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m05 Q17 Probability/Combination

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Joined: 04 Jul 2013
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Location: India
Concentration: Operations, Technology
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Re: m05 Q17 Probability/Combination [#permalink] New post 01 Aug 2013, 10:07
The answer is 15/32 which comes when we consider the two cases inwhich Kate can actually get a total of more than 10 and less than 15...these are when Kate gets 3 tosses in her favour and 2 against her ie a total of 11...similarly when she gets 4 in her favour and 1 against ie total of 13....
Note a total of 12 and 14 is not possible....so a total of 15 possibilities of 32 are there....


Thanks....

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Re: m05 Q17 Probability/Combination [#permalink] New post 11 Aug 2013, 01:08
Expert's post
d2touge 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}

[Reveal] Spoiler: OA
B

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So to answer this, we find the total combinations when you flip the coin 5 times. --> 2^5 = 32.

The explanation tells us to find the combinations when kate wins 3 times and 4 times. --> 5C3 and 5C4. we get 10 and 5 respectively.

Lastly, we simply find the probability (5/32) + (10/32) = 15/32. and that is the answer..

***Why do we only find the combinations when Kate wins 3 times and 4 times? Why not find 1 and 2 times?


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.

Hope it helps.
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Re: m05 Q17 Probability/Combination   [#permalink] 11 Aug 2013, 01:08
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