GMAT Pre Questions ---Probability

[maypoon]

Dear all,

Please help me to deal with the following questions related to "Probabilty".
Thank you so much!


Q1. From a group of 3 boys & 4 girls, 4 children are to be randomly selected. What's the probabilty that equal numbers of boys & grils will be selected?

A. 1/10
B. 4/9
C. 1/2
D. 3/5
E. 2/3

Ans D


Q.2. There're 11 women & 9 men in a certain club. If the club is to select a committe of 2 women & 2 men, how many different such committees are possible?

A. 120
B. 720
C. 1060
D.1250
E.1980

Q3. A shipment of 8 television sets contains 2 black -and white sets & 6 colors sets . If 2 television sets are to be chosen at random from this shipment , what 's the probabilty that at least 1 of the 2 sets chosen will be black -and -white sets?

A. 1/7
B. 1/4
C.5/14
D.11/28
E.13/26

Ans. E


Best Wishes,
May

[earth@work]

Q1.
Soln: Choosing 2 boys out of 3 boys=3C2=3!/2! = 3 ways
Choosing 2 girls out of 4 girls = 4C2=6 ways
Total Favourable ways = 3*6=18
Now, total ways = 7C4=7!/4!*3! = 35
P(2B& 2G) = 18/35 ..... my answer does not match ans D

Q.2.
Soln: Choosing 2 women out of 11 = 11C2 = 55
choosing 2 men out of 9 =9C2 = 36
total no . committees = 55*36 = 1980 - Ans E

Q3.
Prob(atleast 1 B/W tv) = 1- Prob(NO b/w tv)
Prob of no b/w tv = prob of both Colour tv
P(colour tv) = 6C2/8C2 =15/28
P(atleast one b/w tv) = 1-15/28 = 13/28 (this is again not matching your answer choice E 13/26 )
COuld u pls confirm the answers for 1 & 3

[betamax]

I got the same.
1. 18/35
2. 1980
3. 13/28
I would like to see the solutions to these.

[maypoon]

Dear all,

I also have no idea about the ans of Q1 & Q3.

The ans of mine of Q3 is also 13/28.

And for the Q1, I have totally no idea !!

Anyway, thank you so much for your help!

If I find out the solution..I will post it later !!

Best Wishes,
May

[BuckeyeT]

May,

I'm getting the same 18/35, albeit by a slightly different method.

Q1: Probability of equal boys and girls on 4 picks. First, how many possible selection choices are there? Write out each possible choice possibility that meets the requirement:
C1 = B,B,G,G
C2 = B,G,B,G
C3 = B,G,G,B
C4 = G,B,G,B
C5 = G,B,B,G
C6 = G,G,B,B

So to determine the probability of these choices, we need to determine the probability of any choice type and multiply by 6.

C1 = B,B,G,G
P(C1) = P(1st pick boy)*P(2nd pick boy)*P(3rd pick girl)*P(4th pick girl)
P(C1) = (3/7)*(2/6)*(4/5)*(3/4) = 3/35

P(C1) * 6 = (3/35)*6 = 18/35

[ashishsj]

The actual question is:

From a group of 3 boys & 3 girls, 4 children are to be randomly selected. What's the probabilty that equal numbers of boys & grils will be selected?

Ans: 3/5