Math, asked by DanishKhan4575, 1 year ago

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

Answered by amitnrw
1

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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