Probabilty question balls

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abhi.genx7: Senior | Next Rank: 100 Posts; Posts: 85; Joined: Sat Jan 30, 2010 3:49 am; Location: Hyderbad, India; GMAT Score:660

by abhi.genx7 » Thu Oct 21, 2010 3:29 am

Q2 ) A bag contains 3 white balls, 3 black balls & 2 red balls. One by
one three balls are drawn out without replacement. What is the
probability that the third ball is red?

Answer Choices
0.15
0.25
0.35
0.40
0.45

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Source: Beat The GMAT — Problem Solving | Thu Oct 21, 2010 3:29 am

abhi.genx7: Senior | Next Rank: 100 Posts; Posts: 85; Joined: Sat Jan 30, 2010 3:49 am; Location: Hyderbad, India; GMAT Score:660

by abhi.genx7 » Thu Oct 21, 2010 3:34 am

I'm weak at probability .
Can anyone explain a good method to attack such problems ?

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GMATGuruNY: GMAT Instructor; Posts: 15539; Joined: Tue May 25, 2010 12:04 pmFollowed by: 1906 members; Location: New York, NY; Thanked: 13060 times; GMAT Score:790

by GMATGuruNY » Thu Oct 21, 2010 5:31 am

abhi.genx7 wrote:Q2 ) A bag contains 3 white balls, 3 black balls & 2 red balls. One by
one three balls are drawn out without replacement. What is the
probability that the third ball is red?

Answer Choices
0.15
0.25
0.35
0.40
0.45

P(R) on the nth pick = P(R) on the first pick.

Since P(R) = 2/8 = .25 on the first pick, P(R) = .25 on the 3rd pick. The probability is the same for each pick.

The correct answer is B.

Here's proof:

P(WWR) = 3/8 * 2/7 * 2/6 = 1/28
Thus, P(BBR) = 1/28 (The probability is the same because we have the same number of B as we have of W.)

P(WRR)= 3/8 * 2/7 * 1/6 = 1/56
The probability will be the same for P(RWR), P(BRR), and P(RBR).
Thus, P(RWR) + P(BRR) + P(RBR) = 3*(1/56) = 3/56

P(WBR)= 3/8 * 3/7 * 2/6 = 3/56
Thus, P(BWR) = 3/56

Since all of the above are ways to get R on the 3rd pick, we add the fractions:
1/28 + 1/28 + 1/56 + 3/56 + 3/56 + 3/56 = 14/56 = .25.

Last edited by GMATGuruNY on Thu Oct 21, 2010 5:45 am, edited 1 time in total.

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shovan85: Community Manager; Posts: 991; Joined: Thu Sep 23, 2010 6:19 amFollowed by: 24 members; Location: Bangalore, India; Thanked: 146 times

by shovan85 » Thu Oct 21, 2010 5:31 am

abhi.genx7 wrote:I'm weak at probability .
Can anyone explain a good method to attack such problems ?

Dude I m speaking honestly I was (hopefully not now) weak in probability. Even I was so poor I was about to leave this part of GMAT. After joining BTG I was forced to solve some problem on probability here. People always posting the Probability problems.

The best way is to grab any book or start with a problem. try yourself. Fail, no problem but match your way of solving to the actual solution. Dont just say yes to what others are explaining but fight with them why it is happening so? Get the clear picture. Dont think if I miss such a silly part what others will think, end of the day u r paying 250$ to GMAT.

This was a philosophical approach for you how to behave to a probability question. Now the actual part

1. Draw a diagram: says X balls Y Plates.. what ever draw a whole container where u keep all these items.
2. Check all possible scenarios.
3. When one is happening AND other is happening MULTIPLY.
4. When one happens OR other happens ADD.

I think this much is the total ways to deal with a problem. Mind you this is my idea of solving it may not help you.

Let me apply my set of RULES to the Question in another post.

Last edited by shovan85 on Thu Oct 21, 2010 6:46 am, edited 1 time in total.

