## Probability Quiz-9

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. A coin is tossed n times. The probability that head will turn up an odd number of times, is
•  1/2
•  (n+1)/(2 n)
•  (n-1)/(2 n)
•  (2n-1-1)/2n
Solution

Q2.If the letters of the word ‘REGULATIONS’ be arranged at random, the probability that there will be exactly 4 letters between R and E is
•  6/55
•  3/55
•  49/55
•  None of these
Solution
(a) There are 11 letters in the word “REGULATIONS” which can be arranged in 111 ways
Other than R and E there are 9 letters out of which 4 can be chosen in 9 C4 ways.
These four letters can be arranged between R and E in 4! Ways.
Also, R and E can interchange their positions in 2! Ways.
∴ Number of ways in which there are exactly four letters between R and E= 9 C4×4 !×2 !
Considering this group of 6 letters as one letters and the remaining 5 letters can be arranged in 6 ! ways ∴ Number of arrangements of the letters of the word “REGULATIONS” in which there are exactly four letters between R and E= 9 C4×4 !×2 !×6 !
Hence, required probability =( 9 C4×4 ! ×2 ! ×6 !)/(11 !)=6/55
Q3.  The probability of getting at least one tail in 4 throws of a coin is
•   15/16
•  1/16
•  1/4
•  None of these
Solution
(a) Required probability=1-(1/2)4=15/16

Q4. If A is a finite set, then the probability that the mapping is a bijection, is
•  1/nn
•  1/n!
•  (n-1) !/nn-1
•  (n !)/nn-1
Solution
(c) Since A is a finite set, therefore every injective map from A to itself is bijective also ∴Required probability =(n !)/nn =(n-1)!/n(n-1)

Q5.There are 4 white and 4 black balls in a bag and 3 balls are drawn at random. If balls of same colour are identical, the probability that none of them is black, is
•  1/4
•  1/14
•  1/2
•  None of these
Solution
(a) Three balls can be selected in the following ways:
White : 3 2 1 0
Black : 0 1 2 3
∴ Total number of ways =4,
Clearly, there is only one favourable ways in which three balls are white
Hence, required probability =1/4
NOTE
It should be noted that the number of ways of selecting 3 white balls from 4 white balls is not equal to 4C3, because all white balls are identical and all black balls are also identical

Q6. When three dice are thrown the probability of getting 4 or 5 on each of the dice simultaneously, is
•  1/72
•  1/108
• 1/24
•   None of these
Solution
(d) Total number of cases=63=216
Favorable cases are (4,4,5),(4,5,4),(5,4,4),(4,5,5),(5,4,5),(5,5,4),(5,5,5),(4,4,4,).
∴Total number of cases=8
∴Required probability=8/216=1/27

Q7.A coin is tossed 2n times. The chance that the number of times one gets head is not equal to the number of times one gets tail, is
•  (2n)!/(n!)2 (1/2)2n
•  1-(2n)!/(n!)2
•  1-(2n)!/(n!)2 .1/4n
•  None of these
Solution
(c) The required probability =1- probability of equal number of heads and tails.
Out of 2n tossed n times heads and n times tails. =1-2n Cn (1/2)n (1/2)(2n-n ) =1-(2n)!/n!n! (1/2)2n=1-(2n)!/(n!)2 ∙1/4n

Q8.The probability that a number n chosen at random from 1 to 30, to satisfy n+(50/n)>27 is
•  7/30
•  3/10
•  3/5
•  1/5
Solution
(d) Total outcomes=30
Now, n+(50/n)>27
⇒n2-27n+50>0
⇒(n-2)(n-25)>0
Favourable outcomes are 1,26,27,28,29,30 Number of favourable outcomes=6
∴ Required probability=6/30=1/5

Q9.Given two events A and B. If odds against A are 2 : 1 and those in favour of A∪B are as 3 : 1, then
•  1/2≤P(B)≤3/9
•  5/12≤P(B)≤3/4
•  1/2≤P(B)≤3/5
•  None of these
Solution
(b) 5/12 ≤ P(B) ≤ 3/4

Q10. In a binomial distribution the mean is 15 and variance is 10. Then parameter n is
•  28
•  16
•  45
• 25
Solution
(c) Given mean, np=15 and variance np(1-p)=10
∴ 1-p=10/15=2/3 ⇒p=1/3
∴ n=15×3=45

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