Mr. A lists all the positive divisors of the number N =(2010)2. If Mr.A randomly selects two divisors from the list then the probability that exactly one of the selected divisor is a perfect square, is

Question

Mr. A lists all the positive divisors of the number N =(2010)2. If Mr.A randomly selects two divisors from the list then the probability that exactly one of the selected divisor is a perfect square, is (A) (26/81) (B) (13/81) (C) (1/3) (D) (2/9)

Tardigrade Admin · Accepted Answer

N =(2010)2=22 ⋅ 32 ⋅ 52 ⋅(67)2=( a ⋅ b ⋅ c ⋅ d text say ) text Total number of divisors =3 ⋅ 3 ⋅ 3 ⋅ 3=81 text Number of divisors which are perfect square is = 4 C 0+ 4 C 1+ 4 C 2+ 4 C 3+ 4 C 4=16 <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/mathematics/43dfc9ba7b414baea6ef70194bdfa77b-.png" /> Let E: two randomly chosen divisors have exactly one perfect square. P ( E )=( 16 C 1 ⋅ 65 C 1/ 81 C 2)=(26/81)

Q. Mr. A lists all the positive divisors of the number $N = (2010)^{2}$ .
If Mr.A randomly selects two divisors from the list then the probability that exactly one of the selected divisor is a perfect square, is

$\frac{26}{81}$

$\frac{13}{81}$

$\frac{1}{3}$

$\frac{2}{9}$

Q. Mr. A lists all the positive divisors of the number N=(2010)2. If Mr.A randomly selects two divisors from the list then the probability that exactly one of the selected divisor is a perfect square, is

8126​

8113​

31​

92​

N=(2010)2=22⋅32⋅52⋅(67)2=(a⋅b⋅c⋅d say ) Total number of divisors =3⋅3⋅3⋅3=81 Number of divisors which are perfect square is =4C0​+4C1​+4C2​+4C3​+4C4​=16 Let E : two randomly chosen divisors have exactly one perfect square. P(E)=81C2​16C1​⋅65C1​​=8126​

Q. Mr. A lists all the positive divisors of the number $N = (2010)^{2}$ .
If Mr.A randomly selects two divisors from the list then the probability that exactly one of the selected divisor is a perfect square, is

$\frac{26}{81}$

$\frac{13}{81}$

$\frac{1}{3}$

$\frac{2}{9}$