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# Since 1 appears exactly three times, we can solve for the ot

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Joined: 08 Mar 2013
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Kudos [?]: 1 [0], given: 7

Since 1 appears exactly three times, we can solve for the ot [#permalink]  06 Apr 2013, 22:24
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I'm working thru PS problems and came upon this one:

John wrote a phone number on a note that was later lost. John can remember that the number had 7 digits, the digit 1 appeared in the last three places and 0 did not appear at all. What is the probability that the phone number contains at least two prime digits?

Here's my issue with this problem and similar ones. Hope somebody can help me grasp this.

The posted solution is this:
Since 1 appears exactly three times, we can solve for the other four digits only. For every digit we can choose out of 8 digits only (without 1 and 0). Since we have 4 prime digits (2, 3, 5, 7) and 4 non-prime digits (4, 6, 8, 9), the probability of choosing a prime digit is ½.
We need at least two prime digits:
One minus (the probability of having no prime digits + having one prime digit):
There are 4 options of one prime digit, each with a probability of (1/2)4.
There is only one option of no prime digit with a probability of (1/2)4.
So: [1- ((1/2)4+(1/2)4*4)] = 11/16.

I understand how they calculated this, but here's what I don't get: say for one prime digit scenario, the solution calculates different arrangements, as in prob of 4*(1/2)^4. I get the 1/2^4 as being the probability of getting one prime number and three non-prime numbers. But why take into account the arrangements by multiplying by 4? Whether the prime number comes in position 1, 2, 3 or 4, the possibility of getting one prime and three non-prime is the same, 1/16 regardless of the order, no? Same with probability of no primes, 1/16, regardless of the order.

So wouldn't the probability of having at least two primes be 1-(1/16+1/16)=1-2/16=1-1/8=7/8?
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Kudos [?]: 1522 [0], given: 219

Re: Probability of a number when order doesn't matter [#permalink]  07 Apr 2013, 00:19
Dixon wrote:

I understand how they calculated this, but here's what I don't get: say for one prime digit scenario, the solution calculates different arrangements, as in prob of 4*(1/2)^4. I get the 1/2^4 as being the probability of getting one prime number and three non-prime numbers. But why take into account the arrangements by multiplying by 4? Whether the prime number comes in position 1, 2, 3 or 4, the possibility of getting one prime and three non-prime is the same, 1/16 regardless of the order, no? Same with probability of no primes, 1/16, regardless of the order.

So wouldn't the probability of having at least two primes be 1-(1/16+1/16)=1-2/16=1-1/8=7/8?

You have to multiply by 4 $$\frac{1}{2}^4$$ in order to consider all the combinations.
P,N,N,N - N,P,N,N - N,N,P,N - N,N,N,P - each of those has a probability of $$\frac{1}{2}^4$$ as you said, but there are 4 combinations in which we can find only one prime. Thats why you multiply by 4.
"Whether the prime number comes in position 1, 2, 3 or 4, the possibility of getting one prime and three non-prime is the same, 1/16 regardless of the order, no? Same with probability of no primes, 1/16, regardless of the order."
So, the answer to your question is : the order is important.
You dont multiply the probability of no primes by 4, because there is only one way to get it N,N,N,N!

Hope its clear
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Re: Probability of a number when order doesn't matter   [#permalink] 07 Apr 2013, 00:19
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