## Probability Quiz-5

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.Fifteen coupons are numbered 1 to 15. Seven coupons are selected at random, one at a time with replacement. The probability that the largest number appearing on a selected coupon be 9, is
•  (1/15)7
•  (8/18)7
•  (3/5)7
•  None of these
Solution
(c) Probability of each case =9/15=3/5
Required probability (with replacement)=(3/5)7

Q2.The mean and variance of a random variable X having a binomial distribution are 4 and 2 respectively. Then, P(X>6) is equal to
•  1/256
•  3/256
•  9/256
•  7/256
Solution
(c) Given, np=4,npq=2
⇒ p=q=1/2
∴ n=4×2=8
∴ P(X>6)=8C7 (1/2)7 (1/2) +8 C8(1/2)8
=8/256+1/256=9/256
Q3.A bag contains 7 red and 2 white balls and another bag contains 5 red and 4 white balls. Two balls are drawn, one from each bag. The probability that both the balls are white, is
•   2/9
•  2/3
•  8/81
•  35/81
Solution
(c) Required probability =2C1×4C1/9C1 ×9C1 =8/81

Q4. A pair of dice is rolled together till a sum of either 5 or 7 is obtained. The probability that 5 comes before 7 is
•  2/5
•  1/5
•  3/5
•  None of these
Solution
(a) Explaination not available

Q5.A and B are two independent events such that P(A) 1/2 and P(B)=1/3,then P(neither A nor B)is equal to
•  2/3
•  1/6
•  5/6
•  1/3
Solution
(d) Since, A and B are independent events.
∴ P(A∩B)=P(A).P(B)=1/2×1/3=1/6
Now, P(A∪B)=P(A)+P(B)-P(A∩B) =1/2+1/3-1/6=2/3
∴ P(Ac∩Bc )=1-P(A∪B) =1-2/3=1/3

Q6. A five digit number is chosen at random. The probability that all the digit are distinct and digits at odd places are odd and digits at even place are even, is
•  1/60
•  2/75
• 1/50
•  1/75
Solution
(d) Total number of 5 digit number =9×10×10×10×10=90000
Number of favourable numbers =5×5×4×4×3=12000
Thus, required probability =1200/90000=1/75

Q7.A and B are the independent events. The probability that both occur simultaneously is 1/6 and the probability that neither occur is 1/3. The probability of occurrence of the events A and B is
•  1/2,3/2
•  1/2,1/3
•  Not possible
•  None of these
Solution
(b) Given, P(A∩B)=1/6
⇒P(A)P(B)=1/6 ...(i)
and P(Ac∩Bc )=1/3
⇒P(Ac )P(Bc )=1/3
⇒{1-P(A)}{1-P(B)}=1/3
⇒1-1/3+P(A)P(B)=P(A)+P(B)
⇒2/3+1/6=P(A)+P(B) [from Eq.(i)]
⇒P(A)+P(B)=5/6 ...(ii)
On solving Eqs. (i) and (ii),
we get P(A)=1/2,P(B)=1/3 or P(A)=1/3,P(B)=1/2

Q8.If A and B are two independent events, then A and Bc are
•  Not independent
•  Also independent
•  Mutually exclusive
•  None of these
Solution
(b) We have, P(A∩B)=P(A)P(B)
Now, P(A∩Bc)=P(A)-P(A∩B)
⇒P(A∩Bc )=P(A)-P(A)P(B)=P(A)P(Bc)
∴A and Bc are independent events

Q9.The probability that the same number appear on throwing three dice simultaneously, is
•  1/36
•  5/36
•  1/6
•  4/13
Solution
(a) Total number of favorable cases =6
Total number of cases =216
∴ Required probability =6/216=1/36

Q10. If A and B are two independent events, the probability that both A and B occur is 1/8 and the probability that neither of them occurs is 3/8. The probability of the occurrence of A, is
•  1/2,1/4
•  1/3,1/4
•  1/4,1/6
• 1/5,1/2
Solution
(a) Since A and B are independent events
∴P(A∩B)=1/8 and P(Ac∩Bc)=3/8
⇒P(A)P(B)=1/8 and P(Ac )P(Bc )=3/8
Now, P(Ac∩Bc )=3/8
⇒1-P(A∪B)=3/8
⇒1-{P(A)+P(B)-P(A∩B)}=3/8
⇒1-{P(A)+P(B)}+1/8=3/8
⇒P(A)+P(B)=3/4
The quadratic equation whose roots are P(A) and P(B) is x2-x{P(A)+P(B)}+P(A)P(B)=0
⇒x2-3/4 x+1/8=0
⇒8 x2-6 x+1=0
⇒x=1/2,1/4
∴P(A)=1/2,1/4

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