Probability students in a room

[Louise Johnson]

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?

 

[galactus]

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:

                                      LEFT

                                 YES                             NO
-                   -----------------------------------------------------------
                YES               2                            4                       6
     DEAF        -----------------------------------------------------------
                 NO               8                            136                  144 
                   ------------------------------------------------------------
                                  10                             140                  150

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.

 

[mark07]

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:

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

 

[Louise Johnson]

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