Question

Can someone please answer these two questions by Friday?

Problem 1 An urn contains 10 red and 5 black and 3 green balls. Two balls are picked (one after the other) from the urn. Given that the second ball is red, find the probability that the first was green.

Problem 2 Charles is taking a multiple choice probability exam, and for each question, there are 3 possible answers, out of which only one is correct. Since the time is short, for any question, with probability 3/4 Charles decides to do the calculations and with probability 1/4 he just choses one of the 3 answers randomly. Whenever he decides to do the calculations, with probability 4/5 he will get the correct answer and with probability 1/5, he gets an answer which matches one of the 2 wrong answers. Suppose that he got a particular question wrong. What is the probability he actually did the calculations?

Answer 1

P(2nd ball was red) =P(1st green and 2nd red)+P(1st red and 2nd red)+P(1st black and 2nd red)

=(3/18)*(10/17)+(10/18)*(9/17)+(5/18)*(10/17)=5/9

P( first was green given 2nd ball was red )=P(1st green and 2nd red)/P(2nd ball was red)

=(3/18)*(10/17)/(5/9)=3/17

2)

P(got it right) =P(do the calculation and got it right)+P(not do the calculation and do it right)

=(3/4)*(4/5)+(1/4)*(1/3)=41/60

therefore P( did the calculations |got it right)

=P(do the calculation and got it right)/P(got it right)

=(3/4)*(4/5)/(41/60)

=36/41