GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 22 Feb 2019, 11:07

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

## Events & Promotions

###### Events & Promotions in February
PrevNext
SuMoTuWeThFrSa
272829303112
3456789
10111213141516
17181920212223
242526272812
Open Detailed Calendar
• ### Free GMAT RC Webinar

February 23, 2019

February 23, 2019

07:00 AM PST

09:00 AM PST

Learn reading strategies that can help even non-voracious reader to master GMAT RC. Saturday, February 23rd at 7 AM PT
• ### FREE Quant Workshop by e-GMAT!

February 24, 2019

February 24, 2019

07:00 AM PST

09:00 AM PST

Get personalized insights on how to achieve your Target Quant Score.

# M05-17

 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics
Author Message
TAGS:

### Hide Tags

Math Expert
Joined: 02 Sep 2009
Posts: 53066

### Show Tags

15 Sep 2014, 23:25
00:00

Difficulty:

75% (hard)

Question Stats:

56% (01:52) correct 44% (02:05) wrong based on 147 sessions

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. $$\frac{11}{16}$$ _________________ Math Expert Joined: 02 Sep 2009 Posts: 53066 Re M05-17 [#permalink] ### Show Tags 15 Sep 2014, 23:25 2 3 Official Solution: 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. $$\frac{11}{16}$$

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:

$$C_5^3 = 10$$

Number of combinations for winning 4 times:

$$C_5^4 = 5$$

Remember, $$C_n^k = \frac{n!}{k!*(n-k)!}$$.

Probability equals $$\frac{5+10}{32} = \frac{15}{32}$$.

Approach 2: write out the combinations:

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

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

Total: $$\frac{15}{32}$$.

The combinations can be summed because they have equal probabilities of $$\frac{1}{32}$$ each.

_________________
Manager
Joined: 20 Jan 2014
Posts: 142
Location: India
Concentration: Technology, Marketing

### Show Tags

02 Jan 2015, 09:17
Bunuel wrote:
Official Solution:

Approach 1: Find the total number of combinations = $$2^5 = 32$$. F: 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.

I don't understand this: Even if tail comes one time then Kate can have 11$_________________ Consider +1 Kudos Please Board of Directors Joined: 17 Jul 2014 Posts: 2587 Location: United States (IL) Concentration: Finance, Economics GMAT 1: 650 Q49 V30 GPA: 3.92 WE: General Management (Transportation) M05-17 [#permalink] ### Show Tags 02 Jan 2015, 09:22 him1985 wrote: Bunuel wrote: Official Solution: Approach 1: Find the total number of combinations = $$2^5 = 32$$. F: 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 $10. Answer: B I don't understand this: Even if tail comes one time then Kate can have 11$

read the question attentively; the coin is flipped 5 times!
in case the tail comes once, and head 4 times, then she will have <10$if Tails comes up 5 times, she will have 15$ - not what we are looking for
if tails comes up 2 times, she will have 9$the only way to get: 10<X<15 where X is Kate's$ amount
is when we have at least 3 tails or 4 tails
Intern
Joined: 30 Dec 2011
Posts: 5
Schools: Haas '15

### Show Tags

15 Jan 2015, 20:08
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
Intern
Joined: 09 Feb 2015
Posts: 20

### Show Tags

16 Apr 2015, 02:41
Dear Bunuel!
Please explain why the total number of combinations = 25^5 ?
I don't understand. Tks
Math Expert
Joined: 02 Sep 2009
Posts: 53066

### Show Tags

16 Apr 2015, 03:22
2
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.
_________________
Intern
Joined: 22 Apr 2015
Posts: 11
GMAT 1: 760 Q50 V44

### Show Tags

04 Sep 2015, 04:55
1
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, Intern Joined: 21 Jul 2015 Posts: 34 Re: M05-17 [#permalink] ### Show Tags 04 Sep 2015, 06:30 1 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
Intern
Joined: 24 Jun 2012
Posts: 10

