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# M05-17

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Re: M05-17 [#permalink]
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I couldn't wrap my mind around how to apply combinatorics to this.

Instead, I figured that K and D both have 50% chance on each flip, therefore K success on the first 4 = 2 win: 2 loss. Probability of winning the tiebreaker (5th flip) is 50%

therefore she has 50% chance of winning 3/5 flips.

Lastly, K cannot win all 5 or she goes over the max. Prob 5/5 = (1/2)^5 = 1/32

Prob $0<K<$5 = 1/2 - 1/32 = 15/32
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Re: M05-17 [#permalink]
Dear Bunuel!
Please explain why the total number of combinations = 25^5 ?
I don't understand. Tks
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Re: M05-17 [#permalink]
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Mrtinhnv wrote:
Dear Bunuel!
Please explain why the total number of combinations = 25^5 ?
I don't understand. Tks

They flip a coin 5 times. Each time it's either heads or tails, so 2 outcomes for each flip, there are 5 flips, so the total number = 2*2*2*2*2 = 2^5.

Hope it's clear.
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Re: M05-17 [#permalink]
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Bunuel 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. $$\frac{5}{16}$$ B. $$\frac{15}{32}$$ C. $$\frac{1}{2}$$ D. $$\frac{21}{32}$$ E. $$\frac{11}{16}$$ Kate looses$1 every time Head is flipped
Kate gains $1 every time Tail is flipped Total combinations of coin flipping 5 times are 32 ways Kate H H H H H : Kate has$5 at end of 5 coin flips (Kate lost $5 for 5 head flip) T T T T H : Kate has$13 (she gains $4 for 4 tails but looses$1 for 1 heads)
T T T H H : Kate has $12 (she gains$3 for 3 tails but looses $2 for 2 heads) T T H H H : Kate has$9 (she gains $2 for 2 tails but looses$3 for 3 heads)
T T H H H : Kate has $9 (she gains$2 for 2 tails but looses $3 for 3 heads) . . . . T T T T T : Kate has$15 (she gains $1 for 5 tails) We are asked what is probability that Kate has more than$10 but less than $15 This is possible if there are atlest 3 or 4 Tails are flipped in 5 total coin flips Number of ways 3 Tails flipped are 10 Number of ways 4 Tails flipped are 5 Probability = # of favourable outcomes / Total outcomes = (10+5)/32 LBS Moderator Joined: 30 Oct 2019 Posts: 830 Own Kudos [?]: 790 [5] Given Kudos: 1576 Re: M05-17 [#permalink] 3 Kudos 2 Bookmarks Kate wins denoted by K (say) and David's win denoted by D So it can be the permutation of these two cases - 1. KKKDD 2. KKKKD Now the first case is = $$\frac{5!}{3!2!} = 10$$ And the second case = $$\frac{5!}{4!} = 5$$ Total occurrence = $$2^5 = 32$$ Therefore, probability = $$\frac{10+5}{32} = \frac{15}{32}$$ Answer: C Senior Manager Joined: 05 Aug 2017 Posts: 361 Own Kudos [?]: 471 [3] Given Kudos: 277 Location: India Concentration: Strategy, Marketing WE:Engineering (Energy and Utilities) Re: M05-17 [#permalink] 3 Kudos 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?

Given:
Initially Kate has $10 and David has$10
Heads: Kate to david $1 Tails: David to Kate$1
Coin flipped 5 times

Find: Probability $10<Kate<$15

Case 1) Let suppose kate wins 1 game and David wins 4 game: KDDDD : After 5 toss, kate will have 7 and david will have 13: This case is not possible.
Case 2) Let suppose kate wins 2 game and David wins 3 game: KKDDD : After 5 toss, kate will have 9 and david will have 11: This case is not possible.
Case 3) Let suppose kate wins 3 game and David wins 2 game: KKKDD : After 5 toss, kate will have 11 and david will have 9: This case is possible. No of possible case : 5!/3!*2!= 10
Case 4) Let suppose kate wins 4 game and David wins 1 game: KKKKD : After 5 toss, kate will have 13 and david will have 7: This case is possible. No of possible case : 5!/4!=5
Case 5) Let suppose kate wins 4 game and David wins 1 game: KKKKK : After 5 toss, kate will have 15 and david will have 7: This case is not possible.

