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# m05 Q17 Probability/Combination

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m05 Q17 Probability/Combination [#permalink]

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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) $$\frac{11}{16}$$ [Reveal] Spoiler: OA B Source: GMAT Club Tests - hardest GMAT questions 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? Kudos [?]: 6 [2], given: 0 Manager Joined: 28 Jul 2008 Posts: 99 Kudos [?]: 9 [2], given: 0 Re: m05 Q17 Probability/Combination [#permalink] ### Show Tags 21 Sep 2008, 10:12 2 This post received KUDOS in the conditions of KJJJJ and KKJJJ I mean 2 kates and 3 johns, kate will have less than 10. got it ? Kudos [?]: 9 [2], given: 0 CIO Joined: 02 Oct 2007 Posts: 1216 Kudos [?]: 989 [0], given: 334 Re: m05 Q17 Probability/Combination [#permalink] ### Show Tags 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? _________________ Welcome to GMAT Club! Want to solve GMAT questions on the go? GMAT Club iPhone app will help. Please read this before posting in GMAT Club Tests forum Result correlation between real GMAT and GMAT Club Tests Are GMAT Club Test sets ordered in any way? Take 15 free tests with questions from GMAT Club, Knewton, Manhattan GMAT, and Veritas. GMAT Club Premium Membership - big benefits and savings Kudos [?]: 989 [0], given: 334 Senior Manager Joined: 18 Aug 2009 Posts: 413 Kudos [?]: 145 [0], given: 16 Schools: UT at Austin, Indiana State University, UC at Berkeley WE 1: 5.5 WE 2: 5.5 WE 3: 6.0 Re: m05 Q17 Probability/Combination [#permalink] ### Show Tags 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? _________________ Never give up,,, Kudos [?]: 145 [0], given: 16 CIO Joined: 02 Oct 2007 Posts: 1216 Kudos [?]: 989 [0], given: 334 Re: m05 Q17 Probability/Combination [#permalink] ### Show Tags 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 _________________ Welcome to GMAT Club! Want to solve GMAT questions on the go? GMAT Club iPhone app will help. Please read this before posting in GMAT Club Tests forum Result correlation between real GMAT and GMAT Club Tests Are GMAT Club Test sets ordered in any way? Take 15 free tests with questions from GMAT Club, Knewton, Manhattan GMAT, and Veritas. GMAT Club Premium Membership - big benefits and savings Kudos [?]: 989 [0], given: 334 Manager Joined: 06 Oct 2009 Posts: 68 Kudos [?]: 36 [1], given: 5 Re: m05 Q17 Probability/Combination [#permalink] ### Show Tags 26 Jul 2010, 05:26 1 This post received KUDOS Only in 2 cases Kate will have more than$10 and less than $15. Either 3 heads/2 tails or 4 heads/1 tail Probability of having 3h and 2t is 5c3*(1/32)=10/32 Probability of having 4h and 1t is 5c4*(1/32)=5/32 So probability of either of the above is 10/32 + 5/32 = 15/32 (Ans) _________________ +1 kudos me if this is of any help... Kudos [?]: 36 [1], given: 5 Intern Joined: 27 May 2010 Posts: 6 Kudos [?]: 23 [7], given: 0 Re: m05 Q17 Probability/Combination [#permalink] ### Show Tags 26 Jul 2010, 09:39 7 This post received KUDOS 1 This post was BOOKMARKED Coin is flipped 5 times hence the sample space contains 2^5 ie 32 outcomes. We get this by general formula number of outcomes ^ number of repetitions. Now Kate can have more than 10 and less than 15 only if there are 4 Tails and 1 Head or 3 Tails and 2 Heads, this can be found out by simple manual calculation as below: All 5 Tails, Kate: 15 David: 5 4 Tails, 1 Head, Kate: 13 David: 7 3 Tails, 2 Head, Kate: 11 David: 9 less than 2 tails then Kates total goes below 10 which is not to be considered. Now cases with 4 tails and 1 head are 5!/4!*1! = 5 --> Simple arrangement rule Now cases with 3 tails and 2 heads are 5!/3!*2! = 10 --> Simple arrangement rule Favorable cases=10+5 =15 Hence probability = 15/32 Kudos [?]: 23 [7], given: 0 Intern Joined: 23 Jun 2010 Posts: 1 Kudos [?]: [0], given: 0 Re: m05 Q17 Probability/Combination [#permalink] ### Show Tags 27 Jul 2010, 01:35 I think I am missing something here!! Kate can have only one of the following SIX amounts after flipping the coin: 1- 5$ (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]