### Show Tags

19 Sep 2015, 10:00
Hello . Is this approach correct?

two possibilities need to happen for Kate for the given condition

Scenario 1 - Win Win Win Lose Lose (or) Scenario 2 - Win Win Win Win Lose

Scenario 1 : possible combinations for 3 wins out of 5 = 5!/3!*2! = 10
so 10 * ( 1/2*1/2*1/2*1/2*1/2) = 10/32 - (each 1/2 is the prob for a lose or a win)

similarly for Scenario 2 : Possible ways are 5!/4!*1! = 5 ways
so 5 * (1/2*1/2*1/2*1/2*1/2) = 5/32

so we need scenario 1 OR scenario 2
so 10/32 + 5/32 = 15/32
Senior Manager
Joined: 31 Mar 2016
Posts: 383
Location: India
Concentration: Operations, Finance
GMAT 1: 670 Q48 V34
GPA: 3.8
WE: Operations (Commercial Banking)

### Show Tags

23 Aug 2016, 05:01
I think this is a high-quality question and I agree with explanation.
Current Student
Joined: 26 Aug 2015
Posts: 33
Concentration: Strategy, Economics
GMAT 1: 570 Q40 V28
GMAT 2: 740 Q49 V41

### Show Tags

26 Aug 2016, 09:08
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

Hi, could you clarify something for me?

If they both have 50% of winning 2 and loosing 2, why do you conclude that she has 50% chance of winning 3/5? It seems to work but I don't understand why. Its like ignoring the first 4 flips and just using the 50% of the fifth (which, if correct, would be very smart). It can be said that she has 50% of winning 2 / 4, then she needs to win the next, wouldn't that require an AND condition and a multiplication of 1/2*1/2?

Greetings.
_________________

Send some kudos this way if I was helpful! !

Manager
Joined: 17 Aug 2015
Posts: 96
GMAT 1: 650 Q49 V29

### Show Tags

30 Aug 2016, 14:08
it is not a question of combination but of permutation

SO total possibility is 2 to the power 5 ( one coin can be either H or T SO five times tossed = 2*2*2*2*2*2= 32
Kate should get between 10 and 15 dollar 10<k<15
if either HHTTT ( OVERALL ONE DOLLAR PLUS MEANS 11 DOLLARS

if 02 head and 3 tails total possibility ( here order matters so that is why i said it is not a combination question= factorial 5/ factorial 2 * factorial 3 = 10

second scenario for four tails and one head- means 03 dollars net gain so , she will be having 13 dollars

OR HTTTT, THTTT, TTHTT, TTTHT, TTTTH = 05 CASES
============================
S0 TOTAL = 10+5 = 15

PROABILITY 15/32
Current Student
Joined: 22 Jan 2016
Posts: 12
Location: Thailand
Concentration: Finance, Statistics
GMAT 1: 640 Q46 V31
GMAT 2: 620 Q50 V25
GMAT 3: 650 Q50 V28
GPA: 2.89
WE: Accounting (Consulting)

### Show Tags

20 Sep 2016, 10:05
Can you explain why my method is wrong:

{(3c5)(2c5) + (4c5)(1c5)} / (5c10)
Retired Moderator
Joined: 25 Feb 2013
Posts: 1217
Location: India
GPA: 3.82

### Show Tags

16 Jan 2017, 10:34
For Kate this is only possible when below flips occur-

1) TTTTH or (no of ways this is possible = 5!/4!)
2) TTTHH (no of ways this is possible = 5!/3!*2!)

Total no of outcomes in 5 flips (either head or tails) = 2^5 = 32

Probabilities for each scenario -
(5!/4!)/32 + (5!/3!2!)/32 = 5/32 + 10/32 = 15/32
Intern
Joined: 17 Aug 2016
Posts: 48

### Show Tags

06 Feb 2017, 10:48

For Kate to have more than 10, then the number of T needs to be bigger than number of H in the series. But since we want Kate to have less than 15 T cannot be 5.

So probability T>H minus probability 5T. Since we have an odd number of flips (no the possibility for a tie), then P(T>H)=1/2. P(5T)=1/32

1/2-1/32= 15/32
Intern
Joined: 06 Jun 2018
Posts: 45
Location: India
Schools: IIMA PGPX"20 (D)
GMAT 1: 640 Q47 V30
GMAT 2: 540 Q42 V22
GPA: 3.7

### Show Tags

09 Oct 2018, 07:24
Bunuel wrote:
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.

Bunuel - 2 sides and 5 flips so it should 10 ? I know I am asking basic question but can you help to put the same 32 or 10 [ my logic] in terms of nCr terms ? I understand that better.
Re: M05-17   [#permalink] 09 Oct 2018, 07:24
Display posts from previous: Sort by

# M05-17

 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics

Moderators: chetan2u, Bunuel

 Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.