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A collection of 36 cards are numbered in two sets. Each card has a uni
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12 Nov 2018, 19:40
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60% (02:17) correct 40% (02:44) wrong based on 139 sessions
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A collection of 36 cards are numbered in two sets. Each card has a unique number. The cards are numbered in such a way that the first 18 cards are numbered with all odd integers from 1 to 36. The remaining cards are numbered using the even integers from 35 to 72. If two cards are drawn at random, then what is the probability that the sum of the numbers on those two cards will be odd? A.\(\frac{18 * 18}{^{36}C_2}\)
B. \(\frac{18 * 17}{^{36}C_2}\)
C. \(\frac{18 * 19}{^{36}C_2}\)
D. \(\frac{19 * 17}{^{36}C_2}\)
E. \(\frac{18 * 19}{^{37}C_2}\)
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A collection of 36 cards are numbered in two sets. Each card has a uni
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12 Nov 2018, 23:30
A collection of 36 cards are numbered in two sets. Each card has a unique number. The cards are numbered in such a way that the first 18 cards are numbered with all odd integers from 1 to 36. The remaining cards are numbered using the even integers from 35 to 72. If two cards are drawn at random, then what is the probability that the sum of the numbers on those two cards will be odd? A.\(\frac{18 * 18}{^{36}C_2}\)
B. \(\frac{18 * 17}{^{36}C_2}\)
C. \(\frac{18 * 19}{^{36}C_2}\)
D. \(\frac{19 * 17}{^{36}C_2}\)
E. \(\frac{18 * 19}{^{37}C_2}\) There are \(\frac{36}{2}=18\) odd numbers from 1 to 36 and similarly 18 even numbers in next 36 numbers from 37 to 72 so way two number sum will be odd  one from odd and one from even so 18*18 total  36 hence \(\frac{18 * 18}{^{36}C_2}\) A
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Re: A collection of 36 cards are numbered in two sets. Each card has a uni
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13 Nov 2018, 00:17
How can 36 to 72 be 18 integers. Still ans maybe (eo + oe)/36*35 = (2*18*18)/(36*35) A
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Re: A collection of 36 cards are numbered in two sets. Each card has a uni
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13 Nov 2018, 01:58
Assume set 1 to be 1st 18 cards with unique odd integer values from 1 to 36 Assume set 2 to be remaining 18 cards with unique even integer values from 35 to 72 For sum of 2 numbers to be odd, one number must be odd and other must be even.
Here, if the 1st card drawn is odd and 2nd card drawn is even, then the probability that sum will be odd = (18/36)*(18/35) (since, there are totally 36 cards, i can choose any 1 of 18 cards from 1st set, and in the remaining 35 cards, i can choose any 1 of 18 cards from 2nd set)
There is one more possibility that, 1st card drawn is even and 2nd card drawn is odd, then the probability that sum will be odd = (18/36)*(18/35) (same reason as above)
so, total probability that sum will be odd = (18/36)+(18/36)=(18/36)*(18/35)*2 =(18*18)/36C2 NOTE : 36C2= (36*35)/2, since formula for combination is nCr= (n!)/((r!)*(nr)!)
OPTION :A



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Re: A collection of 36 cards are numbered in two sets. Each card has a uni
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14 Nov 2018, 05:14
Solution Given:• A collection of 36 cards, numbered in 2 sets • Each card has a unique number • First 18 cards are numbered with all the odd integers from 1 to 36 • Remaining 18 cards are numbered using the even integers from 35 to 72 To find:• The probability that the sum of the numbers on the two cards chosen at random is odd Approach and Working: • For the sum of two numbers to be odd
o One of them must be odd and the other must be even • Total number of cards with an odd number = 18 • Total number of cards with an even number = 18
Thus, probability for the sum to be odd = \(\frac{^{18}C_1 * ^{18}C_1}{^{36}C_2}\) Hence the correct answer is Option A. Answer: A
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Re: A collection of 36 cards are numbered in two sets. Each card has a uni
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15 Nov 2018, 03:05
EgmatQuantExpert wrote: A collection of 36 cards are numbered in two sets. Each card has a unique number. The cards are numbered in such a way that the first 18 cards are numbered with all odd integers from 1 to 36. The remaining cards are numbered using the even integers from 35 to 72. If two cards are drawn at random, then what is the probability that the sum of the numbers on those two cards will be odd? A.\(\frac{18 * 18}{^{36}C_2}\)
B. \(\frac{18 * 17}{^{36}C_2}\)
C. \(\frac{18 * 19}{^{36}C_2}\)
D. \(\frac{19 * 17}{^{36}C_2}\)
E. \(\frac{18 * 19}{^{37}C_2}\) To get an odd sum of number upon drawing two cards one has to be odd and other even.. From given info we know that from no of odd cards 18 and even is 18 so P of getting odd sum : 18*18/36c2= option A
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A collection of 36 cards are numbered in two sets. Each card has a uni
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22 Jan 2019, 07:32
chetan2u sir can you help? I considered 2 cases: 1) even, odd 2) odd, even As there are 18 even and 18 odd, total 36 first case: \(\frac{18}{36}*\frac{18}{35}=\frac{9}{35}\) Second case: \(\frac{18}{36}*\frac{18}{35}=\frac{9}{35}\) prob to occur first OR second case \(\frac{9}{35}+\frac{9}{35}=\frac{18}{35}\) Where do I go wrong? Regards L



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Re: A collection of 36 cards are numbered in two sets. Each card has a uni
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22 Jan 2019, 07:43
LevanKhukhunashvili wrote: chetan2u sir can you help? I considered 2 cases: 1) even, odd 2) odd, even As there are 18 even and 18 odd, total 36 first case: \(\frac{18}{36}*\frac{18}{35}=\frac{9}{35}\) Second case: \(\frac{18}{36}*\frac{18}{35}=\frac{9}{35}\) prob to occur first OR second case \(\frac{9}{35}+\frac{9}{35}=\frac{18}{35}\) Where do I go wrong? Regards L You have not gone wrong anywhere. It is only that the answer is not in a simified form.. (18*18)/(36C2)=18*18/(36!/2!(362)!)=18*18/(36*35/2)=18*18/(18*35)=18/35..
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Re: A collection of 36 cards are numbered in two sets. Each card has a uni
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22 Jan 2019, 08:02
Hello Chetan2u Sir, Here you mentioned that 18 numbers from 37 to 72, but in question stem says 18 numbers from 35 to 72. Is it a typo error in question? chetan2u wrote: A collection of 36 cards are numbered in two sets. Each card has a unique number. The cards are numbered in such a way that the first 18 cards are numbered with all odd integers from 1 to 36. The remaining cards are numbered using the even integers from 35 to 72. If two cards are drawn at random, then what is the probability that the sum of the numbers on those two cards will be odd? A.\(\frac{18 * 18}{^{36}C_2}\)
B. \(\frac{18 * 17}{^{36}C_2}\)
C. \(\frac{18 * 19}{^{36}C_2}\)
D. \(\frac{19 * 17}{^{36}C_2}\)
E. \(\frac{18 * 19}{^{37}C_2}\) There are \(\frac{36}{2}=18\) odd numbers from 1 to 36 and similarly 18 even numbers in next 36 numbers from 37 to 72 so way two number sum will be odd  one from odd and one from even so 18*18 total  36 hence \(\frac{18 * 18}{^{36}C_2}\) A




Re: A collection of 36 cards are numbered in two sets. Each card has a uni
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22 Jan 2019, 08:02






