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# m05 questions 17 - someone please explain

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13 Jul 2008, 13:43
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Can someone explain this questions to me. I dont understand how they get the 32 and the combinations?
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Senior Manager
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13 Jul 2008, 17:58
Quote:
Can someone explain this questions to me. I dont understand how they get the 32 and the combinations?

OK, I’ll try.

Let’s begin with general case.

Imagine that a fair coin is tossed N times. Outcome of each toss is either tails or heads.
After N tosses we will get a sequence of heads and tails – something like that:

TTHHHHTTHTHH… etc.

Now the question is: What is the probability that from N tosses, there will be exactly K tails?

To answer this question, we need, first, to calculate the number of all possible outcomes of the experiment – namely, the number of different sequences of heads and tails. It’s easy to see that this number equals 2^N (we have N tosses, and each toss has two possible outcomes).

Next, we need to calculate how many of those sequences have exactly K number of tails – and the answer is C(K,N). I’ve omitted the reasoning behind the calculation here.

So, overall, the probability of getting exactly K tails from N tosses is C(K,N)/2^n.

Now, let’s return to our problem. Kate will have more than 10\$ but less than 15\$ if she wins 3 or 4 times. This means we need to calculate probability(3 tails from 5) = C(3,5)/(2^5) and probability(4 tails from 5) = C(4,5)/(2^5) and then sum these probabilities. And the solution provided with the problem does exactly this.

Well, this is it. I hope this explanation was helpful.

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14 Jul 2008, 11:09

I got it in two ways:

1) The outcome can be greater than 10 and smaller than or equal to 15 if- (kate,david)={(3,2),(4,1),(5,0)}

The outcome of it not happening is - (kate,david)= {(2,3),(1,4),(0,5), exactly the opossit.

The two are symmetrical events that constitute 100% of the possible events, so each one of them has 50% chance.

2) the chance of (kate,david)={(3,2),(4,1),(5,0)} is:

(binomial distribution)

(5c3)*(0.5)^5 + (5c4)*(0.5)^5 + (5c5)*(0.5)^5= 10/32 + 5/32 + 1/32= 16/32 =0.5

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14 Jul 2008, 16:27
dariusbanasik wrote:
Can someone explain this questions to me. I dont understand how they get the 32 and the combinations?

At the end of 5 coin flips, Kate needs to have a total amount in the set {11, 12, 13, 14}
For every loss, she loses a dollar; for every win, she gains a dollar

From a total of 5 flips:

If she wins 1 flip, she loses 4 flips -> net amount 10 + 1 - 4 = \$7
If she wins 2 flips, she loses 3 flips -> net amount 10 + 2 - 3 = \$9

If she wins 3 flips, she loses 2 flips -> net amount 10 + 3 - 2 = \$11
If she wins 4 flips, she loses 1 flip -> net amount 10 + 4 - 1 = \$13

If she wins all 5 flips -> net amount 10 + 5 = \$15

Clearly, she needs to win 3 or 4 times to meet the condition \$10 < x < \$15

To calculate the probability of 3 wins, imagine the positions of the 5 flips as XXXXX
Total outcomes = 2*2*2*2*2 = 32, as each X could be a win or loss
Total ways of winning exactly 3 times = 5C3 = 10

Therefore,
P(3 wins) = 5C3 / 2*2*2*2*2 = 10/32
P(4 wins) = 5C4 / 2*2*2*2*2 = 5/32
P(3 or 4 wins) = 10/32 + 5/32 = 15/32
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Re: m05 questions 17 - someone please explain   [#permalink] 14 Jul 2008, 16:27
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