Total outcome: 32

Probability : 10+5/32=15/32

A. 5/16
B. 15/32
C. 1/2
D. 21/32
E. 11/16
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Re: M05-17 [#permalink]
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Bunuel 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. $$\frac{5}{16}$$ B. $$\frac{15}{32}$$ C. $$\frac{1}{2}$$ D. $$\frac{21}{32}$$ E. $$\frac{11}{16}$$ It can also be solved without using P&C So Kate and David have equal chances of winning. But the question has a restriction that Kate cannot win all 5 times. So the probability would be less than 1/2 Now comparing Answer choices we see A. 5/16 >>too less (since the probability we are trying to find is eliminating only one possibility out of 5>>) B. 15/32 >> Keep it C. 1/2 >> cannot be since we are eliminating a possibility D. 21/32 >> Greater than 1/2 E. 11/18 >> Again Greater than 1/2 Hence Ans is B VP Joined: 14 Aug 2019 Posts: 1344 Own Kudos [?]: 857 [0] Given Kudos: 381 Location: Hong Kong Concentration: Strategy, Marketing GMAT 1: 650 Q49 V29 GPA: 3.81 Re: M05-17 [#permalink] Hi IanStewart Bunuel chetan2u : Please help me to identify the value which I get is actually what? Out of 5 times coin flipped : Kate can have$11 or $13 , more than 10$ and less than 15$. The values that she can have are: 5$, 7$, 9$, 11$, 13$,15$( if Kate wins 0, 1,2,3,4,5 times respectively) Favored cases = 2/6 ( total cases) = 1/3 In other words, what would be the question exactly, if the answer is 1/3. please suggest. GMAT Tutor Joined: 24 Jun 2008 Posts: 4127 Own Kudos [?]: 9491 [2] Given Kudos: 91 Q51 V47 Re: M05-17 [#permalink] 1 Kudos 1 Bookmarks Expert Reply mSKR wrote: Please help me to identify the value which I get is actually what? Out of 5 times coin flipped : Kate can have$11 or $13 , more than 10$ and less than 15$. The values that she can have are: 5$, 7$, 9$, 11$, 13$,15$( if Kate wins 0, 1,2,3,4,5 times respectively) Favored cases = 2/6 ( total cases) = 1/3 You might imagine using the same approach to answer the question "what is the probability I will win the lottery?" You might then think there are two cases -- you win or you lose -- and that the probability is 1/2. But that's clearly not right. The issue is, the two cases in the lottery question are not equally likely. If we're going to solve a probability question by counting outcomes, those outcomes need to be equally likely to happen. In the question in this thread, Kate only gets$15 when the coin lands on Tails five times, which can only happen in one way (TTTTT). But she gets $11 in the end if the coin lands on Tails three times and on Heads twice, which is a much more likely outcome -- it can happen in ten ways (TTTHH, TTHTH, THTTH, HTTTH, etc). mSKR wrote: In other words, what would be the question exactly, if the answer is 1/3. please suggest. I can quickly modify the question above so 1/3 becomes the right answer, but I think this question setup is very awkward (I'd never write a question like this personally!) : David and Kate each have$10. David and Kate play a game where they flip a coin once, and if the coin lands on Tails, David will give Kate a randomly selected even number of dollars between $2 and$6 inclusive, and if it lands on Heads, Kate will give David a randomly selected even number of dollars between $2 and$6 inclusive. What is the probability Kate will have more than $10 but less than$15 if she plays this game once?

