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Re: Practice questions of Combinations and probability [#permalink]

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05 Apr 2013, 03:51

Thank you for the collection. As I worked through the questions i noticed that some answers are not correct. For example answer 5 should be "B) 66" and not D. However, explaination is correct in my opinion. Cheers,

Re: Practice questions of Combinations and probability [#permalink]

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16 Apr 2014, 02:06

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Re: Practice questions of Combinations and probability [#permalink]

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22 Sep 2014, 14:32

I have a doubt regarding Q4 in the doc: How many 3-digit numbers satisfy the following conditions: The first digit is different from zero and the other digits are all different from each other?

The answer in the doc is 648 but it is derived on the basis - 9 X 9 X 8.

The question does not say the second and third digits are different from the first one. It says they are different from each other. Then should it not be 9 X 10 X 9? Am I mistaken?

I have a doubt regarding Q4 in the doc: How many 3-digit numbers satisfy the following conditions: The first digit is different from zero and the other digits are all different from each other?

The answer in the doc is 648 but it is derived on the basis - 9 X 9 X 8.

The question does not say the second and third digits are different from the first one. It says they are different from each other. Then should it not be 9 X 10 X 9? Am I mistaken?

I see your point. Ambiguous wording. Ignore this question.
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Re: Practice questions of Combinations and probability [#permalink]

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28 Nov 2014, 16:38

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. – Bernoulli’s principle

Hi Narenn This is in regards to your probability question bank 25 .

I dint quite get this - if there are three spots to be filled with 2 prime numbers - would 1/2(probablity of picking the first prime) * 3/7(probablity of picking the second one) would that not be enough - or are we assuming that there can be the two prime number which are same??

Re: Practice questions of Combinations and probability [#permalink]

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09 Nov 2015, 11:04

Narenn wrote:

Hi All,

Below are Combination and Probability practice ques with complete solutions

Thanks,

Narenn

Hi Narenn, I've been reviewing this file, it's been great to enhance my weakest point on Quant (probability). Thanks a lot for that. Just one observation on questions 32:

The total number of diagonals is 171 (not 170) applying (21*(21-3)/2 - 18 = 171. It's not correct applying 20*(20-3)/2 since there will be an extra diagonal between points 1 and 20 that wouldn't exist in a polygon with 20 sides. Also 19*18/2=171

The same for question 33: (18*(18-3))/2 - (15+15+14) = 91. Or 13*14/2=91

Also please give me a little further explanation on question 46. I managed to solve it using the probability of not selecting a couple (1*(8/9)*(6/8)) = 2/3 multiplied by the total number of combination 10C3=120, (2/3)*120=80 . But I don't understand why 5C1*8C1=40 is the number of ways of selecting 3 individuals out of 5 couples.

Hope this helps to improve this great post. Regards

Re: Practice questions of Combinations and probability [#permalink]

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30 Aug 2016, 19:57

Hi there,

Thanks for the post. This helps!

I think there is a problem in Q37.

37. How many 5 digit numbers can be created if the following terms apply: the leftmost digit is even, the second is odd, the third is a non even prime and the fourth and fifth are two random digits not used before in the number?

a) 2520 b) 3150 c) 3360 d) 6000 e) 7500

Solution provided is:

The best answer is A. The first digit has 4 options (2,4,6,8 and not 0), the second has 5 options (1,3,5,7,9) the third has 3 options (3,5,7 and not 2), the fourth has 7 options (10-3 used before) and the fifth has 6 options (10-4 used before). The total is 4*5*3*7*6=2520.

I have problems with explanation in RED.

If on second and third position the number is repeated (3,5, or 7), then fourth position has 8 options and fifth position has 7 options.

So according to me, solution should be 4*2*3*7*6 + 4*3*3*8*7

Please correct me if i am wrong.

Pranav
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Re: Practice questions of Combinations and probability
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30 Aug 2016, 19:57

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