Question

5. Out of (2n + 1) tickets consecutively numbered
from 1 to 2n + 1, three are drawn at random.
Find the probability that the numbers on them
are in A.P.​

Answer 1

Answer:

4n

2

−1

3n

Step-by-step explanation:

Out of 2n+1 numbers,n+1 would be odd and n would be even or vice-versa. If we select any 2 of the n+1 numbers, the third number will be automatically decided so as to have the three in AP and the same reasoning would be valid if we select any 2 of the n numbers (for example, if we select 3 and 11, the third number has to be 7 OR if we select 6 and 14, the third number has to be 10).

Thus, the probability =

2n+1

C

3

n+1

C

2

+

n

C

2

=

2!∗(2n+1)∗2n∗(2n−1)

[(n+1)∗n+n∗(n−1)]∗3!

=

4n

2

−1

3n