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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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13 Jul 2008, 17:58
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Quote:
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. 
14 Jul 2008, 11:09
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My answer is 0.5 (c)
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
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
