Question

Out of 21 tickets marked with numbers 1, 2,…. , 21, three are drawn at random without replacement. The probability that these numbers are in A.P. is -

• A. $\frac { 10 } { 133 }$
• B. $\frac { 9 } { 15 }$
• C. $\frac { 14 } { 261 }$
• D. $\frac { 13 } { 261 }$

Answer 1

Answer: (A)

$\frac { 10 } { 133 }$

Answer 2

The number of ways of selecting three tickets out of 21 is ²¹C₃ = $\frac { 21 \times 20 \times 19 } { 3 \times 2 }$ = 1330. Let d be the common difference of the A.P. We have the following possible cases : When d = 1

In this case the possible A.P’s are 1, 2, 3; 2, 3, 4; 3, 4, 5; …; 19, 20 21

Thus, there are 19 such A.P’s.

When d = 2

In this case the possible A.P’s are 1, 3, 5; 2, 4, 6; 3, 5, 7; …; 17, 19, 21

Thus, there are 17 such A.P’s

When d = 3

In this case the possible A.P’s are 1, 4, 7; 2, 5, 8; 3, 6, 9; …; 15, 18, 21

Thus, there are 15 such A.P’s

……………………………………

…………………………………....

When d = 10

In this case the possible AP is 1, 11, 21 Thus, there is just one AP is this case.

\ number of favourable ways is

19 + 17 + 15 + … + 1 = $\frac { 10 } { 2 }$ (19 + 1) = 100

Hence, probability of the required event is

= $\frac { 100 } { 1330 }$ = $\frac { 10 } { 133 }$

General formula : If three tickets are selected at random out of (2n + 1) tickets numbered 1, 2, 3, …, (2n + 1), then the probability they are in A.P. is

$\frac { n ^ { 2 } } { { } ^ { 2 n + 1 } C _ { 3 } } = \frac { 3 n } { 4 n ^ { 2 } - 1 }$