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M05-17

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New post 16 Sep 2014, 00:25
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Question Stats:

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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. \(\frac{11}{16}\)

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New post 16 Sep 2014, 00:25
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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.


Answer: B
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New post 02 Jan 2015, 10:17
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.

Answer: B


I don't understand this: Even if tail comes one time then Kate can have 11$
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New post 02 Jan 2015, 10:22
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
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New post 15 Jan 2015, 21:08
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
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New post 16 Apr 2015, 03:41
Dear Bunuel!
Please explain why the total number of combinations = 25^5 ?
I don't understand. Tks
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New post 04 Sep 2015, 05:55
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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).

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New post 04 Sep 2015, 07:30
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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
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New post 19 Sep 2015, 11:00
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
OR = addition
so 10/32 + 5/32 = 15/32
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New post 23 Aug 2016, 06:01
I think this is a high-quality question and I agree with explanation.
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Re: M05-17  [#permalink]

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New post 26 Aug 2016, 10:08
MBAUsername wrote:
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.
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New post 30 Aug 2016, 15:08
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
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Re: M05-17  [#permalink]

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New post 20 Sep 2016, 11:05
Can you explain why my method is wrong:

{(3c5)(2c5) + (4c5)(1c5)} / (5c10)
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New post 16 Jan 2017, 11:34
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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
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New post 06 Feb 2017, 11:48
Hey, what about this approach:

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
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Re: M05-17  [#permalink]

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New post 09 Oct 2018, 08:24
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.
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New post 05 May 2019, 10:19
I think this is a high-quality question and I agree with explanation
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New post 03 Sep 2019, 07:47
niks18 wrote:
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


Yup that is what I did too. I thought to type it but then thought someone must have already done that :)
Thank you!
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