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m05 Q17 Probability/Combination

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m05 Q17 Probability/Combination [#permalink] New post 18 Sep 2008, 00:23
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Kate and David each have $10. Together they flip a coin 5 times. Every time the coin lands on heads, Kate gives David $1. Every time the coin lands on tails, David gives Kate $1. After the coin is flipped 5 times, what is the probability that Kate has more than $10 but less than $15.

(A) \frac{5}{16}
(B) \frac{15}{32}
(C) \frac{1}{2}
(D) \frac{21}{32}
(E) \frac{11}{16}

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B

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So to answer this, we find the total combinations when you flip the coin 5 times. --> 2^5 = 32.

The explanation tells us to find the combinations when kate wins 3 times and 4 times. --> 5C3 and 5C4. we get 10 and 5 respectively.

Lastly, we simply find the probability (5/32) + (10/32) = 15/32. and that is the answer..

***Why do we only find the combinations when Kate wins 3 times and 4 times? Why not find 1 and 2 times?
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Re: m05 Q17 Probability/Combination [#permalink] New post 21 Sep 2008, 10:12
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in the conditions of

KJJJJ

and

KKJJJ


I mean 2 kates and 3 johns, kate will have less than 10. got it ?
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Re: m05 Q17 Probability/Combination [#permalink] New post 23 Sep 2008, 07:44
Hi. Think about it this way.
If K wins only 1 time that means she loses 4 time (the coin is flipped 5 times in total). So this way K will have

10 + 1 - 4 = 7 dollars (less than we need)

If K wins 2 time (which means she loses 3 times), she will have

10 + 2 - 3 = 9 dollars (less than we need)

If K wins all 5 times she'll end up with $15 (more than we need).

Now you see why we check only the combinations when K wins 3 (K will have $11) and 4 (K will have $13) times. Hope this helps.
d2touge wrote:
Kate and David each have $10. Together they flip a coin 5 times. Every time the coin lands on heads, Kate gives David $1. Every time the coin lands on tails, David gives Kate $1. After the coin is flipped 5 times, what is the probability that Kate has more than $10 but less than $15.

a) 5/16
b) 15/32
c) 1/2
d) 21/32
e) 11/16

So to answer this, we find the total combinations when you flip the coin 5 times. --> 2^5 = 32.

The explanation tells us to find the combinations when kate wins 3 times and 4 times. --> 5C3 and 5C4. we get 10 and 5 respectively.

Lastly, we simply find the probability (5/32) + (10/32) = 15/32. and that is the answer..

***Why do we only find the combinations when Kate wins 3 times and 4 times? Why not find 1 and 2 times?

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Re: m05 Q17 Probability/Combination [#permalink] New post 03 Jun 2010, 12:36
Hi guys,
I did not get this:

Find the total number of combinations = 2^5=32.
Which formula is being used? Can somebody elaborate on that?
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Re: m05 Q17 Probability/Combination [#permalink] New post 04 Jun 2010, 01:21
You may want to look at it this way.

How many outcomes are there when you flip a coin once? There are 2 outcomes (Heads or Tails).
If you flip a coin twice? - 4 outcomes (2^2). If you flip a coin twice, you can get these 4 outcomes: HH, HT, TH, TT. There are no other outcomes than these 4.
When you flip that coin three times, you will get 8 outcomes (2^3). Here they are: HHH, HHT, HTT, TTT, TTH, THH, HTH, THT.
So, flipping a coin 5 times gives 32 = 2^5 outcomes.

You might want to take a look at this Probability thread from our GMAT Math Book:
math-probability-87244.html

This thread has a lot of valuable resources that you might find helpful:
new-to-the-math-forum-please-read-this-first-77764.html
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Re: m05 Q17 Probability/Combination [#permalink] New post 26 Jul 2010, 05:26
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Only in 2 cases Kate will have more than $10 and less than $15.
Either 3 heads/2 tails or 4 heads/1 tail
Probability of having 3h and 2t is 5c3*(1/32)=10/32
Probability of having 4h and 1t is 5c4*(1/32)=5/32
So probability of either of the above is 10/32 + 5/32 = 15/32 (Ans)
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Re: m05 Q17 Probability/Combination [#permalink] New post 26 Jul 2010, 09:39
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Coin is flipped 5 times hence the sample space contains 2^5 ie 32 outcomes.
We get this by general formula number of outcomes ^ number of repetitions.
Now Kate can have more than 10 and less than 15 only if there are 4 Tails and 1 Head or 3 Tails and 2 Heads, this can be found out by simple manual calculation as below:
All 5 Tails, Kate: 15 David: 5
4 Tails, 1 Head, Kate: 13 David: 7
3 Tails, 2 Head, Kate: 11 David: 9
less than 2 tails then Kates total goes below 10 which is not to be considered.

