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

(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.

^{n}-4

^{n}

^{n}-4

^{n})/10

^{n}=(4

^{n}-2

^{n})/5

^{n}

**Q2.**If A and B are independent events such that P(A)>0,P(B)>0, then

(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

(a) Required probibility=(

^{5}C

_{1}×

^{8}C

_{1})/(

^{13}C

_{2})+(

^{5}C

_{2})/(

^{13}C

_{2})=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

(d) Let E=E=Events of numbers divisible by 2 and 3 [ie, divisible by 6] =(6,12,…,96) n(E)=16 ∴Required probability =(

^{16}C

_{3})/(

^{100}C

_{3}) =((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

(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 =Î£P

_{1}X

_{i}=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

(a) If any number the last digits can be 0,1,2,3,4,5,6,7,8,9.

^{n}ways.

^{n}

^{n}/10

^{n}=(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?

(d) ∴ Required probability=(

^{10}C

_{1}+

^{6}C

_{1})/(

^{10}C

_{1}) =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

(b) Required number of ways(b)

**Q9.**Which one of the following is not correct?

(c) (a) P (no boy in family of 4) =P ( all girls in it)=(1/2)

^{4}=1/16

^{10}C

_{8}(1/2)

^{8}(1/2)

^{2}+

^{10}C

_{9}(1/2)

_{9}(1/2)+

^{10}C

_{10}(1/2)

_{10}

^{3}C

_{2}(1/2)

^{2}(1/2)

^{1}=3/8

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

(d) Total number=90