1q1
1/6
7C3 /10C4
(1) Various injective functions f are defined from set A≡ {1,2,3,4} to {1.2.3.4.5.6.7.8.9.10} and a function among them is choosen randomly. If it is found to be increasing , then the probability that f(4) = 8 is
(A) 1/12 (B) 1/3 (C) 1/4 (D) 1/6
(2) Three dice are thrown simultaneously. The probability that 4 appears on two dice given that 5 has occured on atleast one die is
(A) 3/91 (B) 9/91 (C) 81/91 (D) 11/91
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UP 0 DOWN 0 0 7
7 Answers
132) Cases Occuring-
(5,X,X) (X,5,X) (X,X,5).....P(1) = 3*(1/6*5/6*5/6)
(5,5,X) (5,X,5) (X,5,5)......P(2) = 3*(1/6*1/6*5/6)
(5,5,5)........P(3) = 1/6*1/6*1/6
Favourable cases -
(5,4,4) (4,5,4) (4,4,5)......P(f)= 3*1/6*1/6*1/6
Req. Prob. = P(f)/{P(1)+P(2)+P(3)} = 3/91
1Injective means one-one function
That means exactly four mappings from A to B
Total cases=Choosing four elements from 10=10C4
Given that f(4)=8 and it is increasing function,we define favorable cases as choosing 3 elements out of 7 elements which are less than 8
That implies favorable cases=7C3
Probability =favorable casesTotal cases=7C3
10C4