Now cases with 4 tails and 1 head are 5!/4!*1! = 5 --> Simple arrangement rule
Now cases with 3 tails and 2 heads are 5!/3!*2! = 10 --> Simple arrangement rule

Favorable cases=10+5 =15

Hence probability = 15/32
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Re: m05 Q17 Probability/Combination [#permalink] New post 27 Jul 2010, 01:35
I think I am missing something here!!

Kate can have only one of the following SIX amounts after flipping the coin:
1- 5$ (If tails = 0 & heads = 5)
2- 7$ (If tails = 1 & heads = 4)
3- 9$ (If tails = 2 & heads = 3)
4- 11$ (If tails = 3 & heads = 2)
5- 13$ (If tails = 4 & heads = 1)
6- 15$ (If tails = 5 & heads = 0)

So in between 10 & 15$, only amounts no 4 & 5 apply, which is (2/6+2/6) = 1/3! What am I missing??
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Re: m05 Q17 Probability/Combination [#permalink] New post 27 Jul 2010, 05:59
tariqbakeer wrote:
I think I am missing something here!!

Kate can have only one of the following SIX amounts after flipping the coin:
1- 5$ (If tails = 0 & heads = 5)
2- 7$ (If tails = 1 & heads = 4)
3- 9$ (If tails = 2 & heads = 3)
4- 11$ (If tails = 3 & heads = 2)
5- 13$ (If tails = 4 & heads = 1)
6- 15$ (If tails = 5 & heads = 0)

So in between 10 & 15$, only amounts no 4 & 5 apply, which is (2/6+2/6) = 1/3! What am I missing??


Tariqbakeer
Did you mean that 1/6 + 1/6 = 2/6 ?

If you are considering only the outcomes you are correct that only 2 of the outcomes satisfy the conditions from the question.

However, you must consider the probability of each outcome.

Start with just to coin.

There are 3 outcomes possible:

Kate loses 2 dolllars
Kate wins 2 dollars
Kate wins 1 and loses 1

However each one is not equally likely

Here is a list of the possible out comes

(H=head T=tails)
HH
HT
TH
TT

So we see that the chances of:
2 losses is 1/4
2 wins is 1/4
1 win and 1 loss is 2/4

In this question we have 32 possible orders, but as you stated only 6 different outcomes.
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Re: m05 Q17 Probability/Combination [#permalink] New post 27 Jul 2010, 06:15
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If this is confusing just ignore it. I think in words more than symbols.

After reading the given answer I see in this question that it is the correct answer. However, I can't seem to remember the notation for these kind of questions. So I am trying to assemble a method that I can remember.

I am thinking something like this

\frac{Ways to Satisfied Conditions}{Possible Outcomes}=Probability of Conditions

For a question where multiple outcomes satisfy the conditions. Something like this might be better.

\frac{Ways to GetOutcome#1 + Ways to Get Outcome #2}{Possible Outcomes}

For this question I write
\frac{Different Ways To Get 4 Wins For Kate+Different Ways To Get 3 Wins For Kate}{Possible Outcomes}
then
\frac{10+5}{32}=15/32

And for those who want a further challenge, I suggest this question

A dice is rolled 5 times, what are the chances of rolling a 6, at least 3 times but not more than 4 times?

I am thinking \frac{5C3 + 5C4}{6^5}
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Re: m05 Q17 Probability/Combination [#permalink] New post 27 Jul 2010, 06:41
Another way to solve this using the 6 outcomes listed previously and understanding of the 32 possibilities

tariqbakeer wrote:

Kate can have only one of the following SIX amounts after flipping the coin:
1- 5$ (If tails = 0 & heads = 5)
2- 7$ (If tails = 1 & heads = 4)
3- 9$ (If tails = 2 & heads = 3)
4- 11$ (If tails = 3 & heads = 2)
5- 13$ (If tails = 4 & heads = 1)
6- 15$ (If tails = 5 & heads = 0)


only one way to lose $5 that would be all heads or HHHHH
only one way to win $5 that would be all tails or TTTTT

So out of 32 we have eliminated 2, 30 remain.

I can't explain why but I knew the chance of #3 and #4 above were equal, as well as the chance of #2 and #5.

So then divide 30/32 by 2

15/32
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Re: m05 Q17 Probability/Combination [#permalink] New post 28 Jul 2010, 17:35
Hi, what does "5C3" stand for?
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Re: m05 Q17 Probability/Combination [#permalink] New post 29 Jul 2010, 06:10
It means C_5^3 = \frac{5!}{3!*2!} = 10
clutterman wrote:
Hi, what does "5C3" stand for?

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Re: m05 Q17 Probability/Combination [#permalink] New post 28 Jul 2011, 05:59
There are two winning scenarios. Kate should win either 3 or 4 times.