Then there are six equally likely outcomes (-$6, -$4, -$2, +$2, +$4, +$6), two of which produce the result the question asks for, so the answer is 2/6 = 1/3.
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Re: M05-17 [#permalink]
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I think this is a high-quality question and I agree with the explanation.
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Schools: Cox (A$) GMAT 1: 730 Q49 V41 GPA: 3.2 Re: M05-17 [#permalink] I think this is a high-quality question and I agree with explanation. Intern Joined: 20 Oct 2020 Posts: 22 Own Kudos [?]: 52 [2] Given Kudos: 1203 Location: Mexico GMAT 1: 690 Q48 V35 GPA: 3.8 Re: M05-17 [#permalink] 1 Kudos 1 Bookmarks Easy if done by letters. Total number of outcomes is 2*2*2*2*2 = 32 Outcomes where K's money is between 11 and 14 ( 11 < k < 14) w=win L=loss WWWLL + WWWWL total number of letters/repetition of letters 5! /(3!*2!) + 5! /(4!*1!) = ? 10 + 5 = 15 Answer = 15/32 Intern Joined: 26 Apr 2021 Posts: 25 Own Kudos [?]: 17 [0] Given Kudos: 39 Location: India Concentration: Strategy, Finance GMAT 1: 710 Q47 V40 GPA: 3.9 Re: M05-17 [#permalink] Can someone please help me understand what's wrong with my below approach: Possible amounts with Kate after 5 coin flips are: - 15 (if Kate wins all 5) - 13 (if Kate wins 4 and loses 1) - 11 (if Kate wins 3 and loses 2) - 9 (if Kate wins 2 and loses 3) - 7 (if Kate wins 1 and loses 4) - 5 (if Kate loses all 5) Now only 11 and 13 (2) are the relevant ones out of 6 possibilities. So, it can be 2/6 or 1/3. Right? Intern Joined: 20 Oct 2020 Posts: 22 Own Kudos [?]: 52 [1] Given Kudos: 1203 Location: Mexico GMAT 1: 690 Q48 V35 GPA: 3.8 Re: M05-17 [#permalink] 1 Kudos Anjoom wrote: Can someone please help me understand what's wrong with my below approach: Possible amounts with Kate after 5 coin flips are: - 15 (if Kate wins all 5) - 13 (if Kate wins 4 and loses 1) - 11 (if Kate wins 3 and loses 2) - 9 (if Kate wins 2 and loses 3) - 7 (if Kate wins 1 and loses 4) - 5 (if Kate loses all 5) Now only 11 and 13 (2) are the relevant ones out of 6 possibilities. So, it can be 2/6 or 1/3. Right? There are more than six possibilities... Lets take 13 as an example (4 wins and 1 loss) there is actually 5 ways for this outcome to come true...you are saying in your approach that there is only one way for getting this result. -LWWWW -WLWWW -WWLWW -WWWLW -WWWWL Now lets take 11 (3 wins and 2 losses) for this there are actually 10 possible outcomes that you are discounting as one again -LLWWW, WLLWW, WWLLW, WWWLL, LWLWW, LWWLW, LWWWL, WLWWL, WLWLW, WWLWL, Your reasoning would be correct if all of your 6 options had the same number of ways to get it but it is not the case Math Expert Joined: 02 Sep 2009 Posts: 94776 Own Kudos [?]: 646253 [0] Given Kudos: 86843 Re: M05-17 [#permalink] Expert Reply I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand. Manager Joined: 08 Nov 2022 Posts: 76 Own Kudos [?]: 36 [0] Given Kudos: 56 Re: M05-17 [#permalink] bsattin1 wrote: Bunuel, my approach to this problem was as follows: There is an equal probability that Kate or David will end up with more money by the end (and note there is no situation where Kate and David both end up with$10). Therefore, we can subtract the probability that Kate wins all 5 flips from 1/2 to get the probability:

combinations = 2^5 = 32
Probability of winning all 5 flips = 1/32
Probability Kate ends up with $11-$14 = 1/2 - 1/32 = 15/32

If this game had been flipping an even number of coins, you'd have to do 1/2 - 1/64 - (probability of same number heads and tails).

Cheers,

Why are we Subtracting $$\frac{1}{2}$$ in this equation?
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Re: M05-17 [#permalink]
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sanchitb23 wrote:
bsattin1 wrote:
Bunuel, my approach to this problem was as follows:

There is an equal probability that Kate or David will end up with more money by the end (and note there is no situation where Kate and David both end up with $10). Therefore, we can subtract the probability that Kate wins all 5 flips from 1/2 to get the probability: combinations = 2^5 = 32 Probability of winning all 5 flips = 1/32 Probability Kate ends up with$11-\$14 = 1/2 - 1/32 = 15/32

If this game had been flipping an even number of coins, you'd have to do 1/2 - 1/64 - (probability of same number heads and tails).

Cheers,

Why are we Subtracting $$\frac{1}{2}$$ in this equation?

In a given scenario, Kate wins in half of the cases, and David wins in the other half. Out of the cases in which Kate wins (1/2), we want to exclude the instance where she wins all five times (1/2^5). Therefore, the probability we are looking for is 1/2 - 1/32.
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