Find all School-related info fast with the new School-Specific MBA Forum

 It is currently 03 May 2015, 22:32

### 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 June
Open Detailed Calendar

# Kate and Danny each have $10. Together, they flip a fair coin 5 times.  Question banks Downloads My Bookmarks Reviews Important topics Author Message TAGS: Manager Joined: 22 Jul 2009 Posts: 193 Followers: 4 Kudos [?]: 210 [0], given: 18 Kate and Danny each have$10. Together, they flip a fair coin 5 times. [#permalink]  22 Sep 2009, 13:36
1
This post was
BOOKMARKED
00:00

Difficulty:

75% (hard)

Question Stats:

63% (03:27) correct 37% (01:58) wrong based on 57 sessions
Kate and Danny each have $10. Together, they flip a fair coin 5 times. Every time the coin lands on heads, Kate gives Danny$1. Every time the coin lands on tails, Danny gives Kate $1. After the five coin flips, what is the probability that Kate has more than$10 but less than $15? (A) 5/16 (B) 1/2 (C) 12/30 (D) 15/32 (E) 3/8 Source: Manhattan GMAT Archive (tough problems set).doc [Reveal] Spoiler: OA _________________ Please kudos if my post helps. Last edited by Bunuel on 17 Feb 2015, 11:34, edited 1 time in total. Renamed the topic, edited the question and added the OA.  Manhattan GMAT Discount Codes Kaplan GMAT Prep Discount Codes GMAT Pill GMAT Discount Codes Manager Joined: 22 Jul 2009 Posts: 193 Followers: 4 Kudos [?]: 210 [0], given: 18 Re: Kate and Danny each have$10. Together, they flip a fair coin 5 times. [#permalink]  22 Sep 2009, 13:40
Decided to post this question as Manhattan's explanation was unnecessarily cumbersome.

Question asks for the probability of the coin landing tails up either 3 or 4 times = P(3t) + P(4t)

Binomial distribution formula: nCk p^k (1-p)^(n-k)

P(3t) = 5C3 (1/2)^3 (1/2)^2 = 10 (1/2)^5
P(4t) = 5C4 (1/2)^4 (1/2)^1 = 5 (1/2)^5

=> P(3t) + P(4t) = 15/32
_________________

Please kudos if my post helps.

Senior Manager
Affiliations: SPG
Joined: 15 Nov 2006
Posts: 326
Followers: 11

Kudos [?]: 367 [3] , given: 20

Re: Kate and Danny each have $10. Together, they flip a fair coin 5 times. [#permalink] 01 Mar 2011, 04:28 3 This post received KUDOS Q: what is the probability of getting 3 or 4 heads if a fair coin is flipped 5 times. total number of possibilities $$= 2 * 2 * 2 * 2 * 2 = 2^5$$ $$P(3) = \frac{5C3}{2^5} = \frac{10}{32}$$ $$P(4) = \frac{5C4}{2^5} = \frac{5}{32}$$ Required probability $$= P(3) + P(4) = \frac{15}{32}$$ _________________ press kudos, if you like the explanation, appreciate the effort or encourage people to respond. Download the Ultimate SC Flashcards Intern Joined: 17 Oct 2011 Posts: 2 Concentration: Entrepreneurship, Marketing GMAT Date: 11-25-2011 GPA: 3.9 Followers: 0 Kudos [?]: 0 [0], given: 0 Re: Kate and Danny each have$10. Together, they flip a fair coin 5 times. [#permalink]  19 Oct 2011, 08:26
y arent we taking into account the probablility of 2 heads or 1
Manager
Status: Next engagement on Nov-19-2011
Joined: 12 Jan 2011
Posts: 84
Location: New Delhi, India
Schools: IIM, ISB, & XLRI
WE 1: B.Tech (Information Technology)
Followers: 1

Kudos [?]: 10 [0], given: 5

Re: Kate and Danny each have $10. Together, they flip a fair coin 5 times. [#permalink] 19 Oct 2011, 10:20 deep5586 wrote: y arent we taking into account the probablility of 2 heads or 1 For 2 & 1 heads Kate will end up with <$10 and we want her to win . Therefore, only possibilities are 3 or 4 heads.

