probability question

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raunekk: Legendary Member; Posts: 1159; Joined: Wed Apr 16, 2008 10:35 pm; Thanked: 56 times

by raunekk » Thu Aug 21, 2008 7:31 am

Q)A drawer holds 4 red hats and 4 blue hats.What is the probability of getting exactly 3 red hats or exactly 3 blue hats when taking out 4 hats randomly out of the drawer and immediately returning every hat to the drawer before taking out the next?

(a)1/8
(b)1/4
(c)1/2
(d)3/8
(e)7/12

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Source: Beat The GMAT — Problem Solving | Thu Aug 21, 2008 7:31 am

bradley281: Junior | Next Rank: 30 Posts; Posts: 17; Joined: Tue May 27, 2008 4:09 pm; Location: arizona; Thanked: 1 times

by bradley281 » Thu Aug 21, 2008 7:44 am

each time you draw you have 1/2 of a chance of drawing red or blue. since you are returning the hats to the drawer each draw gives you 1/2 of a chance. therefore my guess is.

1/2 x 3 = 1/8

please let me know the answer

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sudhir3127: Legendary Member; Posts: 829; Joined: Mon Jul 07, 2008 10:09 pmFollowed by: 3 members; Location: INDIA; Thanked: 84 times

Re: probability question

by sudhir3127 » Thu Aug 21, 2008 8:14 am

raunekk wrote:Q)A drawer holds 4 red hats and 4 blue hats.What is the probability of getting exactly 3 red hats or exactly 3 blue hats when taking out 4 hats randomly out of the drawer and immediately returning every hat to the drawer before taking out the next?

(a)1/8
(b)1/4
(c)1/2
(d)3/8
(e)7/12

i go for C 1/2

probability of taking a particular hat is 1/2 ( 4/8)

Now assume we have 2 take 3 blues and 1 red

P(BBBR) = (1/2)^4
P(BBRB) = (1/2)^4
P(BRBB) = (1/2)^4
P(RBBB) = (1/2)^4
total = 4 * (1/2)^4

similarly with red

thus the total is 2*4*1/16 = 8/16 = 1/2

hope its clear..

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raunekk: Legendary Member; Posts: 1159; Joined: Wed Apr 16, 2008 10:35 pm; Thanked: 56 times

by raunekk » Thu Aug 21, 2008 8:29 am

hi sudhir3127,

answer is C 1/2

thanks a lot.Tat made things quite clear...

But is it possible to solve this sum by taking the number of ways in which red or blue hats can be selected..i mean..

I did it d following way...can u tell me where i went wrong..

number of ways selecting exactly 3 red or blue balls - 4C3

3 hats are selected so 1 mre hat can be selected in 4 ways

Total number of ways in selectin 4 balls is 8C4 ways..

Thus 4C3*4/8C4.

I know its messed up..But just curious whether we can solve it this ways..

thanks a lot ...

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canuckclint: Master | Next Rank: 500 Posts; Posts: 154; Joined: Tue Aug 26, 2008 12:59 pm; Location: Canada; Thanked: 4 times

by canuckclint » Thu Sep 04, 2008 10:15 pm

raunekk wrote: Total number of ways in selectin 4 balls is 8C4 ways..

Your mistake is here. It is 2^4 = 16 for 4 positions
But if you are trying to solve 4 balls you might as well do it the previous way.

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Stuart@KaplanGMAT: GMAT Instructor; Posts: 3225; Joined: Tue Jan 08, 2008 2:40 pmFollowed by: 614 members; Location: Toronto; Thanked: 1710 times; GMAT Score:800

by Stuart@KaplanGMAT » Thu Sep 04, 2008 10:34 pm

Any time there's a 50/50 chance of something happening, you really have a coin flip question in disguise. The question could have been:

If you flip a fair coin 4 times, what's the probability of getting exactly 3 heads or exactly 3 tails?

There are numerous ways to solve coin flip questions. One of the quickest is to apply the coin flip formula.

The probability of getting exactly k results out of n flips is:

nCk/2^n

Applying the formula to this question, we get:

4C3/2^4 = 4/16 = 1/4

Since we want exactly 3 heads OR exactly 3 tails, we need to double our answer, getting 1/2.

