Answer: D
Let A be the event that ball selected from the first bag is red and ball selected from second bag is green.
Let B be the event that ball selected from the first bag is green and ball selected from second bag is red.
The, P(A)= 5/8 *6/10 = 3/4
and P(B)= 3/8 * 4/10 = 3/20.
Hence required probability = P(A)+P(B)= 3/4+ 3/20 = 9/10

Answer: C
Let probability of the fist event taking place be A and probability of the second event taking place be B.
Then P(A)= 3/(5+3)= 3/8
P(B)= 7/(7+5)= 7/12
The required event cab be defined as that A takes place and B does not take place (A or B takes place and A does not take place or A takes place and B takes place.)
=> {P(A)[1-(P(B)]}+{P(B)[1-P(A)]}+{P(A)P(B)}
=> {3/8 * 5/12}+{5/8 * 7/12}+{3/8 * 7/12}= (15+35+21)/96
=> 71/96.

Answer: A
P(X)= 1/5, P(Y)= 1/4, P(Z)= 1/3
Required probability = [P(A)P(B)(1-P(C))] + [(1-P(A))P(B)P(C)] + [P(A)(C)(1-P(B))] + P(A)P(B)P(C)
=[1/4 * 1/3 * 4/5] + [3/4 * 1/3 * 1/5] + [2/3 * 1/4 * 1/5] + [1/4 * 1/3 * 1/5]
=> 4/60 + 3/60 + 2/60 + 1/60 = 10/60 = 1/6

Answer: D
There are 13 spade and 3 more jack
Probability of getting spade or a jack = (13+3)/52 = 16/52 = 4/13
So probability of getting neither spade nor a jack = 1- 4/13 = 9/13

Answer: A
n(S)= number of ways of sitting 12 persons at round table = (12-1)!= 11!.
Since two persons will be always together, then number of persons = 10+1 = 11. So, 11 persons will be seated in (11-1)! = 10! ways at round table and 2 particular persons will be seated in 2! ways.
n(A)= The number of ways in which two persons always sit together = 10! * 2!
P(A)= n(A)/n(S)= (10! *2!)/11! = 2/11.

Answer: D
Required probability = (4_C1*4_C1*4_C1) / 52_C3= (4*4*4)/22100 = 16/5525