To consider order or not to consider order help!

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KingKing: Newbie | Next Rank: 10 Posts; Posts: 3; Joined: Mon Oct 12, 2009 8:00 am

by KingKing » Mon Oct 12, 2009 8:35 am

Could someone please explain to me why my idea for answering the following question is not correct:

Suppose a box contains 10 balls of which 3 are red, 4 are white and 3 are green. A sample of 4 balls is selected at random without replacement. Find the probability of the event the sample contains 2 red balls

My process of thinking:
(ball 1) can be red in 3 ways
(ball 2 ) can be red in 2 ways
(ball 3) remaining colours in 7 ways
(ball 4) remaining colours in 6 ways

so desired outcomes can happen in 3x2x7x6 = 252 ways
How mayn possible outcomes in total = 10x9x8x7 = 5040 ways

so P(2 red balls in the set) = 252/5040

However the textbook give us the answer (3C2) * (7C2) / (10C4) = 3/10

Now I have read previous topics that discuss this confusion of when to choose "order matters" and "order does not matter"

https://www.beatthegmat.com/combination- ... t8924.html

Ian Stewart mentions that it does not matter which approach you take so as long as you are consistent in calculating numerator and denominator (ie calculating using permutations OR combinations) the difference should be by a constant .... could someone please clarify what I am doing wrong as I am consistent in my calculation

- Thanks

Source: Beat The GMAT — Quantitative Reasoning | Mon Oct 12, 2009 8:35 am

Stuart@KaplanGMAT: GMAT Instructor; Posts: 3224; Joined: Tue Jan 08, 2008 2:40 pmFollowed by: 614 members; Location: Toronto; Thanked: 1710 times; GMAT Score:800

by Stuart@KaplanGMAT » Mon Oct 12, 2009 7:49 pm

Suppose a box contains 10 balls of which 3 are red, 4 are white and 3 are green. A sample of 4 balls is selected at random without replacement. Find the probability of the event the sample contains 2 red balls

My process of thinking:
(ball 1) can be red in 3 ways
(ball 2 ) can be red in 2 ways
(ball 3) remaining colours in 7 ways
(ball 4) remaining colours in 6 ways

so desired outcomes can happen in 3x2x7x6 = 252 ways
How mayn possible outcomes in total = 10x9x8x7 = 5040 ways

so P(2 red balls in the set) = 252/5040

However the textbook give us the answer (3C2) * (7C2) / (10C4) = 3/10

Hi!

The problem with your method is that you assume that the firs two balls need to be red; however, we don't care which two are red, just that two of them are. Therefore, you undercounted the number of ways to get two red balls.

If you had multiplied your result by 4C2 (since that's how many different ways you can get 2 red balls out of 4 balls in total), you'd have gotten:

4C2 * 7*6*3*2 / 10*9*8*7

= (4*3/2*1) * 7*6*3*2 / 10*9*8*7

= 6 * 7*6*3*2 / 10*9*8*7

(mad cancelling out)

= 3/10

That's what Ian meant when he said that you can use many different methods, you just have to make sure that you're consistent.

The textbook solution was to just use combinations. We know that we want 2 red and 2 non-red balls. There are 3 red balls in total (that's the 3C2) and 7 non-red balls in total (that's the 7C2).

We're choosing 2 red balls AND 2 non-red balls; whenever we make MULTIPLE selections, we MULTIPLY them together (hence 3C2 * 7C2 in the numerator).

Since there are 10 balls overall and we're choosing 4 of them, we have 10C4 total selections we could make, hence the denominator.

Stuart Kovinsky | Kaplan GMAT Faculty | Toronto

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vaibhav.iit2002: Master | Next Rank: 500 Posts; Posts: 295; Joined: Tue Jul 15, 2008 10:07 am; Thanked: 4 times; GMAT Score:690

by vaibhav.iit2002 » Sun Oct 18, 2009 5:27 am

Hi Stuart,

I solved this in a different way. Plz let me know where am I wrong.

As per my concepts, whenever we have a problem with selecting few balls from a group, we can use combinations or simply we can say assume the problem to as picking one by one 4 balls, without replacement.

e.g. a container has 4 balls - 2 red, 2 blue.
prob. of picking up 2 red balls from container:

either we do 2C2/4C2 = 1/6

OR

prob. of 1st red ball = 2/4
prob. of 2nd red ball = 1/3

hence prob = 2/4x1/3 = 1/6

both ans are same. Fine

but when i apply same concept here in given ques. it fails,
prob. = 3/10 X 2/9 X 7/8 X 6/7 = 1/20

where am i wrong, plz help

vaibhav.iit2002: Master | Next Rank: 500 Posts; Posts: 295; Joined: Tue Jul 15, 2008 10:07 am; Thanked: 4 times; GMAT Score:690

by vaibhav.iit2002 » Sun Oct 18, 2009 6:52 am

Got it.
ans will be 1/20 x 4!/2!.2! = 3/10

life is a test: Master | Next Rank: 500 Posts; Posts: 138; Joined: Mon Mar 02, 2009 12:02 pm; Thanked: 15 times

by life is a test » Mon Oct 19, 2009 2:55 am

i worked out ans to be 1/15

Can we use the following way to work out the ans?

3/10 *2/9 = 6/90 = 1/15.

Stuart@KaplanGMAT: GMAT Instructor; Posts: 3224; Joined: Tue Jan 08, 2008 2:40 pmFollowed by: 614 members; Location: Toronto; Thanked: 1710 times; GMAT Score:800

by Stuart@KaplanGMAT » Mon Oct 19, 2009 11:38 am

life is a test wrote:i worked out ans to be 1/15

Can we use the following way to work out the ans?

3/10 *2/9 = 6/90 = 1/15.

Well, that's not the right answer, so my short reply is "no, we can't".

You've ignored the fact that we're selecting 4 balls, not 2. If the question had been:

Suppose a box contains 10 balls of which 3 are red, 4 are white and 3 are green. 2 balls are selected at random without replacement. Find the probability that both balls are red.

then your method would be perfect.

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