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
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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.
\(C(4,2) = \frac{4!}{(4-2)!2!}=\frac{4!}{2!2!}=6\)
Case 2:ODD+ODD
We have to find the number of ways of selecting 2 Odd integers.
\(C(5,2) = \frac{5!}{(5-2)!2!}=\frac{5!}{3!2!}=10\)
Total favourable case\(= 10+6 = 16\)
Total Number of ways of selecting 2 integers \(= C(9,2) = \frac{9!}{(9-2)!2!}=\frac{9!}{7!2!}=36\)
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.
\(C(4,1)*C(5,1)= \frac{4!}{(4-1)!1!}*\frac{5!}{(5-1)!1!}=\frac{4!}{3!1!}*\frac{5!}{4!1!}=20\)
Total favourable case\(=20\)
Total Number of ways of selecting 2 integers \(= C(9,2) = \frac{9!}{(9-2)!2!}=\frac{9!}{7!2!}=36\)
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.
\(C(5,2) = \frac{5!}{(5-2)!2!}=\frac{5!}{3!2!}=10\)
Total Number of ways of selecting 2 integers \(= C(9,2) = \frac{9!}{(9-2)!2!}=\frac{9!}{7!2!}=36\)
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 \(= C(9,2) = \frac{9!}{(9-2)!2!}=\frac{9!}{7!2!}=36\)
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 \(= C(9,2) = \frac{9!}{(9-2)!2!}=\frac{9!}{7!2!}=36\)
Probability of product of the two integers is negative\(= \frac{15}{36}=\frac{5}{12}=0.42\)