Thursday

October 2, 2014

October 2, 2014

Posted by **Jacob** on Sunday, March 6, 2011 at 6:08pm.

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 5-committee 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 out-number 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**, Sunday, March 6, 2011 at 7:15pm1a

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**, Sunday, March 6, 2011 at 7:25pm2a) (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 5-person committee.

I don't know question b.

- Math -
**MathMate**, Sunday, March 6, 2011 at 7:46pmC(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**, Sunday, March 6, 2011 at 8:10pmSo, for b it is (4,3)(4,4) + (4,2)(4,1)/(8,5)= 28/56= 1/2?

- Math -
**MathMate**, Sunday, March 6, 2011 at 9:30pmExactly!

- Math -
**Jacob**, Sunday, March 6, 2011 at 9:48pmIf 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**, Sunday, March 6, 2011 at 10:13pmIt's correct, but remember to divide the number by (20,10) to give the probability requested.

- Math -
**Jacob**, Sunday, March 6, 2011 at 10:24pmThanks! 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**, Sunday, March 6, 2011 at 10:43pmHave you done the inclusion-exclusion 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)=1-0.2=0.4+0.6-N(EG)

=> N(EG)=0.4+0.6-0.8=0.2

and

N(E~G)=0.4-0.2=0.2

N(G~E)=0.6-0.2=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.

- Math-correction -
**MathMate**, Sunday, March 6, 2011 at 10:47pmN(E∨G)=1-0.2=0.4+0.6-N(E∧G)

=> N(E∧G)=0.4+0.6-0.8=0.2

- Math -
**Jacob**, Sunday, March 6, 2011 at 10:56pmThank 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**, Sunday, March 6, 2011 at 11:29pmIt 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**, Monday, March 7, 2011 at 12:35amAh, ok thank you!

- Math :) -
**MathMate**, Monday, March 7, 2011 at 7:40am:)

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