Latest Elitmus Aptitude Question SOLUTION: If a student can solve one of 4 problems. If he attempts 4 questions in succession. What is the probability that at least one of them is right?

p(s)=1/4
p(s')=3/4
prob of not solvng ny problem is 3/4*3/4*3/4*3/4=81/256
so prob tht he slv 1ques is=1-81/256=175/256

Prob. that he can solve at least one = 1 - prob(solving none)^n

Prob(solving 1 problem out of 4) = 1/4
Prob(solving none) = 1 - 1/4 => 3/4
Therefore Prob(atleast one) = 1 - (3/4)^4
=> 1 - 81/256
=> 175/256

Ans : 175/256

p(s)=1/4
p(s')=3/4
Prob that atleast one of them is right= (1/4*3/4*3/4*3/4)+(1/4*1/4*3/4*3/4)+(1/4*1/4*1/4*3/4)+(1/4*1/4*1/4*1/4)
= 27/256+ 9/256+ 3/256+ 1/256
=40/256

p(s)=1/4
p(s')=3/4
student can solve only one problem
so the cases are
when 1st problem is solved or 2nd is solver or 3rd is solved or 4th is solved
= (1/4*3/4*3/4*3/4)+(3/4*1/4*3/4*3/4)+(3/4*3/4*1/4*3/4)+(3/4*3/4*3/4*1/4)
=27/64

p(s)=1/4
p(~s)=3/4
prob of not solvng ny problem is 3/4*3/4*3/4*3/4=81/256
so prob tht he slv 1ques is=1-81/256=175/256

when one of the 4 is right
p=1/4*3/4*3/4*3/4=27/256
when 2 of them is right
p=1/4*1/4*3/4*3/4=9/256
when 3 of 4 is right
p=1/4*1/4*1/4*3/4=3/256
when all are right
p=1/256
so
p(S) = 1/256 +3/256 +9/256 + 27/256 =40/256 =5/32

It may be 1.. I'm nt sure.

->>> way 1
---------

p(solving at least one of the problem right) = 1- p(solving none of the problem right)
p(solving) = 1/4
p(not solving)=1-1/4=3/4
p(solving none of the problem right)= p(all 4 problems wrong)
=p(not solving)^4
=(3/4)^4
=81/256

p(solving at least one of the problem right) = 1 - 81/256
= 175/256
---->> way 2
-------------------
p(s)=1/4
p(s')=3/4
p(solving at least one of the problem right) = p(1 right problem) + p(2 right problem)
+p(3 right problem) + p(4 right problem)
p(1 right problem) = 4C1 * p(s) * 3C3 * p(s')^3 = 4 * (1/4) * (3/4)^3 = 108/356
p(2 right problem) = 4C2 * p(s)^2 * 2C2 * p(s')^2 = 6 * (1/4)^2 * (3/4)^2 = 54/256
p(3 right problem) = 4C3 * p(s)^3 * 1C1 * p(s') = 4 * (1/4)^3 * (3/4) = 12/256
p(4 right problem) = 4C4 * p(s) ^4 = 1/256
p(solving at least one of the problem right) = (108 + 54 +12 + 1)/256
= 175/256

1-(3/4)^4=135/216

Every one has his/her own answer,what kind of this solutin is...

he attempts all 4 que
here it is at least 1 so right ques can be 1 or 2 or 3 or 4
so prp of sucess=1/4*2/4*3/4*4/4=3/32