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Three real numbers are selected independently and at random from the interval [-20,10]. What is the probability that the product of those numbers is greater than zero?
(A) \(\frac{1}{3}\)
(B) \(\frac{10}{27}\)
(C) \(\frac{11}{27}\)
(D) \(\frac{4}{9}\)
(E) \(\frac{13}{27}\)
(A) \(\frac{1}{3}\)
(B) \(\frac{10}{27}\)
(C) \(\frac{11}{27}\)
(D) \(\frac{4}{9}\)
(E) \(\frac{13}{27}\)
15 Aug 2026, 09:43
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For the product to be greater than zero, the following combinations sit (+++) (--+) (-+-) (+--)
First separate positive and negative numbers from the set:
-20,-19....-1...0..1.....10
Number of positive numbers: (10-1)/1+1= 10
Number of negative numbers: (-1-{-20})/1+1=19+1=20
Zero has to be excluded as its neither negative nore positive
Total numbers becomes 30
Probability thus becomes (10/30*10/30*10/30)+ (10/30*20/30*20/30)+(20/30*10/30*20/30)+(20/30*20/30*10/30)=13/27
First separate positive and negative numbers from the set:
-20,-19....-1...0..1.....10
Number of positive numbers: (10-1)/1+1= 10
Number of negative numbers: (-1-{-20})/1+1=19+1=20
Zero has to be excluded as its neither negative nore positive
Total numbers becomes 30
Probability thus becomes (10/30*10/30*10/30)+ (10/30*20/30*20/30)+(20/30*10/30*20/30)+(20/30*20/30*10/30)=13/27
16 Aug 2026, 05:25
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What’s the probability that the product is zero ?
