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# I'm having trouble with AT LEAST in combinatorics. we have 4

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CEO
Joined: 21 Jan 2007
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I'm having trouble with AT LEAST in combinatorics. we have 4 [#permalink]

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09 Dec 2007, 13:26
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

I'm having trouble with AT LEAST in combinatorics.

we have 4 slots. There are 8 digits to fill those spots: the digits 2 through 9.

What is the probability that those 4 slots will contain AT LEAST one prime number?

What is the probability that those 4 slots will contain AT LEAST TWO prime numbers?

What is the probability that those 4 slots will contain AT LEAST THREE prime numbers?

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Manager
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Re: combinatorics - at least [#permalink]

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09 Dec 2007, 15:39
bmwhype2 wrote:
I'm having trouble with AT LEAST in combinatorics.

we have 4 slots. There are 8 digits to fill those spots: the digits 2 through 9.

What is the probability that those 4 slots will contain AT LEAST one prime number?

What is the probability that those 4 slots will contain AT LEAST TWO prime numbers?

What is the probability that those 4 slots will contain AT LEAST THREE prime numbers?

Assuming no restrictions (i.e. can repeat the digits).

P(At least 1 Prime) : 1 - P(No Prime)

p(no prime) : 2^-4 = 1/16 ---> selecting 4,6,8,or 9 four times

so 1- (1/16) = 15/16

P(At least 2 Prime) : 1 - [P(No Prime) + P(1 prime)]

p(1 prime) = C(4,1) * (1/2)*(1/2)*(1/2)*(1/2) = 4* 2^-4 = 1/4
*note that the prob of selecting primes equals that of selecting non-primes--1/2 or 4/8

so 1 - [(1/16) + (1/4)] = 11/16

P(At least 3 Prime) : 1 - [P(No Prime) + P(1 prime)+ P(2 prime)]
p(2 prime) = C(4,2) * (1/2)*(1/2)*(1/2)*(1/2) = 6* 2^-4 = 3/8
*note that the prob of selecting primes equals that of selecting non-primes--1/2 or 4/8

so 1 - [(1/16) + (1/4) + (3/8)] = 5/16

If there's a shorter way, i'd like to know. I think it gets much more confusing when there are restrictions such as "exactly 2 primes together".

For instance, what is the probability of getting 3 primes, two of which are back to back - e.g. 2385

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Intern
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09 Dec 2007, 16:11
I find different values.
4C0=1
4C1=4
4C2=6
4C3=4
4C4=1
8C4=70

So
What is the probability that those 4 slots will contain AT LEAST one prime number?
1-(4C0/8C4)= 1- 1/70=69/70

What is the probability that those 4 slots will contain AT LEAST TWO prime numbers?
1-(4C0/8C4 + 4C1x4C3/8C4)= 1- (1+16)/70=53/70

What is the probability that those 4 slots will contain AT LEAST THREE prime numbers?
1-(4C0/8C4 + 4C1x4C3/8C4 + 4C2x4C2/8C4)= 1- (1+16+36)/70=17/70

Any views?

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SVP
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09 Dec 2007, 19:47
alexperi wrote:
I find different values.
4C0=1
4C1=4
4C2=6
4C3=4
4C4=1
8C4=70

So
What is the probability that those 4 slots will contain AT LEAST one prime number?
1-(4C0/8C4)= 1- 1/70=69/70

What is the probability that those 4 slots will contain AT LEAST TWO prime numbers?
1-(4C0/8C4 + 4C1x4C3/8C4)= 1- (1+16)/70=53/70

What is the probability that those 4 slots will contain AT LEAST THREE prime numbers?
1-(4C0/8C4 + 4C1x4C3/8C4 + 4C2x4C2/8C4)= 1- (1+16+36)/70=17/70

Any views?

agree with alex;

we have 4 slots. There are 8 digits to fill those spots: the digits 2 through 9.

1. What is the probability that those 4 slots will contain AT LEAST one prime number?

= 1 - P (no prime)
= 1 - 4c4/8c4
= 1 - 1/70
= 69/70

2. What is the probability that those 4 slots will contain AT LEAST TWO prime numbers?

= 1 - P (no prime) - P (1 prime)
= 1 - 4c4/8c4 - (4c1x4c3)/8c4
= 1 - (1+16)/70
= 53/70

or alternatively:

= [p(2 prime) + p(3 prime) + p(4 or all prime)]/8c4
= (4c2x4c2 + 4c3x4c1 + 4c4x4c0) /8c4
= (36 + 16 +1)/70
= 53/70

3. What is the probability that those 4 slots will contain AT LEAST THREE prime numbers?

= p(3 prime) + p(4 or all prime)]/8c4
= (4c3x4c1 + 4c4x4c0) /8c4
= (16 +1)/70
= 17/70

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Manager
Joined: 03 Sep 2006
Posts: 233

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09 Dec 2007, 21:42
GMAT TIGER, alexperi,

