`5` different balls are placed in `5` different boxes randomly. Find the probability that exactly two boxes remain empty. Given each box can hold any

To draw 2 black, 4 white, and 3 red balls in order is same as arranging two black balls at first 2 places, 4 white balls at next 4 places,...

Each group should have odd numbered balls. Case I: Two groups have three odd numbered balls and third has only one odd numbered ball. Number of such cases = `(7!)/((3!)^(2)xx1!xx2!)xx3^(7)`...

Correct Answer - A Since each ball can be placed in any one of the 3 boxes, therefore there are 3 ways in which a ball can be palced in any...

Correct Answer - C We have, `n(S) = 5^(5)` For computing favorable outcomes, 2 boxes which are to remain empty, can be selected in `.^(5)C_(2)` ways and 5 marbles can be...

Correct Answer - A::B Let the number of red and blue balls be r and b, respectively. Then, the probability of drawing two red balls is `p_(1) = (.^(r )C_(2))/(.^(r+b)C_(2)) =...

Number of ways of drawing 7 balls (second draw) = `.^(10)C_(7)` For each set of 7 balls of the second draw, 3 must be common to the set of 5...

Correct Answer - A::C Given equation is `x^(2) + 2x sin(cos^(-1) y) + 1 =0` Since x is real, `D ge 0`. Therefore, `4(sin(cos^(-1)y))^(2) -4 ge 0` or `(sin(cos^(-1)y))^(2) ge 1`...

Correct Answer - B The required probability is `(n^(2))/(""^(2n)C_(2))((n-1)^(2))/(""^(2n-2)C_(2))((n-2)^(2))/(""^(2n-4)C_(2))...(2^(2))/(""^(4)C_(2))(1^(2))/(""^(2)C_(2))` `=((1xx2xx3xx4xx...xx(n-1)n)^(2))/(((2n)!)/2^(n))=(2^(n)(n!)^(2))/((2n)!)=(2^(n))/(""^(2n)C_(n))`

Correct Answer - B `P=15/25,q=10/25,n=10` `sigma^(2)=npq=10.""3/5.2/5=12/5`

Correct Answer - A P (required) = P (all are white) + P (all are red) + P (all are black) `=1/6xx2/9xx3/12+3/6xx3/9xx4/12+2/6xx4/9xx2/12=6/648+36/648+40/648=82/648`