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
Answer & explanation
Correct answer: option 1
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 $=\frac{100}{1330}=\frac{10}{133}.$