Question

Out of 21 tickets marked with numbers from 1 to 21, three are drawn at random. The chance that the numbers on them are in A.P., is

• A. $\frac { 10 } { 133 }$
• B. $\frac { 9 } { 133 }$
• C. $\frac { 9 } { 1330 }$
• D. None of these

Answer 1

Answer: (A)

$\frac { 10 } { 133 }$

Answer 2

Total number of ways $= { } ^ { 21 } C _ { 3 } = 1330$ . If common

difference of the A.P. is to be 1, then the possible groups are 1, 2, 3; 2, 3, 4; ……19, 20, 21.

If the common difference is 2, then possible groups are 1, 3, 5; 2, 4, 6; ….. 17, 19, 21.

Proceeding in the same way, if the common difference is 10, then the possible group is 1, 10, 21.

Thus if the common difference of the A.P. is to be ≥ 11, obviously there is no favourable case.

Hence, total number of favourable cases

= 19 +17 + 15 + …+ 3 + 1 =100

Hence, required probability $= \frac { 100 } { 1330 } = \frac { 10 } { 133 }$ .