Garp Sample questions

[mastvikas]

can someone please explain me the below 2 sums in detail . They look similar but they are different


Q1 A multiple choice exam has 10 questions with 5 choices per questions . if you need at least 3 correct answer to pass the exam . what is probability that you will pass simply by guessing ?

A 0.8%
B 20.1%
C 67.8%
D 32.2%

Q2 Suppose that a quiz consists of 10 true and false. A student has not studied for the exam and just randomly guesses the answers. What is the probability that student will get at least 3 answer correct ?

A 5.47%
B 33.66%
C 78.62%
D 94.53%

 

[RiskNoob]

Hi mastvikas,

Here is the brief idea:

Q1: Let X the # correct answers, then X ~ binomial with n = 10 (# of questions), p = 1/5 (prob of getting correct answer by just guessing), then you can find the answer by solving 1 - P(X <= 2) = 1 - P(X = 0) - P(X = 1) - P(X = 2)

Q2: Similar to the Q1 once you figure about p.

RiskNoob

 

[ShaktiRathore]

First one is similar to second one,
probability of selecting correct answer out of 5 option=1/5=.2
Prob(pass)=1-Prob(not passing)
Prob(pass)=1-Prob(0 or 1 or 2 correct answers out of 10 questions)
Prob(pass)=1-(10C0*.2^0*.8^10+10C1*.2^1*.8^9+10C2*.2^2*.8^8)
Prob(pass)=1-.8*^8(10C0*.2^0*.8^2+10C1*.2^1*.8^1+10C2*.2^2)
Prob(pass)=1-.8*^8(.8^2+10*.2^1*.8^1+45*.2^2)
Prob(pass)=1-.8*^8(.64+1.6+1.8)
Prob(pass)=1-.8*^8(4.04)=1-.6777~.3223 whci is approx 32.23%

visit http://forum.bionicturtle.com/threads/binomial-distribution.6146/#post-19352 to get more eg on binomial distributon.

thanks

 

[akgvarun]

Question 1:
Bernoulli trial: Prob(answering True by guessing = T) = 1/5 = 0.2
Prob (answering False = F) = 1-P(T) = 4/5 = 0.8

Now we enter binomial dist with 10 i. i. d. (Bernoulli) questions
We need probability of passing the exam, which is answering at least 3 questions correctly. That is, answer 3 true (with 7 false), 4 true (with 6 false), 5 true, 6 true, 7 true, 8 true, 9 true and 10 true. This can be stated as P(T>=3), which is equal to 1 - P(T<3). This can be expanded as 1 - [P(T=0)+P(T=1)+P(T=2)]

So we calculate P(T=0), i.e., 10 False questions. There can only one combination of 10 Fs. Hence P(T=0)=P(F=10)=0.8^10 = 0.1074 or 10.74%
P(T=1) = P(T=1, F=9). There can be 10C1 combinations with one T and 9 F, which works out to 10. Use {n!/r!(n-r)!}, with n=10 and r = 1. Hence P(T=1) = 10*0.2^1*0.8^9 = 0.2684 or 26.84%
P(T=2) = P(T=2, F=8). There can be 10C2 combinations with 2 T and 9 F. Hence P(T=2) = 45*0.2^2*0.8^8 = 0.3020 or 30.20%

Hence P(T>=3) = 1 - P(T<3) =1 - [P(T=0)+P(T=1)+P(T=2)]
= 1 - (0.1074+0.2684+0.3020)
= 1 - 0.6778 = 0.3222 or 32.2%

 

[[email protected]]

Number of question 100 Probability of success 0.25 and Failure 0.75 you will have to score at least 80 questions correct what is the probability? quick way to crack this problem please share?

 

[ShaktiRathore]

We can approximate the given binomial distribution with the normal with the following procedure:

• the standard deviation of a binomial with p = 25% and n =100 is SQRT[25%*75%*100] = 4.33
• the standardized interval [80,100] when mean is 25 is [(80-25)/4.33, (100-25)/4.33] = [12.70,+17.32]
• from formula z = (x - pT)/SQRT[p(1-p)T] ~= N(0,1)
• In short, the exact answer given by the binomial (above) should be approximated by =NORMSDIST(17.32, true)- NORMSDIST(12.7, true) ~=0 %. Indeed, difference of only 0%! actually numbers are not that practical enoogh to get answer a number and getting near 0 in ur example but this is the general method to deal with Qs like this.
• Take above example: T=10, p=20%=> std deviation=sqrt[20%*80%*100]=4
• the standardized interval [3,10] when mean is 2 is [(3-2)/4, (10-2)/4] =[.25,+2]=NORMSDIST(2, true)- NORMSDIST(.25, true)~=37.85% which is closer to the actual answer of 32.23% not quite bad method
• go tolink http://forum.bionicturtle.com/threads/binomial-distribution.6146/#post-19352 good explaination by David

So ....

 

[[email protected]]

Thanks Shakti.. makes sense