Feedback on FRM 2010 Level I Please

syaiful

Member
yup..don't need "right 3".. thanks

can you solve this ?

what is the probability to shoot succesfull with 3 shoot

each fail, add 0.5 m

A= Shoot from 1 m
B = Shoot from 1.5 m
C = Shoot from 2 m

P(A) = 1/2
P(B) = 1/3
P(C) = 1/4

:lol:
 

amitsaraf03

New Member
i think there was something more to the question:

anyways the working would go like this assuming the question is correct:

[(0.5+0.5*(2/3)+0.5+(2/3)*(1/4)]=0.913
 

David Harper CFA FRM

David Harper CFA FRM
Subscriber
lolz - Agree you just need to add the outcome where 0 answers given are correct, which itself has probability of 80%^10 ~= 10.74%. Subtract from your other outcomes = 1 - P(1 of 10) - P(2 of 10) - 10.74% ~= 42.x% - 10.74% ~= 32%
... and we can confirm precise result with 1 - binomial CDF P [X <= 2] = 1 - BINOMDIST (k = 2, n = 10, p = 20%, cdf = true) = 32.22%
 

RomanS

New Member
correct me if i am wrong but concerning the "shooting-question" two things:

1. The question asked for the probability for a hit from one of the 3 distances, right?
2. In that case one would have to go through all possible cases:

x = hit
0 = no hit

3 hits = xxx
2 hits = xx0 or 0xx or x0x
1 hit = x00 or 0x0 or 00x

and add the probabilities. I get 75%...
 

syaiful

Member
Hi Romans,

x = hit
0 = no hit

3 hits = xxx
2 hits = xx0 or 0xx or x0x
1 hit = x00 or 0x0 or 00x

The tricky point is about "add 0.5 m each fail"

i agree with the way you looking the problem.
 

rsinghso

New Member
David, Thanks for the prob(getting pass) Qs explaination,,yup answer would be ~33%..Hopefully I marked the same :)

As for the shooting qs : the shooter quits as soon as he shoots successfully, So

Case 1: Shoots successfully from 1 m = 1/2
Case 2 : Shoots unsuccessfully from 1m and shoots successfully from 1.5m = (1 - 1/2)*(1/3) = 1/6
Case 3: Shoots unsuccessfully from 1m and shoots unsuccessfully from 1.5m and shoots successfully from 2m =
(1 - 1/2)*(1 - 1/3)*(1/4) = 1/12

All these cases are mutually exclusive hence final probability = 1/2 + 1/6 + 1/12 = 75%

How about the question on ethics :

A) no violation of code as the protagonist was unaware
B) voilation of code
C) no voilation as when he took FRM.......
D) another no violation, I don't remember correctly ( I have marked violation of code as the answer,, what's your opinion ? )
 

SURAJM

New Member
Compared with the practice tests of BT, Schweser, Pristine etc. i would definitely say that level 1 was on the easier side. GARP this time put more focus on testing the concepts rather than asking students to do lengthy calculations or memorizing big formulas. There were questions which were almost half page long but in the end required only 1-2 line calculation if you could identify the concept being tested. There were quite a few trick questions as well especially related to FRAs etc.

And i am surprised regarding the time shortage mentioned by others. I think time was sufficient enough to attempt the 100 questions. But the probability of making mistakes in easier questions is quite high. Also given the fact that lot of theory was also tested ( i will guess around 30% theory questions), it is always easy to get those questions wrong.

I remeber the following questions:
1) question related to maintenance margin
2) 2-3 questions related to put call parity and lower upper bound for european and american (yes they asked for american as well) calls
3) questions related to delta, gamma
4) The usual future/ forwards question. They asked one question about convexity adjustment as well
5) 2-3 questions related to FRAs. I was surprised seeing so many questions on FRAs
6) From quant side lot of stress on linear/ multiple regression
7) Var questions (there were atleast 8-10 questions testing different concepts)
 

adash

New Member
reg the shooting question...

i think prob ( hitting at least once) = 1- prob( missing all 3)= 1- prob(missing first)*prob(miss 2nd)* prob(miss 3rd)
= 1- 1/2* 2/3 * 3/4
= 1 - 1/4
= 3/4
 
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