IIT JEE , AIEEE Forum - Mathematics , Algebra

ananth_patri

Out of 15 tickets marked with numbers from 1 to 15 three tickets are drawn at random .what is the chance tht the nos. on them are in AP??...

plz solve fast to get rated.....

allamraju

I think the ans is 7/65.I will post the soln. next.

allamraju

For the three no.s to be in A.P,b=a+c/2 $\Rightarrow$ a+c must be even $\Rightarrow$ Both a,c are even or both a,c are odd.Hence the probability is 8c₂+7c₂/15c₃=28+21/(35)(13)=7/65

mukundmadhav

What I do is, consider A.Ps of common differences 1 through 7
For 7, only one solution
For 6, 3 solutions
For 5, 5 solutions
For 4, 7 solutions
For 3, 9 solutions
For 2, 11 solutions
For 1, 13 solutions

So total of 49 favourable outcomes. And a total of 455 outcomes
So answer's 7/65..

But allamraju's method is obviously much better..

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I think the ans is 7/65.I will post the soln. next.

For the three no.s to be in A.P,b=a+c/2a+c must be evenBoth a,c are even or both a,c are odd.Hence the probability is 8c2+7c2/15c3=28+21/(35)(13)=7/65

For the three no.s to be in A.P,b=a+c/2 $\Rightarrow$ a+c must be even $\Rightarrow$ Both a,c are even or both a,c are odd.Hence the probability is 8c₂+7c₂/15c₃=28+21/(35)(13)=7/65