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Maths-

General

Easy

Question

3 distinct integers are selected at random from 1, 2, 3,…, 20. The probability that the sum is divisible by 5 is

$\frac{49}{285}$
$\frac{29}{285}$
$\frac{11}{285}$
none of these.

The correct answer is: $fraction numerator 29 over denominator 285 end fraction$

1.The no. of ways of choosing 3 distinct integers from 1, 2, 3,...,20 is $scriptbase C end scriptbase presuperscript 20 end presuperscript subscript 3 end subscript$
$equals fraction numerator 20.19.18 over denominator 1.2.3 end fraction equals 20 cross times 57 equals 1140$
Now arrange the given numbers as below.

The sum of three digits is divisible by 5 in the following cases:
Two nos. from first row and one number from 3^rd row or one number from 1^st row and two nos. from 4^th row or three nos. from 5^th row or one number from each (1^st row, 4^th row, 5^th row)
Then the no. of favourable ways is
$equals scriptbase C end scriptbase presuperscript 4 end presuperscript subscript 2 end subscript cross times scriptbase C end scriptbase presuperscript 4 end presuperscript subscript 1 end subscript plus scriptbase C end scriptbase presuperscript 4 end presuperscript subscript 1 end subscript cross times scriptbase C end scriptbase presuperscript 4 end presuperscript subscript 2 end subscript plus scriptbase C end scriptbase presuperscript 4 end presuperscript subscript 3 end subscript plus scriptbase C end scriptbase presuperscript 4 end presuperscript subscript 1 end subscript cross times scriptbase C end scriptbase presuperscript 4 end presuperscript subscript 1 end subscript cross times scriptbase C end scriptbase presuperscript 4 end presuperscript subscript 1 end subscript$
$equals 24 plus 24 plus 4 plus 64 equals 116$
Hence reqd. prob. $equals fraction numerator 116 over denominator 1140 end fraction equals fraction numerator 29 over denominator 285 end fraction.$

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