A hotel room has a night table with one drawer, a dresser with two drawers, and a bureau with 4 drawers. A customer who stayed in the room remembers leaving his phone in one of the drawers with a 50% chance it was in the night table, a 40% chance it was in the dresser, and a 10% chance it was in the bureau – but with each drawer within the dresser or bureau being equally likely.
b) If someone goes into the room and picks one of the seven drawers randomly, what is the chance they will open the correct drawer and find the phone?
c) If someone goes into the room and picks one of the seven drawers randomly and finds the phone, what is the chance they had opened a drawer in the dresser?
The following is what I did
Part b:
P(Finding the phone) = 0.5*(1/7) + 0.4*(2/7) + 0.1*(4/7) = 1.7/7
Part c:
P(Dresser | Finding Phone) = 0.4*(2/7)/(1.7/7) = 8/17
My professor explained that part b should be 1/7 since all the drawers are equally likely. I'm not sure I understand this.
Can anyone here explain why I am wrong?
In part (b) you should keep in mind that probability of phone in a particular drawer of dresser is 20% not 40%. And in a phone in a particular drawer of bearau is 2.5% not ten percent .
P(Finding the phone) $\not=$ 0.5*(1/7) + 0.4*(2/7) + 0.1*(4/7)
But equals to $0.5\frac{1}{7}\ + 0.1\frac{2}{7}\ + 0.025\frac{4}{7}\ =\ \frac{1}{7}$
The explanation of your teacher if it were the exact words as you said was incorrect .