As quick as it was to apply the formula, there's an even BETTER way to solve coin flip questions, involving memorizing a few numbers.

Here are the numbers to remember:

1 3 3 1
1 4 6 4 1
1 5 10 10 5 1

Some of you may recognize those patters as rows of numbers from Pascal's Triangle. The Triangle has a number of uses, but for GMAT purposes its most useful application is to coin flip questions.

The first row applies to 3 flip questions, the second to 4 flip questions and the third to 5 flip questions.

Let's start with the first row, 1 3 3 1, and see how it helps.

"A fair coin is flipped 3 times. What's the probability of getting exactly 2 heads?"

The entries in the row represent the different ways to get 0, 1, 2 and 3 results, respectively. In our question, we want 2 heads, so we go to the 3rd entry in the row, "3".

To find the total number of possibilities, add up the row... 1+3+3+1 = 8

So, our answer is 3/8.

Let's look at a much more complicated question:

"A fair coin is flipped 5 times. What's the probability of getting at least 2 heads?"

If we want at least 2 heads, we want 2 heads, 3 heads, 4 heads OR 5 heads. Pretty much whenever we see "OR" in probability, we add the individual probabilities.

Looking at the 5 flip row, we have 1 5 10 10 5 1. For 2H, 3H, 4H and 5H we add up the 3rd, 4th, 5th and 6th entries: 10+10+5+1=26.

Summing the whole row, we get 32.

So, the chance of getting at least 2 heads out of 5 flips is 26/32 = 13/16.

Stuart Kovinsky | Kaplan GMAT Faculty | Toronto

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mals24: Legendary Member; Posts: 683; Joined: Tue Jul 22, 2008 1:58 pmFollowed by: 2 members; Location: Dubai; Thanked: 73 times

by mals24 » Fri Sep 05, 2008 3:14 am

Hey Stuart thanks for the great tip
It does make 'flip the coin' questions more simpler and fun now

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Stockmoose16: Master | Next Rank: 500 Posts; Posts: 347; Joined: Mon Aug 04, 2008 1:42 pm; Thanked: 1 times

by Stockmoose16 » Wed Sep 17, 2008 12:10 pm

Stuart Kovinsky wrote:Any time there's a 50/50 chance of something happening, you really have a coin flip question in disguise. The question could have been:

If you flip a fair coin 4 times, what's the probability of getting exactly 3 heads or exactly 3 tails?

There are numerous ways to solve coin flip questions. One of the quickest is to apply the coin flip formula.

The probability of getting exactly k results out of n flips is:

nCk/2^n

Applying the formula to this question, we get:

4C3/2^4 = 4/16 = 1/4

Since we want exactly 3 heads OR exactly 3 tails, we need to double our answer, getting 1/2.

Stuart,

I don't understand your logic here. In the numerator, don't you have to account for the fourth flip? Shouldn't the numerator be 4C3* 4C1?

Look at the following example:

Three letters are randomly chosen from the word TUESDAY. How many possible selections are there? How many of these selections have:

(b) exactly 2 vowels?

No. of selections = 3C2* 4C1= 12

How come for the coin flip question, you don't have to do:

4C3 * 4C1 for the numerator?

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Stuart@KaplanGMAT: GMAT Instructor; Posts: 3225; Joined: Tue Jan 08, 2008 2:40 pmFollowed by: 614 members; Location: Toronto; Thanked: 1710 times; GMAT Score:800

by Stuart@KaplanGMAT » Wed Sep 17, 2008 12:36 pm

Stockmoose16 wrote:
Stuart Kovinsky wrote:Any time there's a 50/50 chance of something happening, you really have a coin flip question in disguise. The question could have been:

If you flip a fair coin 4 times, what's the probability of getting exactly 3 heads or exactly 3 tails?

There are numerous ways to solve coin flip questions. One of the quickest is to apply the coin flip formula.

The probability of getting exactly k results out of n flips is:

nCk/2^n

Applying the formula to this question, we get:

4C3/2^4 = 4/16 = 1/4

Since we want exactly 3 heads OR exactly 3 tails, we need to double our answer, getting 1/2.

