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bmwhype2
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I'm having trouble with AT LEAST in combinatorics. we have 4 09 Dec 2007, 13:26
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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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I'm having trouble with AT LEAST in combinatorics. we have 4 09 Dec 2007, 15:39
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bmwhype2
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?
Show more


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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I'm having trouble with AT LEAST in combinatorics. we have 4 09 Dec 2007, 16:11
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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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I'm having trouble with AT LEAST in combinatorics. we have 4 09 Dec 2007, 19:47
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alexperi
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?
Show more


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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I'm having trouble with AT LEAST in combinatorics. we have 4 09 Dec 2007, 21:42
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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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I'm having trouble with AT LEAST in combinatorics. we have 4 09 Dec 2007, 22:04
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alexperi
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
Show more


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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I'm having trouble with AT LEAST in combinatorics. we have 4 09 Dec 2007, 23:44
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Whatever
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
Show more



not sure whether repetition is allowed!!!
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I'm having trouble with AT LEAST in combinatorics. we have 4 10 Dec 2007, 00:13
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GMAT TIGER - gotcha, thanks
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I'm having trouble with AT LEAST in combinatorics. we have 4 10 Dec 2007, 03:55
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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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I'm having trouble with AT LEAST in combinatorics. we have 4 10 Dec 2007, 12:28
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alexperi
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
Show more


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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Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Quantitative Questions Forum
Still interested in this question? Check out the "Best Topics" block above for a better discussion on this exact question, as well as several more related questions.
Thank you for understanding, and happy exploring!