| DIRECTION: Each of these questions contains two statements: Statement-1 (Assertion) and Statement-2 (Reason). Choose the correct answer (Only one option is correct) from the following - |
| Statement-1: Out of 5 tickets consecutively numbered, three are drawn at random, the chance that the numbers on them are in A.P. is\[\frac{2}{15}\] |
| Statement-2: Out of\[(2n+1)\]tickets consecutively numbered, three are drawn at random, the chance that the numbers on them are in A.P. is\[\frac{3n}{4{{n}^{2}}-1}\] |
A) Statement-1 is false, Statement-2 is true.
B) Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1.
C) Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1.
D) Statement-1 is true, Statement-2 is false.
Correct Answer: A
Solution :
For a, b, c in A .P. a + c = 2b \[\Rightarrow \] a + c is eyen, so, a and c are both even or both odd So, a and c can be chosen in \[^{n}{{C}_{2}}{{+}^{n+1}}{{C}_{2}}={{n}^{2}}\] ways \[\therefore \] \[P(E)=\frac{{{n}^{2}}}{{{(2n+1)}_{{{C}_{3}}}}}=\frac{{{n}^{2}}\times 3\times 2\times 1}{(2n+1)2n(2n-1)}=\frac{3n}{4{{n}^{2}}-1}\]\[2n+1=5\] \[2n=4\] \[n=2;P(E)=\frac{3n}{4{{n}^{2}}-1}=\frac{6}{15}=\frac{2}{5}\]
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