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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Please check this;

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

We are arranging 1 win and 4 loses (WLLL). WLLL can be arranged in 5!/4! number of ways.
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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