Stuart,

I don't understand your logic here. In the numerator, don't you have to account for the fourth flip? Shouldn't the numerator be 4C3* 4C1?

Look at the following example:

Three letters are randomly chosen from the word TUESDAY. How many possible selections are there? How many of these selections have:

(b) exactly 2 vowels?

No. of selections = 3C2* 4C1= 12

How come for the coin flip question, you don't have to do:

4C3 * 4C1 for the numerator?

4C3 in the coin flip question does account for all 4 flips - we want exactly 3 of them to be heads, by default the other will be tails.

In the "TUESDAY" question, the "3" in 3C2 doesn't relate to the number of letters we're choosing, it relates to the number of vowels available (we have 3 vowels total, we're choosing 2 of them); similarly, the "4" in 4C1 relates to the number of consonants.

So, we have 3C2*4C1 because we're choosing 2 vowels out of 3 and 1 consonant out of 4.

In our example, we have 4C3 because we're choosing 3 results out of 4 flips - we're not choosing 3 different "heads" out of a total of 4 available heads.

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Stockmoose16: Master | Next Rank: 500 Posts; Posts: 347; Joined: Mon Aug 04, 2008 1:42 pm; Thanked: 1 times

by Stockmoose16 » Wed Sep 17, 2008 12:39 pm

4C3 in the coin flip question does account for all 4 flips - we want exactly 3 of them to be heads, by default the other will be tails.

In the "TUESDAY" question, the "3" in 3C2 doesn't relate to the number of letters we're choosing, it relates to the number of vowels available (we have 3 vowels total, we're choosing 2 of them); similarly, the "4" in 4C1 relates to the number of consonants.

So, we have 3C2*4C1 because we're choosing 2 vowels out of 3 and 1 consonant out of 4.

In our example, we have 4C3 because we're choosing 3 results out of 4 flips - we're not choosing 3 different "heads" out of a total of 4 available heads.

Got it. Thanks for the explanation. It's very helpful.

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Stockmoose16: Master | Next Rank: 500 Posts; Posts: 347; Joined: Mon Aug 04, 2008 1:42 pm; Thanked: 1 times

by Stockmoose16 » Wed Sep 17, 2008 12:55 pm

Stuart Kovinsky wrote:
Stockmoose16 wrote:
Stuart Kovinsky wrote:Any time there's a 50/50 chance of something happening, you really have a coin flip question in disguise. The question could have been:

If you flip a fair coin 4 times, what's the probability of getting exactly 3 heads or exactly 3 tails?

There are numerous ways to solve coin flip questions. One of the quickest is to apply the coin flip formula.

The probability of getting exactly k results out of n flips is:

nCk/2^n

Applying the formula to this question, we get:

4C3/2^4 = 4/16 = 1/4

Since we want exactly 3 heads OR exactly 3 tails, we need to double our answer, getting 1/2.

Stuart,

I don't understand your logic here. In the numerator, don't you have to account for the fourth flip? Shouldn't the numerator be 4C3* 4C1?

Look at the following example:

Three letters are randomly chosen from the word TUESDAY. How many possible selections are there? How many of these selections have:

(b) exactly 2 vowels?

No. of selections = 3C2* 4C1= 12

How come for the coin flip question, you don't have to do:

4C3 * 4C1 for the numerator?
4C3 in the coin flip question does account for all 4 flips - we want exactly 3 of them to be heads, by default the other will be tails.

In the "TUESDAY" question, the "3" in 3C2 doesn't relate to the number of letters we're choosing, it relates to the number of vowels available (we have 3 vowels total, we're choosing 2 of them); similarly, the "4" in 4C1 relates to the number of consonants.

So, we have 3C2*4C1 because we're choosing 2 vowels out of 3 and 1 consonant out of 4.

In our example, we have 4C3 because we're choosing 3 results out of 4 flips - we're not choosing 3 different "heads" out of a total of 4 available heads.

Stuart,

I actually have one follow-up question. In regards to this question:

A drawer holds 4 red hats and 4 blue hats. What is the probability of getting exactly three red hats or exactly three blue hats when taking out 4 hats randomly out of the drawer and immediately returning every hat to the drawer before taking out the next?

