellite73

you are looking for a conditional probability

4. satellite73

and i believe you need baye's formula for this one

5. satellite73

if you put $$A$$ as the probability he gets the answer right, and $$B$$ as the probability he guessed, you are looking for $P(A|B)$ which you compute via $P(A|B)=\frac{P(A\cap B)}{P(B)}$

6. satellite73

ok that was wrong, you are looking for $$P(B|A)$$

7. tomo

yes, P(guess given that he answers correctly)

8. satellite73

using the sets i wrote above, you compute $P(B|A)=\frac{P(A\cap B)}{P(A)}$

9. tomo

so, can we say that P(guess AND correct) = P(guess) * P(correct)

10. satellite73

11. satellite73

oh wait no it isn't sorry

12. tomo

that is assuming the two are independent right

13. satellite73

because they are not independent

14. tomo

so do i have to use the extended bayes formula

15. satellite73

you compute $P(A\cap B)=P(B)\times P(A|B)$ yes, it is baye's formula you need

16. satellite73

both of those are easy to compute, $$P(B)=.25$$ and $$P(A|B)=.2$$ i think

17. satellite73

this line from your question . Since there are five possible answers, the probability he guesses correctly is .25. is wrong i think

18. satellite73

if there are 5 possible answers, and he guesses, the probability he guesses right should be $$.2$$ not $$.25$$

19. tomo

I'm confused. Is P(B AND A) the same as P(A AND B)?

20. satellite73

i think you should get for the numerator $$.25\times .2$$ and for the denominator $.25\times .2+.75\times 1$

21. satellite73

yes, they are the same $$P(A\cap B)=P(B\cap A)$$

22. satellite73

you know how to compute $$P(A)$$ the probability he guesses, and $$P(B|A)$$ the probability he gets it right given that he guesses

23. tomo

where do i find the AND/intersection statement in the equation editor? i can't find it and it would help me explain my thinking.

24. satellite73

i used \cap

25. tomo

thanks, ok I'm confused because I thought that changing the order of B and A in bayes formula changed the B and A in the intersection. $P(B|A) = P(B \cap A)/P(A)$ and $P(A|B) = P(A \cap B)/P(B)$

26. satellite73

let me get to a desktop, it is hard to write this on a lap top

27. satellite73

ok that's better

28. satellite73

both the formulas you wrote above are correct, and the numerators are both the same the point of baye's is that you can use the conditional probability you know (the easy one) to compute the conditional probability you want (the hard one) at the moment, we don't know $$P(A\cap B)$$ so we have to use a conditional probability to find it

29. satellite73

if $$A$$ is you get the right answer, and $$B$$ is you guess, we can know both $$P(B)=.25$$ because it is given to us, and we can compute $$P(A|B)$$ by thinking so we can compute $$P(A\cap B)=P(B)\times P(A|B)$$

30. satellite73

$$P(A|B)$$ means you guess, and you get the answer correct. since one out of the 5 answers is correct, if you guess you have a one in five (aka .2\) chance of getting it right, so $$P(A|B)=.2$$ and therefore $P(A\cap B)=P(B)P(A|B)=.25\times .2$

31. satellite73

we cannot compute $$P(B|A)$$ because it is not obvious, it is what we are looking for so we have to use the one above

32. satellite73

now we need $$P(A)$$ the denominator.

33. satellite73

btw after we do this the hard way i will show you a very quick snappy way to do it

34. tomo

ok, give me one second to compute the answer.

35. satellite73

ok

36. satellite73

then we will do it in our heads with almost no computation at all

37. tomo

P(B|A) = .05/.75 = .0666

38. satellite73

that is not what i get

39. satellite73

your denominator is wrong it should be $$P(A)=.75+.25\times .2=.8$$

40. tomo

oh yeah, I forgot to add the probability of guessing correctly

41. satellite73

$$A$$ is he gets it right, but he gets it right more than 75% of the time he gets it right if he knows it and also if he guesses correctly

42. satellite73

ok now lets do it the snap way

43. tomo

ok

44. satellite73

he answers 100 questions, and he gets 75 right because he knows them and of the remaining 25 he gets 5 right

45. satellite73

therefore he gets 80 right total, and of those 80 he guessed on 5 your answer is therefore $$\frac{5}{80}$$

46. satellite73

wasn't that easy?

47. tomo

interesting. I didn't think of that.

48. satellite73

baye's formula becomes much less mysterious when you use numbers and think

49. tomo

thank you very much for all the help. I really appreciate it. Sometimes this probability can be quite confusing.

50. satellite73

easier than saying $P(B|A)=\frac{P(B)P(A|B)}{P(B)P(A|B)+P(B^c)P(A|B^c)}$

51. satellite73

yw

52. satellite73

hint, on an exam if you get confused use numbers make them big

53. tomo

thanks for the tip.

54. satellite73

yw

55. tomo

A little way up, you said that $P(A|B) = .2$ I was thinking about that and tried doing a probability tree. |dw:1358740038530:dw| why isn't it .25*.2. (.25 for guessing and .2 for guessing it correctly.)? Does $P(A|B)$ mean probability of B given that we know A has occurred? Now that be occurs what is the probability of A?

56. tomo

be should be 'B'

57. tomo

right after P(A|B) i messed up. I mean "Does P(A|B) mean probability of A given that we know B has occurred? Now that B occurs what is the probability of A?"