Kate and Danny each have $10. Together, they flip a fair coin 5 times. : Problem Solving (PS)

powerka

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Kate and Danny each have $10. Together, they flip a fair coin 5 times. [#permalink]

Updated on: 17 Feb 2015, 12:34

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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

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Originally posted by powerka on 22 Sep 2009, 14:36.
Last edited by Bunuel on 17 Feb 2015, 12:34, edited 1 time in total.

Renamed the topic, edited the question and added the OA.

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dimitri92

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Re: Kate and Danny each have $10. Together, they flip a fair coin 5 times. [#permalink]

01 Mar 2011, 05:28

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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 $= P(3) + P(4) = \frac{15}{32}$

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Bunuel

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Kate and Danny each have $10. Together, they flip a fair coin 5 times. [#permalink]

30 Oct 2014, 09:55

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21. Combinatorics/Counting Methods

Theory
Math: Combinatorics
Combinatorics Made Easy!
1 hour Combinatorics Video Lesson
Combinations while dealing with similar and dissimilar things
When Permutations & Combinations and Data Sufficiency Come Together

Questions
Advanced Topic: Probability and Combinations Questions With Solutions
Letter Arrangements in Words
Seating Arrangements in a Row and around a Table
Constructing Numbers, Codes and Passwords
DS Questions
PS Questions

22. Probability

Theory
Math: Probability
Probability Made Easy!
GMAT Tip of the Week: 99 Problems But Probability Ain't One

Questions
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Advanced Topic: Probability and Combinations Questions With Solutions
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For more:
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Re: Kate and Danny each have $10. Together, they flip a fair coin 5 times. [#permalink]

19 Oct 2011, 11:20

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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. :-D

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 :-D

manpreetsingh86

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Re: Kate and Danny each have $10. Together, they flip a fair coin 5 times. [#permalink]

31 Oct 2014, 02:56

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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

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Re: Kate and Danny each have $10. Together, they flip a fair coin 5 times. [#permalink]

06 Feb 2019, 19:32

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powerka wrote:

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

In order for Kate to have more than 10 dollars but less than 15 dollars, either of the following two outcomes must have occurred:

T-T-T-T-H, so Kate would have 13 dollars

T-T-T-H-H, so Kate would have 11 dollars

Let’s calculate the probability of each outcome:

P(T-T-T-T-H) = (1/2)^5 = 1/32

Since T-T-T-T-H can be arranged in 5!/4! = 5 ways, the probability is 5/32.

Next:

P(T-T-T-H-H) = (1/2)^5 = 1/32

Since T-T-T-H-H can be arranged in 5!/(3! x 2!) = 10 ways, the probability is 10/32.

So the overall probability that Kate has more than $10 but less than $15 is 15/32.

Answer: D
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powerka

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Re: Kate and Danny each have $10. Together, they flip a fair coin 5 times. [#permalink]

22 Sep 2009, 14:40

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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

Ramyahari

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Re: Kate and Danny each have $10. Together, they flip a fair coin 5 times. [#permalink]

30 Jul 2018, 22:15

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My explanation is
Total outcomes of tossing a coin =2^5 =32

So the probability is exactly half times head and half time tails...i.e.16/32
And since it is asked for 10-15
Answer is 15/32.

Is my approach right??

Posted from my mobile device

deep5586

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Re: Kate and Danny each have $10. Together, they flip a fair coin 5 times. [#permalink]

19 Oct 2011, 09:26

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y arent we taking into account the probablility of 2 heads or 1

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Hector1S

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Re: Kate and Danny each have $10. Together, they flip a fair coin 5 times. [#permalink]

22 Jun 2018, 23:49

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Can anybody explain why the total no.of possibilities is 32 and not 10.

D

exc4libur

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Re: Kate and Danny each have $10. Together, they flip a fair coin 5 times. [#permalink]

23 Aug 2019, 06:10

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powerka wrote:

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

cases in which $10<total<$15:
WWWLL=+3-2=+1… total=$11… prob: $3W2L=(1/2)^5•{arrangements}\frac{5!}{3!2!}=\frac{10}{32}$
WWWWL=+4-1=+3… total=$13… prob: $4W1L=(1/2)^5•{arrangements}\frac{5!}{4!1!}=\frac{5}{32}$
sum cases: $\frac{10}{32}+\frac{5}{32}=\frac{15}{32}$

Answer (D)

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Re: Kate and Danny each have $10. Together, they flip a fair coin 5 times. [#permalink]

29 Jun 2015, 07:45

Solved it using Pascal's triangle method and able to got the answer in 30 sec. https://en.wikipedia.org/wiki/Pascal's_triangle

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kunalc20

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Re: Kate and Danny each have $10. Together, they flip a fair coin 5 times. [#permalink]

06 Jun 2020, 12:54

manpreetsingh86 wrote:

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

Why do we need to consider the arrangement here
Please help. its all about getting heads and tails . We need to have 4 heads or 3 heads for kate to have money more than 10 and less than 15
doesnt matter at which turn we are getting head or tail

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Re: Kate and Danny each have $10. Together, they flip a fair coin 5 times. [#permalink]

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