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Kate and David each have $10. Together they flip a coin 5
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Updated on: 13 Apr 2012, 13:52
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Difficulty:
65% (hard)
Question Stats:
61% (02:42) correct 39% (02:38) wrong based on 188 sessions
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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. \(\frac{11}{16}\)
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 $15? a. 5/16 b. 15/32 c. 1/2 d. 21/32 e. 11/16
After 5 tries Kate to have more than initial sum of 10$ and less than 15$ must win 3 or 4 times (if she wins 2 or less times she'll have less than 10$ and if she wins 5 times she'll have 15$).
So the question becomes "what is the probability of getting 3 or 4 tails in 5 tries?".
\(P(t=3 \ or \ t=4)=P(t=3)+P(t=4)=C^3_5*(\frac{1}{2})^5+C^4_5*(\frac{1}{2})^5=\frac{15}{32}\)
Answer: B.
To elaborate more:
If the probability of a certain event is \(p\), then the probability of it occurring \(k\) times in \(n\)-time sequence is: \(P = C^k_n*p^k*(1-p)^{n-k}\)
For example for the case of getting 3 tails in 5 tries: \(n=5\) (5 tries); \(k=3\) (we want 3 tail); \(p=\frac{1}{2}\) (probability of tail is 1/2).
So, \(P = C^k_n*p^k*(1-p)^{n-k}=C^3_5*(\frac{1}{2})^3*(1-\frac{1}{2})^{(5-3)}=C^3_5*(\frac{1}{2})^5\)
OR: probability of scenario t-t-t-h-h is \((\frac{1}{2})^3*(\frac{1}{2})^2\), but t-t-t-h-h can occur in different ways:
t-t-t-h-h - first three tails and fourth and fifth heads; h-h-t-t-t - first two heads and last three tails; t-h-h-t-t - first tail, then two heads, then two tails; ...
Certain # of combinations. How many combinations are there? Basically we are looking at # of permutations of five letters t-t-t-h-h, which is \(\frac{5!}{3!2!}\).
Re: Kate and David each have $10. Together they flip a coin 5
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07 Jun 2016, 09:04
Temurkhon wrote:
First option: 1/2 (win)*1/2 (lose)*1/2 (lose)*1/2 (lose)*1/2 (lose)=1/32.
we have 5!/4!*1!=5 such cases. So, (1/32)*5=5/32
B
Could anyone please shed me some light on this part? Why is it that we have to count only the number of cases we can arrange winnings and not just 5! ? Kindly thanks!
Re: Kate and David each have $10. Together they flip a coin 5
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08 Jun 2016, 02:52
fantaisie wrote:
Temurkhon wrote:
First option: 1/2 (win)*1/2 (lose)*1/2 (lose)*1/2 (lose)*1/2 (lose)=1/32.
we have 5!/4!*1!=5 such cases. So, (1/32)*5=5/32
B
Could anyone please shed me some light on this part? Why is it that we have to count only the number of cases we can arrange winnings and not just 5! ? Kindly thanks!
We are arranging 1 win and 4 loses (WLLL). WLLL can be arranged in 5!/4! number of ways.
_________________
Re: Kate and David each have $10. Together they flip a coin 5
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15 Feb 2018, 13:27
Hi All,
This is a layered probability question that requires some additional math (the combination formula, or the equivalent "mapping" of all possibilities).
We're told that the coin will be flipped 5 times; with every "win", Kate gets $1 and with every "loss", Kate loses $1. Kate starts with $10. We're asked for the probability that Kate ends up with MORE than $10 but less than $15 after 5 tosses.
Let's start with the total number of possible outcomes. Since each coin has 2 options (heads or tails), there are 2^5 = 32 possible outcomes for the 5 flips (which will include a certain number of similar outcomes in different orders - for example HHTTT and THTHT).
To end up with MORE than $10, Kate has to win MORE tosses than she loses. However, if she were to win all 5 tosses, she'd have $15 and we want her to end up with LESS than $15. This means that Kate has to win EITHER 3 times or 4 times.
Since it does not matter which of the 5 tosses is won, as long as it's either 3 or 4 of them, we can use the combination formula:
Combinations = N!/[K!(N-K)!]
