Probability Quiz-10
Dear Readers,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. An anti-aircraft gun can take a maximum of four shots at any plane moving away from it. The probabilities of hitting the plane at the 1st, 2nd ,3rd and 4th shots are 0.4, 0.3, 0.2 and 0.1 respectively. What is the probability that at least one shot hits the plane?
Solution
(a) Probability that at least one shot hits the plane =1-P(none of the shot hits the plane)
(a) Probability that at least one shot hits the plane =1-P(none of the shot hits the plane)
=1-0.6×0.7×0.8×0.9
=1-0.3024
=0.6976
Q2.If X has binomial distribution with mean np and variance npq,then (P(X=k))/(P(X=k-1)) is equal to
Solution
(b) (P(X=K))/(P(X=k-1))
(b) (P(X=K))/(P(X=k-1))
=( n Ck pk q(n-k))/( n C(k-1) p(k-1) q(n-k+1) )
=((n-k+1)/k).p/q
Q3. Five different objects A1,A2,A3,A4,A5 are distributed randomly in 5 places marked 1,2,3,4,5. One arrangement is picked at random. The probability that in the selected arrangement, none of the object occupies the place corresponding to its number, is
Solution
(c) We have, Total number of arrangements of 5 objects =5 !
(c) We have, Total number of arrangements of 5 objects =5 !
We know that the total number of de-rangements of n objects in which none of the object occupies its original position is given by
n !{1-1/(1 !)+1/(2 !)-1/(3 !)+1/(4 !)…+(-1)n/(n !)}
Therefore, the total number of de-rangements in which none of the 5 object occupies the place corresponding to it
=5 !{1-1/(1 !)+1/(2 !)-1/(3 !)+1/(4 !)-1/(5 !)}=44
Hence, required probability =44/120=11/30
Q4. If there are 6 girls and 5 boys who sit in a row, then the probability that no two boys sit together is
Solution
(c) Six girls and 5 boys can sit in a row in 11 ! ways
(c) Six girls and 5 boys can sit in a row in 11 ! ways
∴ Total number of elementary events =11 !
Six girl can sit in a row in 6 ! and in each such arrangement there are 7 places between them in which 5 boys can be seated in 7 C5×5 ! ways.
Therefore, the total number of ways in which no two boys sit together =6 !× 7 C5×5 !
Hence, required probability =(6 !∙ 7 C5×5 !)/(11 !)=(6 !7 !)/(2 !11 !)
Q5.A manufacture of cotter pins knows that 5% of his product is defective. He sells pins in boxes of 100 and guarantees that not more than one pin will be defective in a box. In order to find the probability that a box will fail to meet the guaranteed quality, the probability distribution one has to employ is
Solution
(b) Required probability distribution is poisson distribution.
(b) Required probability distribution is poisson distribution.
Q6. Consider two events A and B such that P(A)=1/4,P(B/A)=1/2,P(A/B)=1/4. For each of the following statements, which is true?
1. P(Ac+Bc )=3/4
2. The events A and B are mutually exclusive.
3. P(A/B) +P(A/Bc)=1
Solution
(a) ∴P(B/A)=(P(A∩B))/(P(A))
(a) ∴P(B/A)=(P(A∩B))/(P(A))
⇒1/2=(P(A∩B))/(1/4)
⇒ P(A∩B)=1/8
Hence, event A and B are not mutually exclusive.
∴ Statement 2 is incorrect.
P(A/B)=(P(A∩B))/(P(B))
⇒P(B)=(P(A∩B))/P(A/B) =(1/8)/(1/4)
⇒P(B)=1/2
∴ Event A and B are independent events.
P(Ac/Bc )=(P(Ac∩Bc))/(P(Bc))=(P(Ac )P(Bc))/(P(Bc))
=3/4∙1/2∙2/1=3/4
Hence, statement 1 is correct.
Again, P(A/B)+P(A/Bc )=1/4+(P(A ∩ Bc))/(P(Bc))
=1/4+(P(A)-P(A∩B))/(P(Bc))
=1/4+(1/4-1/8)/(1/2)=1/4+1/4=1/2
Hence, statement 3 is correct.
Q7.The probability that A can solve a problem is 2/3 and B can solve it is 3/4. If both attempt the problem, what is the probability that the problem gets solved?
Solution
(a) Required probability =1-(1-2/3)(1-3/4)=11/12
(a) Required probability =1-(1-2/3)(1-3/4)=11/12
Q8.For two events A and B,if P(A)=P(A/B)=1/4 and P(B/A)=1/2, then
Solution
(d) Given, P(B/A)=1/2
(d) Given, P(B/A)=1/2
⇒P(B∩A)=1/8
and P(A/B)=1/4
⇒P(B)=1/2
∴P(A∩B)=1/8=P(A).P(B)
∴ Events are independent
Now, P(A'/B)=(P(A'∩B))/(P(B))=(P(B)-P(A∩B))/(P(B))=3/4
and
P(B'/A)=(P(A∩B'))/(P(A))=(P(A)-P(A∩B))/(P(A))=1/2
Q9.A die is thrown. Let A be the event that the number obtained is greater than 3. Let B be the event that the number obtained is less than 5. Then P(A∪B)is
Solution
(d) Since, A={4,5,6}and B={1,2,3,4}
(d) Since, A={4,5,6}and B={1,2,3,4}
∴A∩B={4}
∴P(A∪B)=P(A)+P(B)-P(A∩B)
⇒P(A∪B)=3/6+4/6-1/6=1
Q10. One hundred cards are numbered from 1 to 100. The probability that a randomly chosen card has a digit 5 is
Solution
(c) There are 10 numbers from 50 to 59 such that each has a digit 5 and there are 9 other numbers, 5,15,25,35,45,65,75,85,95 each containing a digit 5
(c) There are 10 numbers from 50 to 59 such that each has a digit 5 and there are 9 other numbers, 5,15,25,35,45,65,75,85,95 each containing a digit 5
So, the favourable number of elementary events =19
Total number of elementary events is 100
Hence, required probability =19/100