(a) 1/8
(b) ¼
(c) ½
(d) 3/8
(e) 7/12

I understand you treat this like a coin flip, but if you didn't recognize that from the beginning, why wouldn't this be exactly like the vowel/consonant question I was referred to above. Meaning, you should multiply the ways of getting exactly 3 red hats times the probability of getting a blue one on the fourth choice? Then multiply by two to account for the "or."

So the answer should be:

4C3*4/2^4 +4C3*4/2^4 =2 ... which obviously makes no sense. Why is this question different from the vowel one above?

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Stuart@KaplanGMAT: GMAT Instructor; Posts: 3225; Joined: Tue Jan 08, 2008 2:40 pmFollowed by: 614 members; Location: Toronto; Thanked: 1710 times; GMAT Score:800

by Stuart@KaplanGMAT » Wed Sep 17, 2008 1:38 pm

Stockmoose16 wrote:I actually have one follow-up question. In regards to this question:

A drawer holds 4 red hats and 4 blue hats. What is the probability of getting exactly three red hats or exactly three blue hats when taking out 4 hats randomly out of the drawer and immediately returning every hat to the drawer before taking out the next?

(a) 1/8
(b) ¼
(c) ½
(d) 3/8
(e) 7/12

I understand you treat this like a coin flip, but if you didn't recognize that from the beginning, why wouldn't this be exactly like the vowel/consonant question I was referred to above. Meaning, you should multiply the ways of getting exactly 3 red hats times the probability of getting a blue one on the fourth choice? Then multiply by two to account for the "or."

So the answer should be:

4C3*4/2^4 +4C3*4/2^4 =2 ... which obviously makes no sense. Why is this question different from the vowel one above?

You're misunderstanding how to apply the combinations formula. To be honest, I have no clue how you derived the expression above.

In the hat question, we're making 4 selections. The chance of getting a red hat on any specific draw is 1/2. The probability of any 4 specific selections is:

1/2 * 1/2 * 1/2 * 1/2 = 1/16

For our particular question, we want either exactly 3R or 3B.

For 3R, we want 3 of our 4 selections to be red, so that's 4C3.

For 3B, we want 3 of our 4 selections to be blue, so that's 4C3.

Since we want 3R or 3B, we add 4C3 + 4C3 = 4 + 4 = 8.

So: # of desired outcomes = 8
Total # of possibilities = 16

Prob = 8/16 = 1/2

By muliplying 4C3 * 4, you're double counting (actually, you're squaring the number of possibilities), since 4C3 already takes into account the "missing" 1.

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cramya: Legendary Member; Posts: 2467; Joined: Thu Aug 28, 2008 6:14 pmFollowed by: 11 members; Thanked: 331 times

by cramya » Wed Sep 17, 2008 1:56 pm

Staurt thanks you so much for sharing the Pascal traingles approach with us.
Here my small addition to that:

If you cant memorize it gies like this

1 3 3 1
1 4 6 4 1

To get to 1 4 6 4 1 all we have to do is carry over the left side1 as is and keep adding the 2 adjacent terms and then again carry the right 1

1(carrying the left 1 as is) 4(left side 1+ first3) 6(left side3 added to right side 3) 4(right side 3 added to right side 1) 1(carrying the right 1 as is)

The same applies in getting the next row of numbers
When I explain this it may look complicated but its very simple to do

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Thouraya: Master | Next Rank: 500 Posts; Posts: 112; Joined: Wed Jan 20, 2010 5:46 am; Thanked: 1 times

by Thouraya » Wed Jun 02, 2010 12:44 am

Hi again Stuart,

Your explanation to te drawers example and your suggestion to treat it as a coins example is very helpful, but Id like to ask what is it that I am doing wrong below if I want to treat the problem as a combinations problem (in the case where we want to choose more than two types, thus can't treat it as a 50 50 coin flip):

4C3 + 4C3 =8 (thats the desired part of choosing 3 red out of 4 OR 3 blue out of 4)

Then, total probability is: 8C4=70

Then, solution: 8/70=4/35

I am aware that I haven't accounted for the last bit of the question about returning the ones I chose first time...

Thank you!

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