For 3 wins, we have 5!/3!2! = 10 possible combinations of 3 wins
For 4 wins, we have 5!/4!1! = 5 possible combinations of 4 wins
So, there are a total of 10+5 = 15 combinations of 5 tosses that "fit" what we're looking for and 32 possible outcomes total.
Kate and David each have $10. Together they flip a coin 5
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24 May 2019, 21:14
EMPOWERgmatRichC wrote:
Hi All,
This is a layered probability question that requires some additional math (the combination formula, or the equivalent "mapping" of all possibilities).
We're told that the coin will be flipped 5 times; with every "win", Kate gets $1 and with every "loss", Kate loses $1. Kate starts with $10. We're asked for the probability that Kate ends up with MORE than $10 but less than $15 after 5 tosses.
Let's start with the total number of possible outcomes. Since each coin has 2 options (heads or tails), there are 2^5 = 32 possible outcomes for the 5 flips (which will include a certain number of similar outcomes in different orders - for example HHTTT and THTHT).
To end up with MORE than $10, Kate has to win MORE tosses than she loses. However, if she were to win all 5 tosses, she'd have $15 and we want her to end up with LESS than $15. This means that Kate has to win EITHER 3 times or 4 times.
Since it does not matter which of the 5 tosses is won, as long as it's either 3 or 4 of them, we can use the combination formula:
Combinations = N!/[K!(N-K)!]
For 3 wins, we have 5!/3!2! = 10 possible combinations of 3 wins
For 4 wins, we have 5!/4!1! = 5 possible combinations of 4 wins
So, there are a total of 10+5 = 15 combinations of 5 tosses that "fit" what we're looking for and 32 possible outcomes total.
there are 2^5 = 32 possible......this means the order is important (matter) so we deal with permutation in nature.
But you used combinations for 5C3 & 5C4.
How we can divide combinations over permutation? as As far as I know, we must either deal with all combinations (in numerator and denominator) or all permutations (in numerator and denominator).
My questions to you stems from a reply to me from another tutor/expert in another question as below:
Re: Kate and David each have $10. Together they flip a coin 5
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26 May 2019, 08:01
Hi Mo2men,
With 32 total outcomes, we need an efficient way to 'count up' all of the options that fit what we're looking for (re: 3 wins or 4 wins). I used the Combination Formula to quickly get those totals (instead of trying to 'map out' every individual win).
From your question, I think you're focusing too much on the individual formulas instead of what the formulas represent in the context of this question. Since this is a PROBABILITY question, we need the total number of ways that are POSSIBLE and the total number of ways that we WANT to occur. I used a Permutation to find the total possibilities and the Combination Formula (twice) to find the number of ways that match what we want to occur.
Kate and David each have $10. Together they flip a coin 5
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23 Jul 2019, 13:23
EMPOWERgmatRichC wrote:
Hi Mo2men,
With 32 total outcomes, we need an efficient way to 'count up' all of the options that fit what we're looking for (re: 3 wins or 4 wins). I used the Combination Formula to quickly get those totals (instead of trying to 'map out' every individual win).
From your question, I think you're focusing too much on the individual formulas instead of what the formulas represent in the context of this question. Since this is a PROBABILITY question, we need the total number of ways that are POSSIBLE and the total number of ways that we WANT to occur. I used a Permutation to find the total possibilities and the Combination Formula (twice) to find the number of ways that match what we want to occur.
GMAT assassins aren't born, they're made, Rich
I got favourable number of outcomes that is 15 but why aren't we doing 10C5, for total number of outcomes. I applied 5C3 +5C4/ 10C5. Can you please explain?
Re: Kate and David each have $10. Together they flip a coin 5
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23 Jul 2019, 17:17
1
Hi kinshuk97gupta,
It might help to think in terms of some simpler examples:
If we flip a coin 1 time, then there are two possible outcomes: H or T --> that's 2^1 If we flip a coin 2 times, then there are four possible outcomes: HH, HT, TH or TT --> that's 2^2
If we flip a coin 3 times, then there are eight possible outcomes: HHH HHT HTH THH TTT TTH THT HTT --> that's 2^3
Based on this pattern, you can probably determine that since we're flipping a coin 5 times, then that's 2^5 = 32 possible outcomes. Since this is a probability question, 32 has to be in the denominator.
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61% (02:42) correct 
