Three integers are chosen at random from the first 20 integers. The probability that their product is even is
A. 2/19
B. 3/29
C. 17/19
D. 4/19
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Three integers are chosen at random from the first 20 integers. The probability that their product is even is
A. 2/19
B. 3/29
C. 17/19
D. 4/19
Three integers are chosen at random from the first 20 integers. The probability that their product is even is
A. 2/19
B. 3/29
C. 17/19
D. 4/19
Three integers are chosen at random from the first 20 integers. The probability that their product is even is
A. \(\frac{2}{19}\)
B. \(\frac{2}{29}\)
C. \(\frac{17}{19}\)
D. \(\frac{4}{19}\)
Three integers are chosen at random from the first 20 integers. The probability that their product is even is
A. \(\frac{2}{19}\)
B. \(\frac{2}{29}\)
C. \(\frac{17}{19}\)
D. \(\frac{4}{19}\)
3 integers out of 20 can be chosen in ^{20}C_{3} ways.
As in first 20 integers, 10 are even.
To have the product of 3 integer even the chosen integers must be even.
∴ 3 integers that are even can be chosen from first 20 integers in ^{10}C_{3} ways.
∴ Probability = \(\frac{^{10}C_2}{^{20}C_2}\) = \(\frac{10\times9\times8}{20\times19\times18} = \frac{2}{19}\)
Our answer matches with option (a)
∴ Option (a) is the only correct choice
Correct Answer – C
Correct Answer – B
N/a
Correct option is C. 17/19
The product of 3 randomly chosen integers are even only if there are even nos.
P(even) = 1 – P(odd)
P(odd) = \(\cfrac{10}{20}\times\cfrac9{19}\times\cfrac8{18}\)
= \(\cfrac2{19}\)
P(even) = \(\cfrac{17}{19}\)