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Probability problem- need a better understanding (m08q05) [#permalink]

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05 Jul 2008, 05:30

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Two sets are defined as follows:

\(A = {2, 3, 4, 4, 4}\) \(B = {0, 1, 2}\)

If a number is taken from set \(A\) at random and another number is taken from set \(B\) at random, what is the probability that the sum of these numbers is a prime integer?

Re: Probability problem- need a better understanding [#permalink]

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05 Jul 2008, 06:19

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Quote:

The total outcomes for the problem is 15. However when we calculate the sum of 2 sets that generate a prime number, Do we count 4 once or three times.

If you have counted the number of all outcomes as 15 (5*3), you have included the pairs with 4 more than one time. So, when counting the number of favourable outcomes, you need again to count pairs with 4 more than one time.

Alternative approach: P(select 2) = 1/5; and 2 of 3 numbers from the second set will give us prime – 1/5*2/3 to get prime P(select 3) = 1/5; and 2 of 3 numbers from the second set will give us prime – 1/5*2/3 to get prime P(select 4) = 3/5; and 1 of 3 numbers from the second set will give us prime – 3/5*1/3 to get prime

Re: Probability problem- need a better understanding (m08q05) [#permalink]

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24 Jan 2013, 06:28

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To solve such questions, one should keep in mind that the testmakers design such questions to be solved in two mins or less even if one chooses to LIST all possible one by one. The most important thing is to quickly decide the method to use, then LIST away.

In this particular question, there are only 7 desirable outcomes out of a total of 15 possible outcomes. Surely, it would take one about thirty to forty-five seconds to list all 7 prime outcomes.

Total possible outcomes= 5C1 * 3C1 (i.e. picking one item each from two sets of 5 and 3 items respectively= 5*3= 15 Desired prime outcomes from listing= 2,3,3,5,5,5,5 =7

Probability= Desired prime outcomes/ Total possible outcomes = 7/15

Re: Probability problem- need a better understanding (m08q05) [#permalink]

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15 Jan 2010, 07:21

D. But I did it the long way. 5 numbers in the first set and 3 in the second. Yields 15 total possibilities. Then I counted each one that could be prime. Got 7.

Re: Probability problem- need a better understanding (m08q05) [#permalink]

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15 Jan 2010, 08:18

7/15. But I counted each combination of prime numbers and came up with the answer. For large problems, we might need to use the probability of individual combination as suggested few posts above. -Suresh.

Re: Probability problem- need a better understanding (m08q05) [#permalink]

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20 Jan 2011, 07:16

Same here,

Counted all 15 possibilities, No doubt we need to include 4 thrice as it is asking for 'Probability' not 'Permutation & Combination'. Thats the trick ..

In P&C we need to ignore repeating objects but not in 'Probability'

Re: Probability problem- need a better understanding (m08q05) [#permalink]

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20 Jan 2011, 07:33

Good post. I got it to C and D and had the same inquiry. Thanks for the replies :D _________________

I'm trying to not just answer the problem but to explain how I came up with my answer. If I am incorrect or you have a better method please PM me your thoughts. Thanks!

Re: Probability problem- need a better understanding (m08q05) [#permalink]

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21 Jan 2011, 04:56

I have done by long method- When we will take 0 from the B set and match with numbers in set A; only 2, and 3 from the A set can give a prime number. So, probability of picking 2, and 3 is 2/5 and of 0 is 1/3. This gives combined probability of 2/5*1/3=2/15 (Combined probability will decrease so I have used multiplication rule). Next, I have taken 1 from the set B and it can give a prime number when combined with 2,4,4, and 4. So using the same reasoning above its combined probability will be- 4/5*1/3=4/15. In last, 2 from the set B can give only prime number when combined with 1 from the set A. So its combined probability will be-1/5*1/3=1/15.

All the 3 probability I have added to get the final answer because probability should increase- 2/15+4/15+1/15= 7/15.

I think my answer and approach are correct but can somebody will tell me how to solve this problem within 1 minute and I often use to do mistake in counting the number of probable case- How to avoid this.

Re: Probability problem- need a better understanding (m08q05) [#permalink]

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21 Jan 2011, 05:31

rajeshaaidu wrote:

I think my answer and approach are correct but can somebody will tell me how to solve this problem within 1 minute and I often use to do mistake in counting the number of probable case- How to avoid this.

Cheers!!!

Your method shouldn't take longer than 1 min?

Since we only have 3 different integers, I find that calculating the propability for each number is the most logical way for me.

2 is a prime when +0 and +1 => We draw 2 from the first set 1/5 and we draw 0 or 1 from the second 2/3 => 1/5*2/3 = 2/15 3 is a prime when +0 and +2 => We draw 3 from the first set 1/5 and we draw 0 or 2 from the second 2/3 => 1/5*2/3 = 2/15 4 is a prime when +1 => We draw 4 from the first set 3/5 and we draw 1 from the second set 1/3 = 3/5*1/3 = 3/15

2+2+3 = 7/15

This method shouldnt take more than one minute _________________

Re: Probability problem- need a better understanding (m08q05) [#permalink]

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24 Jan 2012, 08:45

picked one digit from set one and added with every digits separately from set 2 to get a prime, and also got the total to get the final probability. _________________

" Make more efforts " Press Kudos if you liked my post

I think my answer and approach are correct but can somebody will tell me how to solve this problem within 1 minute and I often use to do mistake in counting the number of probable case- How to avoid this.

Two sets are defined as follows:

\(A = {2, 3, 4, 4, 4}\) \(B = {0, 1, 2}\)

If a number is taken from set \(A\) at random and another number is taken from set \(B\) at random, what is the probability that the sum of these numbers is a prime integer?

Now, as set B is smaller work with it: 0 can be added to 2 or to 3 to get a prime --> 1+1=2 cases; 1 can be added to 2 or to either of three 4's to get a prime --> 1+3=4 cases; 2 can be added only to 3 to get a prime --> 1 cases.

So total # of favorable outcomes is 2+4+1=7. P=favorable/total=7/15.