I made an educated guess and it worked fine.

Ans- 'D'

MGMAT's anagram helped here as well.

HHHHT = 5!/4!*1! = 5
HHHTT = 5!/3!*2! = 10

Total acceptable cases = 15
Total cases = 32

P = 15/32
_________________

Preparation for final battel:
GMAT PREP-1 750 Q50 V41 - Oct 16 2011
GMAT PREP-2 710 Q50 V36 - Oct 22 2011 ==> Scored 50 in Quant second time in a row
MGMAT---- -1 560 Q28 V39 - Oct 29 2011 ==> Left Quant half done and continued with Verbal. Happy to see Q39

My ongoing plan: http://gmatclub.com/forum/550-to-630-need-more-to-achieve-my-dream-121613.html#p989311

Appreciate by kudos !!

Math Expert
Joined: 02 Sep 2009
Posts: 27170
Followers: 4226

Kudos [?]: 40957 [0], given: 5576

Kate and Danny each have $10. Together, they flip a fair coin 5 times. [#permalink] 30 Oct 2014, 08:55 Expert's post 2 This post was BOOKMARKED Tough and Tricky questions: Probability. Kate and Danny each have$10. Together, they flip a fair coin 5 times. Every time the coin lands on heads, Kate gives Danny $1. Every time the coin lands on tails, Danny gives Kate$1. After the five coin flips, what is the probability that Kate has more than $10 but less than$15?

(A) 5/16
(B) 1/2
(C) 12/30
(D) 15/32
(E) 3/8
_________________
Senior Manager
Joined: 13 Jun 2013
Posts: 275
Followers: 7

Kudos [?]: 171 [2] , given: 12

Re: Kate and Danny each have $10. Together, they flip a fair coin 5 times. [#permalink] 31 Oct 2014, 01:56 2 This post received KUDOS Bunuel wrote: Tough and Tricky questions: Probability. Kate and Danny each have$10. Together, they flip a fair coin 5 times. Every time the coin lands on heads, Kate gives Danny $1. Every time the coin lands on tails, Danny gives Kate$1. After the five coin flips, what is the probability that Kate has more than $10 but less than$15?

(A) 5/16
(B) 1/2
(C) 12/30
(D) 15/32
(E) 3/8

total number of possible outcomes when coin will be flipped 5 times= 2^5=32

now, for kate to have more than $10, she must have min. 3 favorable and max. 4 favorable draws in the 5 throws. (because if she wins in all the 5 draws, then she will have$15)

kate will win, every time tails (T) appears on the coin.

case 1) kate wins in 3 draws and losses in two. so we have 3 T's and 2H's(T,T,T,H,H), which can arrange themselves in 5!/(3!)(2!)=10

case 2) kate wins in 4 draws and losses in the 1 draw. so we have 4 T's and 1H (T,T,T,T,H), which can arrange themselves in 5!/4!=5

thus total no. of favorable ways=10+5=15
and thus required probability=15/32
GMAT Club Legend
Joined: 09 Sep 2013
Posts: 4765
Followers: 295

Kudos [?]: 52 [0], given: 0

Re: Kate and Danny each have $10. Together, they flip a fair coin 5 times. [#permalink] 17 Feb 2015, 11:16 Hello from the GMAT Club BumpBot! Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos). Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________ Re: Kate and Danny each have$10. Together, they flip a fair coin 5 times.   [#permalink] 17 Feb 2015, 11:16
Similar topics Replies Last post
Similar
Topics:
1 Kate and David each have $10. Together they flip a coin 5 6 13 Jul 2010, 08:43 Kate and Danny each have$10. Together, they flip a fair coin 5 times. 0 17 Feb 2015, 11:16
8 KAte and David each have $10. Together they flip a coin 5 2 05 Aug 2008, 10:34 Kate and David each have$10. Together they flip a coin 5 9 06 Mar 2008, 06:33