Answer: B Let X be required events and S be the sample space, then X= {(2,3),(2,6),(4,3),(4,6),(6,3),(6,6),(3,2),(6,2),(3,4),(6,4),(3,6)} n(X)=11, n(S)=36 Hence, required probability = n(X)/n(S)= 11/36

Q. No. 2:

What is the probability that a number selected from numbers 1,2,3,...,30, is prime number, when each of the given numbers is equally likely to be selected?

Answer: D Four cards can be selected from 52 cards in 52_{C4} ways. Now, there are four suits, e.g club, spade, heart and diamond each of 13 cards. So total number of ways of getting all the four cards of the same suit => 13_{C4} +13_{C4} +13_{C4} +13_{C4} = 4* 13_{C4} So required probability = (4*13_{C4})/52_{C4} = 198/20825

Q. No. 4:

Four dice are thrown simultaneously. Find the probability that all of them show the same face.

Answer: A The total number of elementary events associated to the random experiments of throwing four dice simultaneously is 6*6*6*6=6^{4} n(S)=6^{4} Let X be the event that all dice show the same face. X={(1,1,1,1,),(2,2,2,2),(3,3,3,3),(4,4,4,4),(5,5,5,5),(6,6,6,6)} n(X)= 6 Hence required probability = 6/6^{4} = 1/216

Q. No. 5:

Find the probability that in a random arrangement of the letters of the word 'UNIVERSITY' the two I's come together.

Answer: D The total number of words which can be formed by permuting the letters of the word 'UNIVERSITY' is 10!/2! as there is two I's. Hence n(S)= 10!/2! Taking two I's as one letter, number of ways of arrangement in which both I's are together = 9!. So n(X)= 9! Hence required probability = 9!/(10!/2!) = 1/5.

Q. No. 6:

A bag contains 21 toys numbered 1 to 21. A toy is drawn and then another toy is drawn without replacement. Find the probability that both toys will show even numbers.

Answer: B The probability that first toy shows the even number = 10/21. Since, the toy is not replaced there are now 9 even numbered toys and total 20 toys left. Hence, probability that second toy shows the even number = 9/20. Required probability = (10/21)*(9/20) = 9/42