Each of X and Y take not more than 10 pens at random from a box containing unlimited number of pens. What is the probability that the difference in the number of pens they take is not more than 5?

Answer: A Let 'P' be the number of pens X takes, 'Q' be the number of pens B takes According to condition, |P-Q| <=5 'P' and 'Q' can take integral values between 0 and 10. Number of pairs possible for p and q is (11*11)= 121 Number of different values of q for a given value of p, |p-q|<=5 p=0, |-q|<=5, -5<=q<=5. q can take {0,1,2,3,4,5}--> total 6 values p=1, q will similarly take 7 values.. Total number of favourable cases would be = 6+7+8+9+10+11+10+9+8+7+6 = 91. Required probability = 91/121

Q. No. 2:

Sum of digits of a 5 digit number is 41. Find the probability that such a number is divisible by 11?

Answer: D In order to get the sum as 41, the following 5 digit combination exist: 99995--->number of 5 digits =5 99986--->number of 5 digit =20 99977--->number of 5 digit =10 99887--->number of 5 digit =30 98888--->number of 5 digit =5 Now, 70 such number exists. Now for a 5 digit number of form (pqrst) to be divisible by 11(p+r+t)-(q+s)=11, also (p+r+t)+(q+s)= 41 p+r+t = 26, q+s=15 (p,r,t)= (9,9,8) and (q,s)= (8,7)------eq(1) or (p,r,t)= (9,9,8) and (q,s)=(9,6)--------eq(2) Using 1st equation we can construct 3!/2! * 2! = 6 numbers. Using 2nd equation we can construct 3!/2! *2! = 6 numbers. Number of favorable cases = 12. Hence, required probability = 12/70 = 6/35

Q. No. 3:

Positive integers (a and b)are chosen from a set of integers. What is the probability that the units digit of the product (297)^{a} * (323)^{b} is 1?

Answer: D Units digit of (297)^{a} = {7,9,3,1} and,Units digit of (323)^{b} = {3,9,7,1}Number of combinations of these, which would give 1 in the unit digit of the productwould be:- (7*3), (9*9), (3*7), (1*1) , Favorable cases =4.Total number of cases = (4*4)= 16Required probability = 4/16 = 1/4

Q. No. 4:

In a game there are 70 people in which 40 are boys and 30 are girls, out of which 10 people are selected at random. One from the total group, thus selected is selected as a leader at random. What is the probability that the person, chosen as the leader is a boy?

Answer: A The total groups contains boys and girls in the ratio 4:3 If some person are selected at random from the group, the expected value of the ratio of boys and girls will be 4:3 If the leader is chosen at random from the selection, the probability of him being a boy = 4/7.

Q. No. 5:

Two urn contains 5 white and 7 black balls and 3 white and 9 black balls. One ball is transferred to the second urn and then one ball is drawn from the second urn. Find the probability that the first ball transferred is black, given that the ball drawn is black?

Answer: C P(A) = probability that the ball transferred from urn first to second is black = 7/12. P(B) = probability that the ball drawn from second urn is black. Case I--> If white ball goes to urn II, P(C)= 5/12 * 9/13 case II--> If black ball goes to urn II P(D) = 7/12 * 10/13 Thus, P(B) = P(C) + P(D) = (45/156) + (70/156) = 115/156. P(A/B)= Probability of event A when B has occurred = Probability that ball drawn from IInd urn is black. => P(A intersects B)/ P(B) = (7/12 *10/13)/ 115/156 = 70/115 = 14/23.

Q. No. 6:

Summation(n) is written for n=1 to n=199 on cards. What is the probability of drawing a card with an even number written on it?

Answer: C Sigma(1)= 1---->odd Sigma(2)= 3---->odd Sigma(3)= 6---->even
Sigma(4)= 10---->even Sigma(5)= 15----->odd Sigma(6)= 21----->odd sigma(7)= 28----->even.....and so on we see that we have 2 odds with evens. This will go on till the cards having Summation(196) gives us 98 even and 96 odds numbers. for n= 197 and n=198 It is odd and for 199 it is even, thus, required probability = 96+1/199= 97/199