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If two integers between 5 and 3, inclusive, are chosen at random, whi
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12 Jun 2018, 23:14
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If two integers between 5 and 3, inclusive, are chosen at random, which of the following is most likely to be true? A. The sum of the two integers is even B. The sum of the two integers is odd C. The product of the two integers is even D. The product of the two integers is odd E. The product of the two integers is negative
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Re: If two integers between 5 and 3, inclusive, are chosen at random, whi
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12 Jun 2018, 23:30
Bunuel wrote: If two integers between 5 and 3, inclusive, are chosen at random, which of the following is most likely to be true?
A. The sum of the two integers is even
B. The sum of the two integers is odd
C. The product of the two integers is even
D. The product of the two integers is odd
E. The product of the two integers is negative Questions involving number properties can often be solved logically, without using equations. This is a Logical approach. We have a total of 9 numbers, 5 odd and 4 even A. occurs when both integers are odd or both are even this is 5C2 + 4C2 = 10 + 6 = 16 B. occurs when one is odd and the other even this is 5C1*4C1 = 5*4 = 20 C. occurs when at least one is even = 4C2 + 4C1*5C1 = 6 + 20 = 26 D. occurs when both are odd = 5C2 = 10 E. occurs when one is positive and the other negative. = 5C1*2C1 = 5*2 = 10 So (C) is our answer.
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If two integers between 5 and 3, inclusive, are chosen at random, whi
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12 Jun 2018, 23:31
The list of numbers are {5, 4, 3, 2, 1, 0, 1, 2, 3}, since it is asked as most likely to be true. I calculated the possible combinations to arrive at an answer.
Even numbers of the set (4, 2, 0, 2) = 4 numbers Odd numbers of the set (5, 3, 1, 1, 3) = 5 numbers Negative numbers of the set (5, 4, 3, 2, 1) = 5 numbers Positive numbers (1, 2, 3) = 3 numbers
A. The sum of the two integers is even
For the sum to be even both number should be even or both should be odd.
4C2 + 5C2 = 6+10 = 16
B. The sum of the two integers is odd
The sum to be odd one should be even and one should be odd. 4C1 * 5C1 = 20
C. The product of the two integers is even
Product to be even both should be even or one should be even.
4C2 + (4C1*5C1) = 26
D. The product of the two integers is odd
product to be odd both should be odd. 5C2 = 10
E. The product of the two integers is negative
product to be negative. one should be negative and other should be positive. 5C1 * 3C1 = 15
Hence OA  C



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Re: If two integers between 5 and 3, inclusive, are chosen at random, whi
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16 Jun 2018, 10:12
Bunuel wrote: If two integers between 5 and 3, inclusive, are chosen at random, which of the following is most likely to be true?
A. The sum of the two integers is even
B. The sum of the two integers is odd
C. The product of the two integers is even
D. The product of the two integers is odd
E. The product of the two integers is negative The integers are: 5, 4, 3, 2, 1, 0, 1, 2, 3 We have 5 negative numbers, 1 zero, and 3 positive numbers. We have 5 evens and 4 odds. Since even x even = even and even x odd = even, it’s most likely to be true that we will get an even product. Answer: C
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Re: If two integers between 5 and 3, inclusive, are chosen at random, whi
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27 Jun 2018, 10:27
DavidTutorexamPAL wrote: Bunuel wrote: If two integers between 5 and 3, inclusive, are chosen at random, which of the following is most likely to be true?
A. The sum of the two integers is even
B. The sum of the two integers is odd
C. The product of the two integers is even
D. The product of the two integers is odd
E. The product of the two integers is negative Questions involving number properties can often be solved logically, without using equations. This is a Logical approach. We have a total of 9 numbers, 5 odd and 4 even A. occurs when both integers are odd or both are even this is 5C2 + 4C2 = 10 + 6 = 16 B. occurs when one is odd and the other even this is 5C1*4C1 = 5*4 = 20 C. occurs when at least one is even = 4C2 + 4C1*5C1 = 6 + 20 = 26 D. occurs when both are odd = 5C2 = 10 E. occurs when one is positive and the other negative. = 5C1*2C1 = 5*2 = 10 So (C) is our answer. Hi for C, I calculated as below occurs when at least one is even 4C1 * 8C1 (select 1 even int and select 1 from remaining 8 integers ) =32 I too got answer C , but what's wrong with my method ?



