Out of a classroom of 6 boys and 4 girls the teacher picks a president for the student board, a vice president and a secretary.

What is the probability that only girls will be elected?

What is the probability that only boys will be elected?

What is the probability that at least one girl will be elected?

1) 4/10 * 3/9 * 2/8 = 2/5 * 1/3 * 1/4 = 2/60 = 1/30

2) 6/10 * 5/9 * 4/8 = 3/5 * 5/9 * 1/2 = 3/18 = 1/6

3) 1 - P(all boys selected) = 1- (1/6) = 5/6

I must be stupid. I was wondering how I got it wrong so I went back to my numbers and somehow I multiplied by 7 when I had 8:
4/10)(3/9)(2/8 but mutliplied by 7) = 4/105 ... WTF?

Is third right, though? 1-all boys possibility?

4/10*3/9*2/8 24/720 --> 12/360 --> 6/180 --> 3/90 --> 1/30

6/10*5/9*4/8 --> 120/720--> 12/72 --> 1/6

1-Probability that none are girls: --> just use the above 1-6/10*5/9*4/8 = 5/6.

1 - p(no girls selected) is the correct strategy, p(no girls selected) isn't as simple as 6/10

6/10 is the probability of one boy to be "pulled out of the hat", the next boy is 5/9 and so forth

p(no girls selected) is therefore p(all boys selected) = 6C3/10C3 = 1/6
as shown above

1.4C3/10C3
2. 6C3/10C3
3. 1-6C3/10C3
_________________

http://gmatclub.com/forum/countdown-beginshas-ended-85483-40.html#p649902

I have another opinion on the third question. Wonder, if it is right?

1) I got 1/30 too
2) got 1/6 as well

But, in my opinion, the third question should be solved this way:
the question asks what is the probability of AT LEAST one girl will be selected.

we have three mutually exclusive possibilities to select (they are all satisfy the asked condition):
GBB (girl,boy,boy)
GGB (girl,girl,boy)
GGG (girl,girl,girl)

the probability of GGG is already known - 1/30.
the probability of GGB is: 4/10*3/9*6/8 - 1/10.
the probability of GBB is: 4/10*6/9*5/8 - 1/6.

As these are mutually exclusive, we shall add them together: 1/30+1/10+1/6 = 3/10.

What do you think of this?

ANS.......
u r right but only mistake u r doing is that u have to multiply two of possiblities by 3....reason:-
possiblities of taking two girls is GGB,GBG,BGG ie 1/10*3=3/10.....
possiblities of taking two girls is GBB,BGB,BBG ie 1/6*3=1/2....
As these are mutually exclusive, we shall add them together: 1/30+3/10+1/2 = 5/6.
_________________

1) Absolute modulus : http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
2)Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html
3) effects of arithmetic operations : https://gmatclub.com/forum/effects-of-arithmetic-operations-on-fractions-269413.html


GMAT online Tutor

Yup, this is a permutation problem since three people will be selected for three different posts.

A) 4P3/10P3

B) 6P3/10P3

C) 1- (6P3/10P3)

IMO.. it shouldn't matter whether it is calculated with combinations or permutations. Reason being, if you consider the order in choosing the girls (numerator), you do the same in the denominator too. Hence it balances it out...

the results are coming out to be the same....

Cheers!
JT
_________________

Cheers!
JT...........
If u like my post..... payback in Kudos!!

|Do not post questions with OA|Please underline your SC questions while posting|Try posting the explanation along with your answer choice|
|For CR refer Powerscore CR Bible|For SC refer Manhattan SC Guide|


~~Better Burn Out... Than Fade Away~~

Out of a classroom of 6 boys and 4 girls the teacher picks a president for the student board, a vice president and a secretary.

What is the probability that only girls will be elected?

What is the probability that only boys will be elected?

---------------------------------------------------------------------------------------------------------------------------------------------------------------
(1) What is the probability that only girls will be elected?

Just for practice, since the most efficient way to solve this problem is the Probability approach.

(1) probability approach:

\(\frac{4}{10} * \frac{3}{9} * \frac{2}{8} = \frac{1}{30}\)

But I need to practice so I solved this problem with the other 3 approaches:

(2) Reversal Probability approach:
P = 1 - q (q=probability of not picking only girls, this means, picking 1, 2 or 3 boys):

--> 1 boy + 2 girls = BGG (and combinations):
\(1 - \frac{6}{10} * \frac{4}{9} * \frac{3}{8} * \frac{3!}{2!} = \frac{3}{10}\)

-->2 boys + 1 Girl = BBG (And combinations):
\(\frac{6}{10} *\frac{5}{9} * \frac{4}{8} * \frac{3!}{2!} = \frac{1}{2}\)

--> 3 boys:
\(\frac{6}{10} * \frac{5}{9} * \frac{4}{8} * \frac{3!}{3!} = \frac{1}{6}\)

Then, we add all the probabilities:

\(\frac{3}{10} + \frac{1}{2}+ \frac{1}{6} = \frac{29}{30} = q\)

\(1-q = 1- \frac{29}{30} = \frac{1}{30}\) = P

(3) Combinatorial approach:

\(\frac{C^4_3}{C^10_3} = \frac{4}{120} = \frac{1}{30}\)

(4) Reversal combinatorial approach:

\(1 - \frac{{C^6_1 * C^4_2 + C^6_2 * C^4_1 + C^6_3}}{C^10_3} = 1- \frac{{36 + 60 + 20}}{120} = 120/120 - \frac{116}{120} = \frac{1}{30}\)

(2) What is the probability that only boys will be elected?

(1) Probability approach:

\(\frac{6}{10} * \frac{5}{9} * \frac{4}{8} = \frac{4}{24} = \frac{1}{6}\)

(2) Reversal probability approach --> P = 1 - q

1girl + 2 boys: \(\frac{4}{10} *\frac{6}{9} *\frac{5}{8} * \frac{3!}{2!} = \frac{1}{6} * 3 = \frac{1}{2}\)

2 girls + 1 boy: \(\frac{4}{10} *\frac{3}{9} * \frac{6}{8} *\frac{3!}{2!} = \frac{1}{10} * 3 = \frac{3}{10}\)

3 girls: \(\frac{4}{10} *\frac{3}{9} * \frac{2}{8} = \frac{1}{30}\)

(3) Combinatorial approach:

\(\frac{C^6_3}{C^10_3} = \frac{20}{120} = \frac{1}{6}\)

(4) Reversal Comb approach:

\(\frac{C^4_1 * C^6_2 + C^4_2 * C^6_1 + C^4_3}{C^10_3}\)

1 girl + 2 boys: \(C^4_1 * C^6_2 = 60\)
2 girls + 1 boy: \(C^4_2 * C^6_1 = 36\)
3 girls:\(C^4_3 = 4\)

60 + 36 + 4 = 100

\(\frac{120}{120} - \frac{100}{120} = \frac{20}{120} = \frac{1}{6}\)
_________________

Encourage cooperation! If this post was very useful, kudos are welcome
"It is our attitude at the beginning of a difficult task which, more than anything else, will affect It's successful outcome" William James