JEE MAIN - Mathematics (2020 - 6th September Morning Slot - No. 15)
Out of 11 consecutive natural numbers if three numbers are selected at random (without repetition), then the probability that they are in A.P. with positive common difference, is :
$${{10} \over {99}}$$
$${{5} \over {33}}$$
$${{15} \over {101}}$$
$${{5} \over {101}}$$
Explanation
Out of 11 consecutive natural numbers either
6 even and 5 odd numbers or 5 even and 6
odd numbers.
Let, E = Even
O = Odd
Case-1 :
E, O, E, O, E, O, E, O, E, O, E
2b = a + c $$ \Rightarrow $$ Even
$$ \Rightarrow $$ Both a and c should be either even or odd.
P = $${{{}^6{C_2} + {}^5{C_2}} \over {{}^{11}{C_3}}}$$ = $${5 \over {33}}$$
Case -2 :
O, E, O, E, O, E, O, E, O, E, O
P = $${{{}^5{C_2} + {}^6{C_2}} \over {{}^{11}{C_3}}}$$ = $${5 \over {33}}$$
Total probability = $${1 \over 2} \times {5 \over {33}}$$ + $${1 \over 2} \times {5 \over {33}}$$ = $${5 \over {33}}$$
Let, E = Even
O = Odd
Case-1 :
E, O, E, O, E, O, E, O, E, O, E
2b = a + c $$ \Rightarrow $$ Even
$$ \Rightarrow $$ Both a and c should be either even or odd.
P = $${{{}^6{C_2} + {}^5{C_2}} \over {{}^{11}{C_3}}}$$ = $${5 \over {33}}$$
Case -2 :
O, E, O, E, O, E, O, E, O, E, O
P = $${{{}^5{C_2} + {}^6{C_2}} \over {{}^{11}{C_3}}}$$ = $${5 \over {33}}$$
Total probability = $${1 \over 2} \times {5 \over {33}}$$ + $${1 \over 2} \times {5 \over {33}}$$ = $${5 \over {33}}$$
Comments (0)
