three tickets are taken out of 21 tickets which are numbered from 1 to 21. find the probability that the numbers printed on three tikets may be in A.P
Answers
Three tickets taken out of 21 numbered from 1 to 21 then Probability of having numbers printed on three tickets in A.P = 10/133
Step-by-step explanation:
Total numbers = ²¹C₃ = 1330
Cases in AP
1 2 3 , 1 3 5 , 1 4 7 , 1 5 9 , 1 6 11 , 1 7 13 , 1 8 15 , 1 9 17 , 1 10 19 , 1 11 21
- 10 cases
2 3 4 , 2 , 4 , 6 , ..........................................................................................2 , 11 , 20
9 cases
3 4 5 , 3 5 7 , ...............................................................................................3 , 12 , 21
9 cases
Similalry
8 + 8 ( for 4 & 5 starting)
7 + 7 ( for 6 & 7 starting)
6 + 6 ( for 8 & 9 starting)
5 + 5 ( for 10 & 11 starting)
4 + 4 ( for 12 & 13 Starting)
3 + 3 ( for 14 & 15 starting)
2 + 2 ( for 16 & 17 starting)
1 + 1 ( for 18 & 19 starting)
Total Cases
= 10 + 9 + 9 + 8 + 8 + 7 + 7 + 6 + 6 + 5 + 5 + 4 + 4 + 3 + 3 + 2 + 2 + 1 + 1
= (1 + 2 + ..............+ 10) + (1 + 2 + ...............+ 9)
= 55 + 45
= 100
Probability = 100/1330
=> Probability = 10/133
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