This is how I solved it:
scenarios 1, Kate wins 4 times = 3*(1/2)^3 + (1/2)^2 = 5/8
scenarios 2, Kate wins 3 times = 4*(1/2)^4+ (1/2) = 3/4
5/8*3/4 = 15/32

I believe I solved it in a stupid way but that somehow worked. Can someone tell whether I just guessed or this method works?
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Re: m05 Q17 Probability/Combination [#permalink] New post 28 Jul 2011, 09:19
I did not know where to start his, but after reading the thread, some of my statistics knowledge came back.

Here is a really nice trick using Pascal's triangle.

--------1
-------1--1
-----1--2--1
----1--3-3--1
---1-4--6-4--1
-1-5-10-10-5--1

There are six combinations, so we use the sixth row. Or another angle to look at it is there are five flips, and then we count the first row as the zero row. Either way we arrive at the 1-5-10-10-5-1.

Now write out the combinations and money Kate gets.

HHHHH $5
HHHHT $7
HHHTT $9
HHTTT $11
HTTTT $13

TTTTT $15

We can assign the row to the probabilities as long as they're in this order.

Also, it is out of 32 because that is the number you get when you add up the row.

HHHHH 1/32
HHHHT 5/32
HHHTT 10/32
HHTTT 10/32
HTTTT 5/32
TTTTT 1/32

Finally, the only combos that fit the criteria for the answer are HHTTT and HTTTT.

Add up the probabilities to get 15/32.

Too bad if I had this problem on the test, I would get it wrong! At least now I am beginning to warm up.
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Re: m05 Q17 Probability/Combination [#permalink] New post 28 Jul 2011, 11:26
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I answered this problem logically rather than using a formula.

With 5 coin tosses, there is a 50% chance Kate will have more than $10, however the problem states what is the possibility of having more than $10 but less than $15. In order for Kate to have $15, she would need to win all 5 tosses (1 out of 32 possibility).

So...1/2 - 1/32 = 15/32.

Even if you couldn't figure out the 1/32 possibility, 15/32 is the only answer just below 1/2. None of the other answers are even close.
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Re: m05 Q17 Probability/Combination [#permalink] New post 01 Aug 2012, 19:48
GZR4DR wrote:
I answered this problem logically rather than using a formula.

With 5 coin tosses, there is a 50% chance Kate will have more than $10, however the problem states what is the possibility of having more than $10 but less than $15. In order for Kate to have $15, she would need to win all 5 tosses (1 out of 32 possibility).

So...1/2 - 1/32 = 15/32.

Even if you couldn't figure out the 1/32 possibility, 15/32 is the only answer just below 1/2. None of the other answers are even close.


This is kind of a silly question: can you or someone please explain how there is a 50% chance that Kate will have more than $10?

I get the \frac{1}{32} part for her winning all 5 tosses.
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Re: m05 Q17 Probability/Combination [#permalink] New post 01 Aug 2012, 23:49
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dpvtank wrote:
GZR4DR wrote:
I answered this problem logically rather than using a formula.

With 5 coin tosses, there is a 50% chance Kate will have more than $10, however the problem states what is the possibility of having more than $10 but less than $15. In order for Kate to have $15, she would need to win all 5 tosses (1 out of 32 possibility).

So...1/2 - 1/32 = 15/32.

Even if you couldn't figure out the 1/32 possibility, 15/32 is the only answer just below 1/2. None of the other answers are even close.


This is kind of a silly question: can you or someone please explain how there is a 50% chance that Kate will have more than $10?

I get the \frac{1}{32} part for her winning all 5 tosses.


After each toss Kate either gains $1 or losses $1.

After 5 tosses she'll have either more than $10 or less than $10 (she cannot have exactly $10). Since there is no reason one case to have more chances to appear than another then the probability that Kate will have more than $10 is 1/2 and the probability that Kate will have less than $10 is 1/2 too.

Hope it's clear.
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Re: m05 Q17 Probability/Combination [#permalink] New post 02 Aug 2012, 09:48
That makes perfect sense. Thanks a lot!

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Re: m05 Q17 Probability/Combination [#permalink] New post 18 Nov 2012, 00:42
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Approach 1: Find the total number of combinations = 2^5=32. Find the number of combinations when Kate wins: out of 5 games, she can win 3 or 4 times only as 5 victories would put her over the $14 mark and less than 3 victories, below the $10.

The number of combinations for winning 3 times:

5C3=10

Number of combinations for winning 4 times:

5C4=5

Remember, nCk=n! / k!∗(n−k)!.

Probability equals 5+1032=1532.

Approach 2: write out the combinations:

[3] : 123, 124, 125, 134, 135, 145, 234, 235, 245, 345

[4] : 1234, 1235, 1245, 1345, 2345

Total: 1532.

The combinations can be summed because they have equal probabilities of 132 each.


The correct answer is B
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Re: m05 Q17 Probability/Combination   [#permalink] 18 Nov 2012, 00:42
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