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 52C4 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 => 13C4 +13C4 +13C4 +13C4 = 4* 13C4 So required probability = (4*13C4)/52C4 = 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=64 n(S)=64 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 = 6/64 = 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