If the problem is Easy Respect it, if the problem is tough Attack it

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shovan85: Community Manager; Posts: 991; Joined: Thu Sep 23, 2010 6:19 amFollowed by: 24 members; Location: Bangalore, India; Thanked: 146 times

by shovan85 » Thu Oct 21, 2010 5:42 am

abhi.genx7 wrote:Q2 ) A bag contains 3 white balls, 3 black balls & 2 red balls. One by
one three balls are drawn out without replacement. What is the
probability that the third ball is red?

Answer Choices
0.15
0.25
0.35
0.40
0.45

1. Draw a diagram:
3W, 3B, 2R in a bag. Literally draw the bag and mention the same inside it. Then check what you are asked
third one has to be Red _ _ R. (Must)

2. Check all possible scenarios:
XRR : 3rd pick is bound to be RED, First one can be anything so X, If you pick second as RED then there will be a change in the situation
RXR : 3rd pick is bound to be RED, Second one can be anything so X, If you pick First as RED then there will be a change in the situation
XXR: First two can be anything except RED and 3rd pick is bound to be RED.

Dont agree with me... you think why no more scenarios and though 1 and 2 seem same why I am showing 2 scenarios.

3. When one is happening AND other is happening MULTIPLY.

These 3 events are happening one after another. Means in XRR: X happened AND R happened and R happened.
XRR = [6/8 * 2/7 * 1/6]
RXR = [2/8 * 6/7 * 1/6 ]
XXR = [6/8 * 5/7 * 2/6]

You figure out why 6/8 why 2/7 why 1/6...etc.

4. When one happens OR other happens ADD.

Now from the above 3 scenarios one is going to happen not all 3. Right!!
Either XXR OR XRR OR RXR

Thus adding the values from 3 you will get 0.25

Hope this helps

If the problem is Easy Respect it, if the problem is tough Attack it

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abhi.genx7: Senior | Next Rank: 100 Posts; Posts: 85; Joined: Sat Jan 30, 2010 3:49 am; Location: Hyderbad, India; GMAT Score:660

by abhi.genx7 » Thu Oct 21, 2010 6:22 am

I just don't know what to say .
So many innovative tips .
Thanks a lot

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aimscore: Senior | Next Rank: 100 Posts; Posts: 36; Joined: Wed Aug 04, 2010 9:35 am; Thanked: 2 times

by aimscore » Thu Oct 21, 2010 7:22 am

Hi all,

GMATGuruNY:
I solved the problem like shovan85 did and it took me around 4 mins to solve

P(R) on the nth pick = P(R) on the first pick.
Will this statement hold true even if the balls are put bag in the bag keepng the no of balls same at all times?
In fact are there any other scenarios you can think of, where this statement will hold true.

Shovan85:
Were you able to solve this sum in 2 mins ? I am very worried abt my timing although I have been extensively practicing. It isnt improving

any tips/suggestions

Thanks

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shovan85: Community Manager; Posts: 991; Joined: Thu Sep 23, 2010 6:19 amFollowed by: 24 members; Location: Bangalore, India; Thanked: 146 times

by shovan85 » Thu Oct 21, 2010 7:42 am

aimscore wrote:Hi all,

Shovan85:
Were you able to solve this sum in 2 mins ? I am very worried abt my timing although I have been extensively practicing. It isnt improving any tips/suggestions

Thanks

As I had mentioned earlier I am not good at Probability (and also Combinatorics). So I take my chances on these two area for solving. Although it takes around 3-4 mins .. its OK i guess for some questions.
Sorry I will not be able to help you on this

.

The way Guru has solved is quite fast but I believe it requires a strong foundation on Probability. I am damn sure and would have answered the question in 3 seconds if replacements were allowed. But without replacement decreases the total number of chances (The denominator) so cannot take a chance.

If the problem is Easy Respect it, if the problem is tough Attack it

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