$S = {1, 2, 3, . . .20}$ if $3$ numbers are chosen at random from S, the probability for they are in A.P. is?

\frac{3}{38}

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\frac{35}{33}

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\frac{33}{35}

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\frac{1}{38}

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Solution

The correct option is A $\frac{3}{38}$
Total number of possible three numbers= ${20_{C}}_{3}$ =1140
a,b,c are in AP
2b=a+c=even
Therefore a,c are either both even or both odd
There are 10 odd and 10 even numbers in 1..20
Favourable case= ${10_{C}}_{2}$ + ${10_{C}}_{2}$ =90
Probability=90/1140=3/38

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