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29 Nov 2021, 04:57
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A
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
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Question Stats:
50% (01:54) correct
50%
(01:23)
wrong
based on 4
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\(A = \{-10, \ -9, \ -8, \ -7, \ -6, \ -5, \ -4\}\)
\(B = \{-\frac{1}{2}, \ -\frac{1}{3}, \ -\frac{1}{5}, \ 0, \ \frac{1}{5}, \ \frac{1}{3}, \ \frac{1}{2}\}\)
If a number is to be randomly selected from set A above and a number is to be randomly selected from set B above, what is the probability that the product of the two numbers will be a negative integer?
A. \(\frac{8}{49}\)
B. \(\frac{12}{49}\)
C. \(\frac{13}{49}\)
D. \(\frac{18}{49}\)
E. \(\frac{21}{49}\)
\(B = \{-\frac{1}{2}, \ -\frac{1}{3}, \ -\frac{1}{5}, \ 0, \ \frac{1}{5}, \ \frac{1}{3}, \ \frac{1}{2}\}\)
If a number is to be randomly selected from set A above and a number is to be randomly selected from set B above, what is the probability that the product of the two numbers will be a negative integer?
A. \(\frac{8}{49}\)
B. \(\frac{12}{49}\)
C. \(\frac{13}{49}\)
D. \(\frac{18}{49}\)
E. \(\frac{21}{49}\)
29 Nov 2021, 04:57
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Official Solution:
\(A = \{-10, \ -9, \ -8, \ -7, \ -6, \ -5, \ -4\}\)
\(B = \{-\frac{1}{2}, \ -\frac{1}{3}, \ -\frac{1}{5}, \ 0, \ \frac{1}{5}, \ \frac{1}{3}, \ \frac{1}{2}\}\)
If a number is to be randomly selected from set A above and a number is to be randomly selected from set B above, what is the probability that the product of the two numbers will be a negative integer?
A. \(\frac{8}{49}\)
B. \(\frac{12}{49}\)
C. \(\frac{13}{49}\)
D. \(\frac{18}{49}\)
E. \(\frac{21}{49}\)
For the product to be a negative integer the numbers chosen must have opposite signs and their product must result in an integer.
All number in A are negative, so from B we can only choose positive numbers.
If from B we choose \(\frac{1}{5}\), then from A we can choose -10 or -5. Two cases.
If from B we choose \(\frac{1}{3}\), then from A we can choose -9 or -6. Two cases.
If from B we choose \(\frac{1}{2}\), then from A we can choose -10, -8, -6, or -4. Four cases.
So, total of eight pairs will give a negative integer.
The total number of pairs is \(7*7=49\), thus the probability is \(\frac{8}{49}\).
Answer: A
\(A = \{-10, \ -9, \ -8, \ -7, \ -6, \ -5, \ -4\}\)
\(B = \{-\frac{1}{2}, \ -\frac{1}{3}, \ -\frac{1}{5}, \ 0, \ \frac{1}{5}, \ \frac{1}{3}, \ \frac{1}{2}\}\)
If a number is to be randomly selected from set A above and a number is to be randomly selected from set B above, what is the probability that the product of the two numbers will be a negative integer?
A. \(\frac{8}{49}\)
B. \(\frac{12}{49}\)
C. \(\frac{13}{49}\)
D. \(\frac{18}{49}\)
E. \(\frac{21}{49}\)
For the product to be a negative integer the numbers chosen must have opposite signs and their product must result in an integer.
All number in A are negative, so from B we can only choose positive numbers.
If from B we choose \(\frac{1}{5}\), then from A we can choose -10 or -5. Two cases.
If from B we choose \(\frac{1}{3}\), then from A we can choose -9 or -6. Two cases.
If from B we choose \(\frac{1}{2}\), then from A we can choose -10, -8, -6, or -4. Four cases.
So, total of eight pairs will give a negative integer.
The total number of pairs is \(7*7=49\), thus the probability is \(\frac{8}{49}\).
Answer: A
18 Nov 2025, 20:34
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😭 I missed the integer at end of the question
