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  • Three numbers are chosen from 1 to 20. Find the probability that they are not consecutive A 186 / 190 B. 187/190 C. 188/190 D. 18/^20C_3

Three numbers are chosen from 1 to 20. Find the probability that they are not consecutive

A. \frac{186}{190}

B. \frac{187}{190}

C. \frac{188}{190}

D. ^{20}C_{3}

 

Answers (1)

The set has 3 consecutive nos. from 1 to 20,

Thus, it is – (1,2,3), (2,3,4), (3,4,5), …….. , (18,19,20)

Now, if we consider 3 nos. as a single digit, there will be 18 nos.

Now, choosing 3 nos. out of 20 can be done in 20C3 ways

This, n(S) = 20C3

Required event – the 3 nos. chosen must be consecutive, thus,

P (nos. are consecutive) = 18/ 30C3

=\frac{\frac{18}{20!}}{3!(20-3)!}

=\frac{18}{\frac{20 \times 19 \times 18 \times 17! }{3\times2\times1\times17!}}

=\frac{18}{\frac{20 \times 19 \times 18 }{6}}

=\frac{6}{20 \times 19 }

=\frac{3}{190 }

Now,

P (nos. that are not consecutive) = 1 – 3/190

                                                = 190 – 3/ 190

                                                = 187 / 190

Thus, option B is the correct answer.
 

Posted by

infoexpert22

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