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Administrator

1. (d)
2. (b)
3. (b)
4. (c)
5. (b)
6. (a)
7. (a)
8. (a)
9. (c)
10. (b)
11. (c)
12. (a)
13. (d)
14. (a)
15. (b)
16. (a)
17. (b)
18. (c)
19. (a)
20. (a)
21. (b)
22. (b)
23. (a)
24. (b)
25. (c)
26. (b)
27. (d)
28. (d)
29. (d)
30. (b)
Ref: MSQE 2014 PEA


Sir,
Plz clarify this doubt
In ques no 6, the occurrence of 4 on the blue die will effect the result of the Event E as (3,4) is one of the event for event E to take place. Dont you think Event E and F are dependent ??
N sir it would be of great help if you could help in solving ques no 7,10 and 14
Thanks :)

Administrator

For Q 5.
Pr(E) = Pr(the sum of the numbers on the two dice is 7) = 1/6
Pr(F) = Pr(the number on the blue die equals 4) = 1/6
Pr(E. F) = Pr(F, G) = 1/36
Since, we also have Pr(E)Pr(F) = 1/36
Therefore, E and F are independent.

Administrator

For Q 7, we have to count the number of ways in which we can fill these seven spots using letters from {A, B, C} so that A cannot be followed by B, B cannot be followed by C, and C cannot be followed by A
_ _ _ _ _ _ _
First spot can be filled by any of the three letters, i.e. in three ways. Once the first spot is filled, we have only two ways to fill the second spot. For example, if the first spot is filled by A, then second spot will be taken by either A or C. Similarly for the third spot and so on. So there are 3(2^6) = 192 ways.

Administrator

For Q 10 the easiest way to do it is to recognise that F(x) <= G(x) for all x means that for any x, X takes smaller values less than x is more likely to occur in G than in F. Or in other words, X takes smaller values than x in G more often than it takes in F. So, expectation of X in F must be greater than or equal to expectation of X in G. More formal proof is attached.
sol.png

Administrator

For Q 14,
One of the ways to do this problem is:
Pr(Head turns up for the first time after even number of tosses)
= Pr(TH) + Pr(TTTH)+ Pr(TTTTTH) + ......
where TH is the event that first toss is tail and second is head, TTTH is the event that first three tosses are tails and forth toss is heads, etc. Thus,
Pr(Head turns up for the first time after even number of tosses)
= [1/(2^2)] + [1/(2^4)] + [1/(2^6)] + ...
This is a geometric series that sums to 1/3.


Amit sir,
Can u plz describe the approach for solving 23rd? can't we solve for real n equal roots of given quadratic equation giving 'alpha' = 12.25 or none of these as answer?

Administrator

This post was updated on .
For Q 23,
You can do the problem directly by observing that (x 1)(x^2  7x + a) = 0 has one root as 1. It will have exactly two roots under the following two cases:
(i) (x^2  7x + a) = 0 has two distinct roots and one of the roots is 1.
(ii) (x^2  7x + a) = 0 has exactly one root that is not equal to 1.
In case (i), for x^2  7x + a to be divisible by (x  1), a must be equal to 6.
Thus, for a = 6
x^2  7x + 6 = (x1)(x6)
we have two roots.
In case (ii), a = (3.5)^2 = 12.25
Thus, we have option (a) 6
as the right answer.


Thanks a lot sir :) Really helpful


please explain question 11 ans 6.


Sir please help me with question 8.Also is there any specific method to solve questions where we are asked to find roots?I really get blown by such questions.Please advice me SIr.


Please explain Q.26.
For these type of questions, I have no idea how to approach.


sir
doesn't after even no of tosses mean on the 3rd than 5th than 7th .....


Yes, I also had the same doubt.!


Sir, thanks for the answer key. Could you please discuss the solution to Q 25? I feel it is (d) All of the above, as the no of persons who shook hands with odd no of persons could possibly be 19, 1 or 20.


Sir, can you explain q 6, how (p1)^2 is coming? Plz explain sir..


Sir, can you please solve question 13 and 27?


HI kalyani, even I think it should be (d) for Q25, since the question says that "each person shakes hands with some of the persons at the party". Cant understand what we're missing out. :/

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