Could you please explain this portion:
- What is the probability that those 4 slots will contain AT LEAST TWO prime numbers?
1-(4C0/8C4 + 4C1x4C3/8C4)

- What is the probability that those 4 slots will contain AT LEAST THREE prime numbers?
1-(4C0/8C4 + 4C1x4C3/8C4 + 4C2x4C2/8C4)

Notably these portions:
for 1 prime number: 4C1x4C3 (Why we have to multiply by 4C3) and
for 2 prime numbers: 4C2x4C2 (Why we have to multiply by the second 4C2)

Thanks

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Manager
Joined: 29 Jul 2007
Posts: 182

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09 Dec 2007, 22:04
GMAT TIGER wrote:
alexperi wrote:
I find different values.
4C0=1
4C1=4
4C2=6
4C3=4
4C4=1
8C4=70

So
What is the probability that those 4 slots will contain AT LEAST one prime number?
1-(4C0/8C4)= 1- 1/70=69/70

What is the probability that those 4 slots will contain AT LEAST TWO prime numbers?
1-(4C0/8C4 + 4C1x4C3/8C4)= 1- (1+16)/70=53/70

What is the probability that those 4 slots will contain AT LEAST THREE prime numbers?
1-(4C0/8C4 + 4C1x4C3/8C4 + 4C2x4C2/8C4)= 1- (1+16+36)/70=17/70

Any views?

agree with alex;

we have 4 slots. There are 8 digits to fill those spots: the digits 2 through 9.

1. What is the probability that those 4 slots will contain AT LEAST one prime number?

= 1 - P (no prime)
= 1 - 4c4/8c4
= 1 - 1/70
= 69/70

2. What is the probability that those 4 slots will contain AT LEAST TWO prime numbers?

= 1 - P (no prime) - P (1 prime)
= 1 - 4c4/8c4 - (4c1x4c3)/8c4
= 1 - (1+16)/70
= 53/70

or alternatively:

= [p(2 prime) + p(3 prime) + p(4 or all prime)]/8c4
= (4c2x4c2 + 4c3x4c1 + 4c4x4c0) /8c4
= (36 + 16 +1)/70
= 53/70

3. What is the probability that those 4 slots will contain AT LEAST THREE prime numbers?

= p(3 prime) + p(4 or all prime)]/8c4
= (4c3x4c1 + 4c4x4c0) /8c4
= (16 +1)/70
= 17/70

Why is P(no prime) = 1/70?
2,3,5,7 are the prime numbers. 4,6,8,9 are not.
favorable outcomes are 4^4 = (2^2)^4 or 2^8 or 256
total outcomes are 8^4= (2^3)^4= 2^12= 4096

256/4096 = 1/16? Where did i go wrong??

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SVP
Joined: 29 Aug 2007
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09 Dec 2007, 23:44
Whatever wrote:
GMAT TIGER, alexperi,

Could you please explain this portion:
- What is the probability that those 4 slots will contain AT LEAST TWO prime numbers?
1-(4C0/8C4 + 4C1x4C3/8C4)

- What is the probability that those 4 slots will contain AT LEAST THREE prime numbers?
1-(4C0/8C4 + 4C1x4C3/8C4 + 4C2x4C2/8C4)

Notably these portions:
for 1 prime number: 4C1x4C3 (Why we have to multiply by 4C3) and
for 2 prime numbers: 4C2x4C2 (Why we have to multiply by the second 4C2)

Thanks

not sure whether repetition is allowed!!!

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Manager
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10 Dec 2007, 00:13
GMAT TIGER - gotcha, thanks

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Intern
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10 Dec 2007, 03:55
Skewed, the logic you used
>P(At least 1 Prime) : 1 - P(No Prime)
>p(no prime) : 2^-4 = 1/16 ---> selecting 4,6,8,or 9 four times
>so 1- (1/16) = 15/16
to me is considering repetitions because you assume a 50% - 50% chance every time you select a prime number.
if there are no repetitions then you have
1 - (4/8 * 3/7 * 2/6 * 1/5)=1- 1/70=69/70

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Manager
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10 Dec 2007, 12:28
alexperi wrote:
Skewed, the logic you used
>P(At least 1 Prime) : 1 - P(No Prime)
>p(no prime) : 2^-4 = 1/16 ---> selecting 4,6,8,or 9 four times
>so 1- (1/16) = 15/16
to me is considering repetitions because you assume a 50% - 50% chance every time you select a prime number.
if there are no repetitions then you have
1 - (4/8 * 3/7 * 2/6 * 1/5)=1- 1/70=69/70

Thx man,

I prefaced my answer by saying that I assumed repetition.
We are just debating semantics here. We understand how to do the problem. I chose to go w/ the less restrictive of the options to encompass a wider solution set.

I do want to ask however what the default GMAT way is when dealing w/ certain combination/permutation/prob questions where there is no info about repetition and context of the question doesn't dictate one way or the other. Is the default to assume repetition or not to assume repetition??

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10 Dec 2007, 12:28
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