Probability students in a room

Question:
Out of 150 students in a room,10 are left-handed and six are deaf.Two of these students are both left-handed and deaf. If a student in this room is chosen at random find the probability that:

a) the student is neither deaf nor left handed
my answer:

\(\displaystyle \L\\\frac{{10}}{{150}} + \frac{6}{{150}} - \frac{2}{{150}} = \frac{7}{{75}} = .093\)

I drew a venn diagram and was wondering if maybe it was correct to use something like C(136,1)/C(150,1) ?

b) The student is not deaf,given that the student is not left-handed.

I am completely lost with this question. What is it asking for is what I need to know? Is it just me or is this a difficult question to interpret?

Your first problem is OK, except subtract from 1. What you have is the probability of choosing someone deaf or left handed.


For the second one, try drawing a chart:



Now, you want not deaf, given they're not left handed.

Go across the not deaf row and down the not lefthanded column.

You get 136/140=34/35

Sorry for the discombobulated display. It does at it pleases.

a) Let A be the set of left-handed students and let B be the set of deaf students. Denote the number of elements in these sets by n(A) and n(B). We are given n(A)=10, n(B)=6, and \(\displaystyle n(A\cap B)=2\). Then

\(\displaystyle \L n(A\cup B) = n(A)+n(B)-n(A\cap B) = 10+6-2 = 14.\)

Probability of a randomly selected student to be out of AUB is

\(\displaystyle \L P((A \cup B)^c)\)

which is the number of students out of AUB divided by total number: (150-14)/150 = 136/150 = 0.91 (approximately).

C(136,1)/C(150,1) = 136/150 as well, so you can do it either way.

b) It's a conditional probability: \(\displaystyle P( B^c | A^c )=?\)

Use the following:

\(\displaystyle \L P(B^c | A^c) = \frac{P(B^c \cap A^c)}{P(A^c)}
= \frac{P((B \cup A)^c)}{P(A^c)} = ?\)

Thank you both for your replies. I have to admit that the latter post by mark I have some difficulty understanding. I think it is most likely my lack of notation knowledge. I thank you all the same as I really look forward to all replies and anything that can help me progress.
Thank you
Louise