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If 2 integers between 5 and 3, inclusive are chosen at random, which
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03 Sep 2018, 09:10
If 2 integers between 5 and 3, inclusive are chosen at random, which of the following is most likely to be true?
A. The sum of two integers is even B. The sum of two integers is odd C. The product of two integers is even D. The product of two integers is odd E. The product of two integers is even negative



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If two integers between 5 and 3, inclusive, are chosen at random, whi
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Updated on: 03 Sep 2018, 22:50
ManasviHP wrote: If 2 integers between 5 and 3, inclusive are chosen at random, which of the following is most likely to be true?
A. The sum of two integers is even B. The sum of two integers is odd C. The product of two integers is even D. The product of two integers is odd E. The product of two integers is even negative
Posted from my mobile device OA: C Total Number of integers between 5 and 3 = 9 (5,4,3,2,1,0,1,2,3) Number of Even Integers between 5 and 3 =4 (4,2,0,2) Number of Odd Integers between 5 and 3 =5 (5,3,1,1,3) A. The sum of two integers is even Sum of Two integers will be even in 2 conditions Case 1: EVEN+EVEN We have to find the number of ways of selecting 2 even integers. \(C(4,2) = \frac{4!}{(42)!2!}=\frac{4!}{2!2!}=6\) Case 2:ODD+ODD We have to find the number of ways of selecting 2 Odd integers. \(C(5,2) = \frac{5!}{(52)!2!}=\frac{5!}{3!2!}=10\) Total favourable case\(= 10+6 = 16\) Total Number of ways of selecting 2 integers \(= C(9,2) = \frac{9!}{(92)!2!}=\frac{9!}{7!2!}=36\) Probability\(= \frac{16}{36} =\frac{4}{9}=0.44\) B. The sum of two integers is odd Sum of Two integers will be even in 1 conditions Case 1: EVEN+ODD We have to find the number of ways of selecting 1 even integers and 1 odd integer. \(C(4,1)*C(5,1)= \frac{4!}{(41)!1!}*\frac{5!}{(51)!1!}=\frac{4!}{3!1!}*\frac{5!}{4!1!}=20\) Total favourable case\(=20\) Total Number of ways of selecting 2 integers \(= C(9,2) = \frac{9!}{(92)!2!}=\frac{9!}{7!2!}=36\) Probability\(= \frac{20}{36} =\frac{5}{9}=0.55\) C.The product of two integers is even The probability of product of two integers being even = 1  Probability of product of two integers being Odd Product of two integers will be odd when both are odd Case :ODD*ODD We have to find the number of ways of selecting 2 Odd integers. \(C(5,2) = \frac{5!}{(52)!2!}=\frac{5!}{3!2!}=10\) Total Number of ways of selecting 2 integers \(= C(9,2) = \frac{9!}{(92)!2!}=\frac{9!}{7!2!}=36\) Probability of product of two integers being Odd \(= \frac{10}{36} = \frac{5}{18}\) Probability of product of two integers being Even\(=1\frac{5}{18} = \frac{13}{18}=0.72\) (Most likely to be true)D. The product of two integers is odd Probability of product of two integers being Odd \(= \frac{10}{36} = \frac{5}{18}=0.27\) E.The product of two integers is even negative Possible combinations such that product of two integers is even negative :9 {(4,1);(4,2);(4,3);(2,1);(2,2);(2,3);(2,1);(2,3);(2,5)} Total Number of ways of selecting 2 integers \(= C(9,2) = \frac{9!}{(92)!2!}=\frac{9!}{7!2!}=36\) Probability of product of two integers is even negative \(= \frac{9}{36} = \frac{1}{4}=0.25\) Edit: E.The product of the two integers is negative Case: ve integer(5 choices) * +ve integer(3 choices) We have find one negative integer and one positive integer Number of Selecting one negative integer and one positive integer\(= C(5,1)*C(3,1) = 5*3=15\) Total Number of ways of selecting 2 integers \(= C(9,2) = \frac{9!}{(92)!2!}=\frac{9!}{7!2!}=36\) Probability of product of the two integers is negative\(= \frac{15}{36}=\frac{5}{12}=0.42\)
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Originally posted by Princ on 03 Sep 2018, 09:59.
Last edited by Princ on 03 Sep 2018, 22:50, edited 1 time in total.



