Probability problem

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Probability problem

by MubbashirAbbas » Sat May 26, 2012 7:24 am
From a group of 5 girls and 4 boys, 7 are to be randomly selected. what is the probability that 4 girls will be selected.

A) 4/9 B) 3/5 C) 5/9 D) NONE OF THESE

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by Anurag@Gurome » Sat May 26, 2012 7:33 am
MubbashirAbbas wrote:From a group of 5 girls and 4 boys, 7 are to be randomly selected. what is the probability that 4 girls will be selected.
Total number of ways to select 7 persons from 5 girls and 4 boys = 9C7 = 9*8/2 = 36

Number of ways to select 4 girls from 5 girls = 5C4 = 5
Number of ways to select (7 - 4) = 3 boys from 4 boys = 4C3 = 4

Hence, number of ways to select 7 persons from a group of 5 girls and 4 boyssuch that 4 girls will be selected = 5*4 = 20

Hence, the required probability = 20/36 = 5/9

The correct answer is C.
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by MubbashirAbbas » Sat May 26, 2012 7:43 am
Why did u put 36 in the denominator in the last step

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by Anurag@Gurome » Sat May 26, 2012 7:51 am
MubbashirAbbas wrote:Why did u put 36 in the denominator in the last step
Because 36 is the total number of possible selections without any restriction.
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by ankush1912 » Sat May 26, 2012 8:58 am
(5c4*4c3)/9c7=(5*4)/36=5/9
since the total no of girls are 5 and the group should consist of 4 girls,so 4 girls can be selected in 5c4 way and we need 3 more people to form the group of 7(7-4=3) hence they can be selected from boys in 4c3 manner.this forms our favorable condition i,e 5c4*4c3
the 9c7 denotes all possible casses of forming the group irrespective of boys n girls and hence forming the total possible conditions.

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by Stuart@KaplanGMAT » Sat May 26, 2012 9:58 am
MubbashirAbbas wrote:Why did u put 36 in the denominator in the last step
Just to elaborate a bit more, here's the probability formula:

probability = (# of desired outcomes)/(total # of possibilities)

and, as Anurag noted, 36 was the total number of possible choices.

Whenever you're doing a probability problem, starting out by writing that equation on your scratch paper will help you see exactly what you need to calculate.
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by MubbashirAbbas » Sat May 26, 2012 11:05 am
You guys used combination formula. Could u specify when should I use permutation and when to use combination. Something like a rule of thumb. I understand that permutation is used when order is important. Thank u all for helping me out.

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by Stuart@KaplanGMAT » Sat May 26, 2012 11:55 am
MubbashirAbbas wrote: I understand that permutation is used when order is important.
That's the exact rule!

If you just care about which objects to select, then you use the combinations formula.

If you care not only about which objects to select, but also the order of selection (or arrangement), then you use the permutations formula.

Here are two examples to illustrate the difference:

1) 7 students are trying out for the school tennis team. If there are 4 spots on the team, how many different teams can be made out of the 7 students?

In this question, we only care about WHICH students we select; in other words, we want to know how many teams of 4 we can make out of the 7 applicants. Since order is irrelevant, we use COMBINATIONS.

2) 7 goats are entered in the county fair. There are ribbons awarded for first, second and third place. How many different arrangements of winners can be made out of the 7 goats?

In this question, we care not only about WHICH goats get selected but ALSO about the ORDER in which they're selected. Since we care about order, we use PERMUTATIONS.
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by aneesh.kg » Sat May 26, 2012 12:56 pm
MubbashirAbbas wrote: Could u specify when should I use permutation and when to use combination. Something like a rule of thumb. I understand that permutation is used when order is important. Thank u all for helping me out.
Thumb Rules of 'Permutations & Combinations':

(i) OR is Addition, AND is addition.

(ii)The number of ways of selecting or choosing or picking 'r' objects from 'n' different objects is given by nCr. The 'C' stands for 'Combinations'.
Key words for using Combinations: selecting, choosing, picking (and other tense forms of these words).
For example:
In how many ways can a 2-member team be chosen from a group of 4 people: A, B, C and D?
4C2

(iii)The number of ways of arranging or distributing 'n' different objects on 'r' places (where r < n) is given by nPr. The 'P' stands for 'Permutations'.
Key words for using Permutations: arranging, distributing(and other tense forms of these words).
For example:
In how many ways can three prizes be distributed among the three winners in a contest in which five people participated?
5P3 or 5C3*3!

(iv) Number of ways of arranging/distributing 'n' different objects on 'n' places is given by nPn or n!

P.S.: More than the rules listed above, it is important that each of the above concepts is well-understood.
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