Elitmus
Exam
Numerical Ability
Probability
A teacher gives three bags to abhishek, amitabh and aiswarya.Each bag contains six balls. abhishek has 2 red balls, amitabh has 5 red balls.When a ball is drawn by them the probability of not being same colour is 5/6.the number of red balls in aiswarya bag is ?
a)2 b)3 c)4 d)5
Read Solution (Total 10)
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- ans- a)2
suppose aish has x red balls.
and given that 1 ball is drawn by them, i.e, 1 ball is drawn from each bag
then prob of not being same color is 5/6
now first let me find out prob of being same color: ((2c1*5c1*xc1) + (4c1*1c1*(6-x)c1)/6*6*6
then prob of not being same color is: 1- (2*5*x + 4*1*(6-x))/6*6*6
so 1- (2*5*x + 4*1*(6-x))/6*6*6 = 5/6
solve this equation, x=2
hence Aish contains 2 red balls in her bag.
- 10 years agoHelpfull: Yes(68) No(6)
- key +> ball can be any ao three bag and probability of getting ball which is not reed;
probability of selecting a ball from any one of three bag = 1/3;
probability of selecting a ball which is not red = 4/6+1/6+(6-x)/6;
where x is no of red ball in ash bag;
1/3[4/6+1/6+(6-x)/6] = 5/6;
=> x= 4; - 10 years agoHelpfull: Yes(7) No(6)
- this type of problems go through options wise
so take first bag(abhi)-2 red balls,assume remains are 4blue
take second bag(amit)-5 red balls,assume ramains one is green
take third bag(ais)-x red balls ,assume remains y blue yellow
in the problem each bag have six balls and also we have not being same color probability i.e 5/6
write a not being same color probability
bag1 (or) bag2 (or) bag3 (drawn one ball only see the problem)=5/6
2/6 (or) 1/6 (or) y/6 = 5/6
red (or) green (or) yellow
(2+1+y)/6 =5/6
if y=2 then only condition is satiefeid(condition is not satiesfied remains values 3,4,5 )
here 2-balls is yellow not red balls,so remains(6(total balls)-2(yellow))=4red balls
ans:C.
pls reply this procedure correct or not - 9 years agoHelpfull: Yes(5) No(0)
- anyone solve dis???
- 10 years agoHelpfull: Yes(1) No(2)
- correct answer is option b) 3
- 10 years agoHelpfull: Yes(1) No(3)
- anyone solve dis???
- 10 years agoHelpfull: Yes(0) No(3)
- anyone solve dis???
- 10 years agoHelpfull: Yes(0) No(2)
- anyone solve dis???
- 10 years agoHelpfull: Yes(0) No(2)
- if you have trouble finding it u can also manipulate with the options.
- 9 years agoHelpfull: Yes(0) No(0)
- I think option
- 9 years agoHelpfull: Yes(0) No(0)
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