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Re: If two integers between 5 and 3, inclusive, are chosen at random, whi
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03 Sep 2018, 20:40
ManasviHP wrote: If 2 integers between 5 and 3, inclusive are chosen at random, which of the following is most likely to be true?
A. The sum of two integers is even B. The sum of two integers is odd C. The product of two integers is even D. The product of two integers is odd E. The product of two integers is even negative Merging topics. Please check the discussion above.
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If two integers between 5 and 3, inclusive, are chosen at random, whi
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03 Sep 2018, 22:55
If two integers between 5 and 3, inclusive, are chosen at random, which of the following is most likely to be true? So we have 5,4,3,2,1,0,1,2,3 choosing two integers = 9C2=36Since the total possible cases in all choices is SAME 36, the choice which gives us the maximum cases in the given restriction is MOST likelyChoices A to D are dependent on Even and Odd, so let us see how many are even and how many odd.. even  4 ; odd  5 A) The sum of the two integers is evenSum is even if both are even or both odd  5C2+4C2=10+6=16B) The sum of the two integers is oddSum is odd if one is even and other odd  5C1*4C1=5*4=20C) The product of the two integers is evenProduct is even when both are even  4C2 and when one is odd and other even  5C1*4C1............Total = 4C2+5C1*4C1=6+5*4=26D) The product of the two integers is oddProduct is odd if both odd  5C2=10E) The product of the two integers is negativethere are 5 negative and 3 positive, so cases = 5*3=15so C gives us the max cases Note  if you have got cases in A as 16, B will be 3616=20 and when you have got cases in C as 26, D will be 36C=3626=10 chetna14, "why C is not 4C1*8C1" is because there are repetitions when both 4C1 and 8C1 give us an even number Example say 4C1 gives us 0 and 8C1 gives us 2 BUT there will be a case when 4C1 will give us 2 and 8C1 will give us 0, so set (0,2) is repeated similarly there are 4C2 cases so 324C2=326=26
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Re: If two integers between 5 and 3, inclusive, are chosen at random, whi
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23 Feb 2019, 11:26
Bunuel wrote: If two integers between 5 and 3, inclusive, are chosen at random, which of the following is most likely to be true?
A. The sum of the two integers is even
B. The sum of the two integers is odd
C. The product of the two integers is even
D. The product of the two integers is odd
E. The product of the two integers is negative We need to choose 2 numbers from {5, 4, 3, 2, 1, 0, 1, 2, 3}The correct answer choice will be the one that has the most possible outcomesA. The sum of the two integers is even There are 2 ways to get an EVEN sum: 1) Both integers are even OR 2) Both integers are odd 1) We can choose 2 even integers in 4C2 ways = 6 ways (since there are 4 even integers to choose from) 2) We can choose 2 odd integers in 5C2 ways = 10 ways (since there are 5 odd integers to choose from) TOTAL possible outcomes = 6 + 10 = 16B. The sum of the two integers is odd One number must be even, AND the other must be odd We can choose 1 odd integer in 5 ways AND we can choose 1 even integer in 4 ways TOTAL possible outcomes = 5 x 4 = 20C. The product of the two integers is even There are 2 ways to get an EVEN product: 1) Both integers are even OR 2) One integer is even AND the other is odd 1) We can choose 2 even integers in 4C2 ways = 6 ways (since there are 4 even integers to choose from) 2) We can choose 1 even integer in 4 ways AND we can choose 1 odd integer in 5 ways. 4 x 5 = 20 TOTAL possible outcomes = 6 + 20 = 26D. The product of the two integers is odd There is one way to get an ODD product: 1) Both integers are odd 1) We can choose 2 odd integers in 5C2 ways = 10 ways TOTAL possible outcomes = 10E. The product of the two integers is negative There is 1 way to get a NEGATIVE product: 1) One integer is negative AND the other is positive We can choose 1 negative integer in 3 ways AND we can choose 1 positive integer in 5 ways. TOTAL possible outcomes = 3 x 5 = 15Answer: C Cheers, Brent
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Re: If two integers between 5 and 3, inclusive, are chosen at random, whi
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25 Feb 2019, 03:25
Bunuel DavidTutorexamPAL Hi, I have a conceptual query. Let's take option a) for example. It can only occur when either both integers are odd or both integers are even Why isn't the total number of possibilities 4C1 X 4C1 + 5C1 X 5C1? Why is it 4C2 + 5C2?



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If two integers between 5 and 3, inclusive, are chosen at random, whi
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25 Feb 2019, 03:45
smlprkh wrote: Bunuel DavidTutorexamPAL Hi, I have a conceptual query. Let's take option a) for example. It can only occur when either both integers are odd or both integers are even Why isn't the total number of possibilities 4C1 X 4C1 + 5C1 X 5C1? Why is it 4C2 + 5C2? Hey smlprkh, The difference is that the question assumes that the integers are different. So your calculation also includes (1,1), for example, while 5C2 does not. Additionally, you do not take into account repetitions. For example, you count (1,3) and (3,1) as separate instances instead of the same one.
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If two integers between 5 and 3, inclusive, are chosen at random, whi
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