probability question

raunekk · Posted: Thu Aug 21, 2008 7:31 am Post subject: probability question

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

bradley281 · Posted: Thu Aug 21, 2008 7:44 am Post subject:

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

sudhir3127 · Posted: Thu Aug 21, 2008 8:14 am Post subject: Re: probability question

raunekk · Posted: Thu Aug 21, 2008 8:29 am Post subject:

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 ...

canuckclint · Posted: Thu Sep 04, 2008 10:15 pm Post subject:

Stuart Kovinsky · Posted: Thu Sep 04, 2008 10:34 pm Post subject:

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.

mals24 · Posted: Fri Sep 05, 2008 3:14 am Post subject:

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

Stockmoose16 · Posted: Wed Sep 17, 2008 12:10 pm Post subject:

Stuart Kovinsky · Posted: Wed Sep 17, 2008 12:36 pm Post subject:

Stockmoose16 · Posted: Wed Sep 17, 2008 12:39 pm Post subject:

Stockmoose16 · Posted: Wed Sep 17, 2008 12:55 pm Post subject:

Stuart Kovinsky · Posted: Wed Sep 17, 2008 1:38 pm Post subject:

cramya · Posted: Wed Sep 17, 2008 1:56 pm Post subject:

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