JEE MAIN - Mathematics (2010 - No. 6)
Four numbers are chosen at random (without replacement) from the set $$\left\{ {1,2,3,....20} \right\}.$$
Statement - 1: The probability that the chosen numbers when arranged in some order will form an AP is $${1 \over {85}}.$$
Statement - 1: The probability that the chosen numbers when arranged in some order will form an AP is $${1 \over {85}}.$$
Statement - 2: If the four chosen numbers form an AP, then the set of all possible values of common difference is $$\left( { \pm 1, \pm 2, \pm 3, \pm 4, \pm 5} \right).$$
Statement - 1 is true, Statement - 2 is true; Statement - 2 is not a correct explanation for Statement - 1.
Statement - 1 is true, Statement - 2 is false.
Statement - 1 is false, Statement -2 is true.
Statement - 1 is true, Statement - 2 is true; Statement - 2 is a correct explanation for Statement - 1.
Explanation
Four numbers can be chosen $${}^{20}{C_4}$$ ways.
When common difference d = 1 then the possible sets are (1, 2, 3, 4) (2, 3, 4, 5) (3, 4, 5, 6) ................. (17, 18, 19, 20) = 17 sets
So when d = 1 then 17 different AP's are possible with 4 numbers.
Now let's create a table of all possible sets -
$$\therefore$$ Total no of AP = 17 + 14 + 11 + 8 + 5 + 2 = 57
$$\therefore$$ Required probability = $${{57} \over {{}^{20}{C_4}}}$$ = $${1 \over {85}}$$
$$\therefore$$ Statement - 1: is true.
$$\therefore$$ Statement - 2: is false as common difference can also be $$ \pm 6$$.
When common difference d = 1 then the possible sets are (1, 2, 3, 4) (2, 3, 4, 5) (3, 4, 5, 6) ................. (17, 18, 19, 20) = 17 sets
So when d = 1 then 17 different AP's are possible with 4 numbers.
Now let's create a table of all possible sets -
| Common Difference (d) |
Possible Sets | No of AP |
|---|---|---|
| d = 1 | (1, 2, 3, 4) (2, 3, 4, 5) (3, 4, 5, 6) ................. (17, 18, 19, 20) | 17 |
| d = 2 | (1, 3, 5, 7) (2, 4, 6, 8) (3, 5, 7, 9) ................. (14, 16, 18, 20) | 14 |
| d = 3 | (1, 4, 7, 10) (2, 5, 8, 11) (3, 6, 9, 12) ................. (11, 14, 17, 20) | 11 |
| d = 4 | (1, 5, 9, 13) (2, 6, 10, 14) (3, 7, 11, 15) ................. (8, 12, 16, 20) | 8 |
| d = 5 | (1, 6, 11, 16) (2, 7, 12, 17) (3, 8, 13, 18) (4, 9, 14, 19) (5, 10, 15, 20) | 5 |
| d = 6 | (1, 7, 13, 19) (2, 8, 14, 20) | 2 |
$$\therefore$$ Total no of AP = 17 + 14 + 11 + 8 + 5 + 2 = 57
$$\therefore$$ Required probability = $${{57} \over {{}^{20}{C_4}}}$$ = $${1 \over {85}}$$
$$\therefore$$ Statement - 1: is true.
$$\therefore$$ Statement - 2: is false as common difference can also be $$ \pm 6$$.
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