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 \(\frac{5!}{3!2!}\).

Hence \(P=\frac{5!}{3!2!}*(\frac{1}{2})^5\).

Check this links for similar problems:
viewtopic.php?f=140&t=96468&p=742767&hilit=+google#p742767
viewtopic.php?f=140&t=56812&hilit=+probability+occurring+times
viewtopic.php?f=140&t=88069&hilit=+probability+occurring+times
viewtopic.php?f=140&t=87673&hilit=+probability+occurring+times

Also you can check Probability chapter of Math Book for more (link in my signature).

Hope it helps.
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Thanks for the explanation

Please check this;

http://gmatclub.com/forum/m05-q17-probability-combination-70384.html

Harriet has won, or that 3 <= # of head <=4 because only in those cases Harriet will have Money > 10 and < 15

So Prpb = 5C3(1/2)^3* (1/2)^2 + 5C4(1/2)^4* (1/2)^1

= (1/2)^5{(5 * 4)/2 + 5} = 15 * (1/2)^5 = 15/32

Answer is D.
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Two options for Kate to get >10 and <15 dollars

Lose (times): 1, 2

Win (times): 4, 3

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

Second option : 1/2 (lose)*1/2 (lose)*1/2 (win)*1/2 (win)*1/2 (win)=1/32.

we have 5!/3!*2!=10 such cases. So, (1/32)*10=10/32

5/32+10/32=15/32

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.
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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.

Final Answer:

GMAT assassins aren't born, they're made,
Rich
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Hi Rich,

I'm confused about something.

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:

https://gmatclub.com/forum/joe-is-among ... 95834.html

I hope I clarified my point and where the confusion happened.

Thanks

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
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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?

Posted from my mobile device

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.

GMAT assassins aren't born, they're made,
Rich
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Bunuel, can you pls explain me why are we using combination here, the order in which the tail or head falls matters right, say for example.,
THHHH
HTHHH
HHTHH
HHHTH
HHHHT
the oder matters here right, so we should use permutation right, pls anyone explain me where im going wrong...

Hi monikakumar,

Most GMAT questions can be approached in more than one way - and you'll find that many questions that can be treated as Combination or Permutation questions can also be solved with the other approach.

In this question, since there are 32 ways to flip the 5 coins - and we're looking for all of the possible outcomes that involve 3 OR 4 tails, mapping out all of the outcomes that fit that situation could take some time (there are actually 15 of them). The "order" of the tosses doesn't ultimately matter though - as long as we end up with either 3 or 4 tails - so doing two separate Combination Formula calculations would likely be faster.

GMAT assassins aren't born, they're made,
Rich
_________________

Given:
1. Kate and David each have $10. Together they flip a coin 5 times.
2. Every time the coin lands on heads, Kate gives David $1. Every time the coin lands on tails, David gives Kate $1.

Asked: After the coin is flipped 5 times, what is the probability that Kate has more than $10 but less than $15?

Kate has more than $10 but less than $15 means Kate wins more than David
Number of times Kate wins = 3 or 4 = 5C3 + 5C4 = 10 + 5 = 15

Total number of ways = 2^5 = 32

The probability that Kate has more than $10 but less than $15 = 15/32

IMO B
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EMPOWERgmatRichC, I could see the order matters in which head or tail falls..my doubt is if order matters, then we should use permutation right, why are we using combination...pls help me on this if im wrong

Hi monikakumar,

In this prompt, after each coin toss, we either add $1 or subtract $1 from Kate's total. Since that's just basic arithmetic, the order of the individual results does NOT matter.

For example, the order TTTHH and HHTTT both end up with Kate gaining a total of $1. Again, the order of the individual tosses does NOT matter as long as the overall result fits what we're looking for (in this case, either 3 or 4 'tails.'), so this math can be handled using the Combination Formula (once for 5c3 and once for 5c4).

GMAT assassins aren't born, they're made,
Rich
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