Question

If 6n tickets numbered 0, 1, 2, ….. , (6n – 1) are placed in a

bag and three are drawn at random, then the probability that

the sum of numbers on the ticket is 6n, is-

• A.
• B.
• C. $\frac { 1 } { 2 }$.
• D. $\frac { 1 } { 2 }$ .

Answer 1

Answer: (B)

Answer 2

The total number of ways of selecting 3 tickets out of 6n number is ⁶ⁿC₃.

The sum of the numbers on three tickets drawn will be 6n in each of the following cases :

Lowest number	Ways of selection	Total number of ways
0	(0, 1, 6n – 1) (0, 2, 6n – 2) ... (0, 3n – 1, 3n + 1)	(3n – 1)
1	(1, 2, 6n – 3) (1, 3, 6n – 4) …(1, 3n – 1, 3n + 1)	(3n – 2)
2	(2, 3, 6n – 5) (2, 4, 6n – 6) … (2, 3n – 2, 3n)	(3n – 4)
3	(3, 4, 6n – 7) (3, 5, 6n – 8) … (3, 3n – 2, 3n – 1)	(3n – 5)
M	M	M
(2n – 2)	(2n – 2, 2n –1, 2n + 3), (2n – 2, 2n, 2n + 2)	2
(2n – 1)	(2n – 1, 2n, 2n + 1)	1

Total number of ways of getting 6n as the sum

= (3n – 1) + (3n – 2) + (3n – 4) + (3n – 5) + …. + 2 + 1

= {3n – 1 + 3n – 2} + {3n – 4 + 3n – 5} + …. +{5 + 4} +

= (6n – 3) + (6n – 9) + ….. + 9 + 3 = 3n²

\required probability = .