Posted by Jacob on .
Find the probability of:
a) Rolling a three OR a six with a single die
Answer:
P(3 U 6)= n(3U6)/n(S)
= 3+6/6
= 9/6 reduced: 3/
b) Dealing an ACE OR a FACE CARD from a standard deck
Answer:
P(A U F)= 4/52 + 12/52 = 4/13
2. A 5committee is to be chosen from 4 teachers and 4 students. What is the probability that
a) there is only one teacher on the committee?
b) the students outnumber the teachers on the committee?
Is question correct? I am not sure...
As for question 2, can someone help me how to do it?
Thank you!

Math 
MathMate,
1a
3∪6 represent 2 successful event out of 6 possible outcomes. If the numbers chosen were 2 and 4 instead of 3 and 6, the probability would remain the same (2 out of 6).
1b.
Correct.
2. Committee of 5 out of 8 persons:
There are C(5,8)=8!/(5!3!)=56 ways to choose 5 persons out of 8.
Possible cases:
1 teacher in C(4,1) ways and 4 students in C(4,4) ways.
Total=C(4,1)*C(4,4)=4*1=4 ways
2 teachers in C(4,2)=6 ways and 3 students in C(4,3)=4 ways.
Total = 6*4=24
3 teachers in C(4,3)=4 ways and 2 students in C(4,2)=6 ways.
Total = 4*6=24
4 teachers in C(4,4)=1 way and 1 student in C(4,1)=4 ways.
Total = 4
Grand total = 4+24+24+4=56 as before.
Now can you put together an answer for 2a and 2b? 
Math 
Jacob,
2a) (4,1)x(4,4)/ (8,5)=4/56
For 2a, why is student 4,4? I don't understand that. I know that for 1 teacher, there is 4 student because there is 5person committee.
I don't know question b. 
Math 
MathMate,
C(4,1) for teachers because we are taking only one teach out of 4 (4 choose 1).
C(4,4) for students because we are choosing 4 students out of 4 (4 choose 4) and there is only one way to choose ALL the students.
Students outnumber teachers when there are 3 students or more in the committee (verses 2 teachers or less). Cases 1 and 2 represent this situation, and you can calculate the probability accordingly. 
Math 
Jacob,
So, for b it is (4,3)(4,4) + (4,2)(4,1)/(8,5)= 28/56= 1/2?

Math 
MathMate,
Exactly!

Math 
Jacob,
If possible, can you help me with this question also?
From a group of 10 girls and 10 boys, 10 will be chosen to form a committee. Find the probability that atleast ONE GIRL is chosen.
I did it like this
At least 1 girl so therefore I separated more than 1.

0 1 2 3 4 5 6 7 8 9 10
n(Zero girl)= (10,0)x(10,10)
= 1
n(atleast 1 girl) = # total possible  (zero)
=(20, 10)  (10,10)
=184,755
Is this correct?
Thank you for helping me! 
Math 
MathMate,
It's correct, but remember to divide the number by (20,10) to give the probability requested.

Math 
Jacob,
Thanks! T___T another question (Last one I promise, lol)
The probability of a household subscribing to The Economist is 0.4 while the probability of subscribing to National Geographic is 0.6, and the probability of subscribing to NEITHER magazine is 0.2. What is the probability that a randomly selected household subscribes to:
a) Either magazine?
b) both magazines?
I have no clue what formula to use this one...I checked my textbook but I don't know 
Math 
MathMate,
Have you done the inclusionexclusion principle?
If you have, then it is:
N=1
N(E)=0.4, N(~E)=0.6
N(G)=0.6, N(~G)=0.4
N(EG)=10.2=0.4+0.6N(EG)
=> N(EG)=0.4+0.60.8=0.2
and
N(E~G)=0.40.2=0.2
N(G~E)=0.60.2=0.4
If not, it is relatively simple to draw a Venn Diagram and figure it out by filling in the numbers, and get the same results. 
Mathcorrection 
MathMate,
N(E∨G)=10.2=0.4+0.6N(E∧G)
=> N(E∧G)=0.4+0.60.8=0.2 
Math 
Jacob,
Thank you! I think I did this before but I forgot...
How do you know whether or not to use those formulas in these kind of questions? rearrange formulas?
P(A U B) = P(A) + P(B)  P(A and B)
and
P(A U B) = P(A) + P(B) 
Math 
MathMate,
It all depends if you suspect/eliminate the possibility of A∩B being null or not. In this case, we know that it is not null, since the sum
P(A)+P(B)+P(A'∩B')
add up to more than 1.
And we know that
P(A)+P(B)P(A∩B)+P(A'∩B')=1 
Math 
Jacob,
Ah, ok thank you!

Math :) 
MathMate,
:)