## PERMUTATION & PROBABILITY MCQ SET-10

Q.1 A farmer, has 9 buffaloes whose ages are 1,2,3,4…9 years respectively. Each buffalo gives milk daily, the quantity of which is proportional to the buffalo’s age. In how many ways can Farmer allocate the buffaloes among his three sons, such that each son gets the same quantity of milk each day ? A) 6 B) 7 C) 8 D) 9 Ans. A ?(1+2+…+9)=45 so,three sets of 15=(1,6,8),(2,4,9),(3,5,7) so,no of ways 3!=6 Q.2 In how many different ways can books A, B, C, D be piled up such that A comes exactly in between C and D ? A) 4 B) 2 C) 1 D) 3 Ans. A Assuming C, A, D as one group, they can be grouped in two ways, CAD or DAC. This group and B again can be arranged in 2 ways. So = 2 × 2 = 4 ways. Q.3 Out of 12 batsmen and 8 bowlers, a cricket team of 11 players has to be selected such that there should be at least 5 batsmen and 5 bowlers. In how many ways this can be done? A) 73,920 B) 72,640 C) 40,420 D) 46,400 Ans. A 5 bowlers and 5 batsmen mean the 11th can either be a batsman or a bowler. So, 6 batsmen and 5 bowlers or 5 batsmen and 6 bowlers. = 12C6 × 8C5 + 12C5 × 8C6 = 73,920 Q.4 An university has to select an examiner from a list of 60 persons. Of 25 are women and 35 men. 15 know French, 45 do not, 20 of them are teachers and 40 are not. What is the probability that the University selects a French knowing woman teacher? A) B) C) D) Ans. B Probability of selecting a teacher = P (T) = 20/60 = 1/3 Probability of selecting a woman = P (W) = 25/60 = 5/12 Probability of selecting a French knowing candidate = P (F) = 15/60 = 1/4 Required probability = P (T) P (W) P (F) = (1/3) (5/12) (1/4) = 5/144.

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