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As per analysis for previous years, it has been observed that students preparing for JEE MAINS find Mathematics out of all the sections to be complex to handle and the majority of them are not able to comprehend the reason behind it. This problem arises especially because these aspirants appearing for the examination are more inclined to have a keen interest in Mathematics due to their ENGINEERING background.

Furthermore, sections such as Mathematics are dominantly based on theories, laws, numerical in comparison to a section of Engineering which is more of fact-based, Physics, and includes substantial explanations. By using the table given below, you easily and directly access to the topics and respective links of MCQs. Moreover, to make learning smooth and efficient, all the questions come with their supportive solutions to make utilization of time even more productive. Students will be covered for all their studies as the topics are available from basics to even the most advanced.

Q1. If n positive integers are taken at random and multiplied together, then the probability that the last digit is 2,4,6 or 8, is
•  8n/5n
•  (8n-2n)/5n
•  (4n-2n)/5n
•  None of these
Solution
(c) If the last digit is 2,4,6 or 8, none of the numbers can end in 0 or 5 and one of the last digits must be even. Now 8n is the number of ways in which 0 and 5 can be excluded and of these we have further to include 4n cases in which the last digit can be selected solely from 1,3,7 or 9 ∴ Favourable number of ways =8n-4n Hence, required probability =(8n-4n)/10n =(4n-2n)/5n

Q2.If A and B are independent events such that P(A)>0,P(B)>0, then
•  A and B are mutually exclusive
•  A and Bcare independent
•  P(A∪B)=P(Ac)P(Bc)
•  P(A/B)=P(Ac/B)
Solution
(b) Expaination not available
Q3.  Out of 13 applicants for a job, there are 5 women and 8 men. It is desired to select 2 persons for the job. The probability that at least one of the selected persons will be a women is
•   25/39
•  14/39
•  5/13
•  10/13
Solution
(a) Required probibility=(5C1 × 8C1)/( 13C2 )+(5C2)/( 13C2)=25/39

Q4. If three natural numbers from 1 to 100 are selected randomly, then probability that all are divisible by both 2 and 3, is
•  4/105
•  4/33
•  4/35
•  4/1155
Solution
(d) Let E=E=Events of numbers divisible by 2 and 3 [ie, divisible by 6] =(6,12,…,96) n(E)=16 ∴Required probability =( 16C3)/( 100 C3 ) =((16×15×14)/(3×2×1))/((100×99×98)/(3×2×1))=4/1155

Q5.Two cards are drawn successively with replacement from a well shuffled deck of 52 cards, then the mean of the number of aces is
•  1/13
•  3/13
•  2/13
•  None of these
Solution
(c) Let X denote the number of aces. Probability of selecting aces, P=4/52=1/13 Probability of not selecting aces, q=1-1/13=12/13 P(X=1)=2×(1/13)×(12/13)=24/169 P(X=2)=2(1/13)2.(12/13)0=2/169 Mean =ΣP1 Xi=24/169+2/169=2/13

Q6. If n integers taken at random are multiplied together, then the probability that the last digit of the product is 1,3,7 or 9, is
•  2n/5n
•  (4n-2n)/5n
•  4n/5n
•   None of these
Solution
(a) If any number the last digits can be 0,1,2,3,4,5,6,7,8,9. Therefore, last digit of each number can be chosen in 10 ways. ∴ The last digit of all numbers can be chosen in 10n ways. If the last digit is to be 1,3,7, or 9, then none of the numbers can be even or end in 0 or 5. Thus, we have a choice of 4 digits viz.1,3,7, or 9 with which each of n numbers should end. So, favourable number of ways =4n Hence, required probability =4n/10n =(2/5)n

Q7.A box contains 10 good articles and 6 with defects, one article is chosen at random. What is the probability that it is either good or has a defect?
•  24/64
•  40/64
•  49/64
•  1
Solution
(d) ∴ Required probability=( 10C1 + 6C1)/( 10 C1 ) =16/16=1

Q8.Two coins and a die are tossed. The probability that both coins fall heads and the die shows a 3 or 6, is
•  1/8
•  1/12
•  1/16
•  None of these
Solution
(b) Required number of ways(b) Consider the following events: A→ Getting head on first coin B→ Getting head on second coin C→ Getting 3 or 6 on die These three events are independent with respective probabilities given by P(A)=1/2,P(B)=1/2,P(C)=2/6=1/3 ∴ Required probability =P(A∩B∩C) =P (A) P (B) P (C) =1/2×1/2×1/3=1/12

Q9.Which one of the following is not correct?
•  The probability that in a family of 4 children, there will be at least one boy, is 15/16
•  Two cards are drawn with replacement from a well shuffled pack. The probability of drawing both aces, is 1/169
•  The probability guessing at least 8 out 10 answers in a true false examination, is 7/64
•  A coin is tossed three times. The probability of getting exactly two heads, is 3/8
Solution
(c) (a) P (no boy in family of 4) =P ( all girls in it)=(1/2)4=1/16 Hence, the probability of having at least one boy =1-1/16=15/16 (b)P (first card is an ace)=1/13 and P (second card is an ace) =1/13 Therefore, P (both cards are aces) =1/13×1/13=1/169 (c) Let guessing correctly one answer as a success. Then, we have p=1/2,q=1/2,n=10 ∴ P(8)+P(9)+P(10) =10C8 (1/2)8 (1/2)2+ 10 C9 (1/2)9 (1/2)+ 10 C10 (1/2)10 =(45+10+1)/1024=7/128 (d) we have, n=3,p=1/2,q=1/2 Where obtaining a head has been reckoned a success. Now, P(2)=3C2 (1/2)2 (1/2)1=3/8 Hence, it is clear that option (c) is not correct.

Q10.The probability of choosing a number divisible by 6 or 8 from among 1 to 90 is

•   1/6
•   1/90
•   1/30
•  23/90
Solution
(d) Total number=90 Number divisible by 6 are {6,12,18,24,30,36,42,48,54,60,66,72,78,84,90} Numbers divisible by 8 are {8,16,24,32,40,48,56,64,72,80,88} Numbers divisible by 6 and 8 are {24,48,72} Total number of numbers divisible by 6 or 8 =15+11-3=23 ∴ Required probability=23/90 ## Want to know more

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