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A: {71,73,79,83,87} B:{57,59,61,67} If one number is select
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19 Nov 2019, 11:20
19 Nov 2019, 11:20
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A: {71,73,79,83,87}; B:{57,59,61,67}
If one number is selected at random from set A, and one number is selected at random from set B, what is the probability that both numbers are prime?
A. \(\frac{9}{20}\)
B. \(\frac{3}{5}\)
C. \(\frac{3}{4}\)
D. \(\frac{4}{5}\)
E. 1
If one number is selected at random from set A, and one number is selected at random from set B, what is the probability that both numbers are prime?
A. \(\frac{9}{20}\)
B. \(\frac{3}{5}\)
C. \(\frac{3}{4}\)
D. \(\frac{4}{5}\)
E. 1
Re: A: {71,73,79,83,87} B:{57,59,61,67} If one number is select
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19 Nov 2019, 20:59
19 Nov 2019, 20:59
2
huda wrote:
A: {71,73,79,83,87}; B:{57,59,61,67}
If one number is selected at random from set A, and one number is selected at random from set B, what is the probability that both numbers are prime?
A. \(\frac{9}{20}\)
B. \(\frac{3}{5}\)
C. \(\frac{3}{4}\)
D. \(\frac{4}{5}\)
E. 1
If one number is selected at random from set A, and one number is selected at random from set B, what is the probability that both numbers are prime?
A. \(\frac{9}{20}\)
B. \(\frac{3}{5}\)
C. \(\frac{3}{4}\)
D. \(\frac{4}{5}\)
E. 1
In set A, there is only one non-prime i.e. 87 so there are 4 possibilities of prime numbers.
Similarly in Set B, 57 is the only non-prime.
The probability becomes \(\frac{4}{5}\) * \(\frac{3}{4}\) = \(\frac{3}{5}\)
OA - B
Re: A: {71,73,79,83,87} B:{57,59,61,67} If one number is select
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21 Nov 2019, 00:20
21 Nov 2019, 00:20
great explanation