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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. Kudos [?]: 75 [0], given: 4 Manager Status: Waiting to hear from University of Texas at Austin Joined: 24 May 2010 Posts: 76 Kudos [?]: 75 [2], given: 4 Location: Changchun, China Schools: University of Texas at Austin, Michigan State Re: m05 Q17 Probability/Combination [#permalink] ### Show Tags 27 Jul 2010, 06:15 2 This post received KUDOS 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 $$\frac{5C3 + 5C4}{6^5}$$ Kudos [?]: 75 [2], given: 4 Manager Status: Waiting to hear from University of Texas at Austin Joined: 24 May 2010 Posts: 76 Kudos [?]: 75 [0], given: 4 Location: Changchun, China Schools: University of Texas at Austin, Michigan State Re: m05 Q17 Probability/Combination [#permalink] ### Show Tags 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 Kudos [?]: 75 [0], given: 4 Intern Joined: 06 Jul 2010 Posts: 1 Kudos [?]: [0], given: 0 Re: m05 Q17 Probability/Combination [#permalink] ### Show Tags 28 Jul 2010, 17:35 Hi, what does "5C3" stand for? Kudos [?]: [0], given: 0 CIO Joined: 02 Oct 2007 Posts: 1216 Kudos [?]: 989 [0], given: 334 Re: m05 Q17 Probability/Combination [#permalink] ### Show Tags 29 Jul 2010, 06:10 It means $$C_5^3 = \frac{5!}{3!*2!} = 10$$ clutterman wrote: Hi, what does "5C3" stand for? _________________ Welcome to GMAT Club! Want to solve GMAT questions on the go? GMAT Club iPhone app will help. Please read this before posting in GMAT Club Tests forum Result correlation between real GMAT and GMAT Club Tests Are GMAT Club Test sets ordered in any way? Take 15 free tests with questions from GMAT Club, Knewton, Manhattan GMAT, and Veritas. GMAT Club Premium Membership - big benefits and savings Kudos [?]: 989 [0], given: 334 Intern Joined: 17 May 2011 Posts: 15 Kudos [?]: 1 [0], given: 6 Re: m05 Q17 Probability/Combination [#permalink] ### Show Tags 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? Kudos [?]: 1 [0], given: 6 Intern Status: UF BSBA '15 - Preparing for the future Joined: 09 May 2011 Posts: 25 Kudos [?]: 5 [0], given: 2 Location: United States GPA: 3.87 WE: Human Resources (Consulting) Re: m05 Q17 Probability/Combination [#permalink] ### Show Tags 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. Kudos [?]: 5 [0], given: 2 Manager Joined: 25 Jun 2008 Posts: 127 Kudos [?]: 26 [3], given: 5 Concentration: General Management, Technology Re: m05 Q17 Probability/Combination [#permalink] ### Show Tags 28 Jul 2011, 11:26 3 This post received KUDOS 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. Kudos [?]: 26 [3], given: 5 Intern Joined: 02 May 2011 Posts: 36 Kudos [?]: 27 [0], given: 55 Location: Canada Concentration: General Management, Social Entrepreneurship GMAT 1: 630 Q45 V33 GMAT 2: 700 Q47 V40 GPA: 3.78 Re: m05 Q17 Probability/Combination [#permalink] ### Show Tags 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]

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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]

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02 Aug 2012, 09:48
That makes perfect sense. Thanks a lot!

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Re: m05 Q17 Probability/Combination [#permalink]

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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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