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

Question

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

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

Source: Manhattan GMAT Archive (tough problems set).doc

dimitri92 · Accepted Answer

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}

Bunuel · Answer

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For more:
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Hope it helps.

powerka · Answer

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

deep5586 · Answer

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

heman2402 · Answer

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 · Answer

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

nitinagg29 · Answer

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

Hector1S · Answer

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

Ramyahari · Answer

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

ScottTargetTestPrep · Answer

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

exc4libur · Answer

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)

kunalc20 · Answer

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

adityaprateek15 · Answer

To solve this question, the key is understanding the impact of each coin flip:

If Kate  wins once , she  also loses four times  (since there are 5 total flips), which results in:
10+1−4=7

So, each win adds +1 to her amount, and each loss subtracts 1.
We’re told that Kate starts with $10 and we want the total after 5 flips to be  more than 10 but less than 15 .
 This means her net gain must be  between 1 and 4 , i.e., she must win more times than she loses.

So, the only conditions are:

1.Kate wins 3 times and looses 2 other times:5c3 x (1/2)^3 x (1/2)^2 = 10/32

2.Kate wins 4 times and loose 1 other time: 5c4 x (1/2)^4 x (1/2)^1 = 5/32

Total = 10+5/32 = 15/32

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

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

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