Latest Elitmus Aptitude Question SOLUTION: In 2 bags, there are to be put together 5 red and 12 white balls, neither bag being empty. How must the balls be divided so as to give a person who draws 1 ball from either bag- (

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(#M40154164) ELITMUS Numerical Ability question Prabability Keep an EYE

In 2 bags, there are to be put together 5 red and 12 white balls, neither bag being empty. How must the balls be divided so as to give a person who draws 1 ball from either bag-

(i) the least chance of drawing a red ball ?
(ii) the greatest chance of drawing a red ball ?

Asked In Elitmus Chandrajeet Yadav (8 years ago)

Solved Utsav Gupta Read Solution (16)

Is this Puzzle helpful?   (29)   (15) Submit Your Solution Probability

Read Solution (16) :

Divyanshu Rohatagi  (8 years ago)

Least Chance.
P= 1/2 * 5/16 + 1/2 * 0 = 5/32


greatest chance.
1/2*1 +1/2*4/16=5/8

Is this solution Helpfull? Yes (47) | No (3)  

Utsav Gupta  (8 years ago) Best Solution

Least chance
-------

Put one white ball in BAG 1 and put the remaining balls in BAG2. Probability of selecting a red ball would be 5/16

Greatest chance
---------------------------------
Put all the red balls in BAG 1 and the white balls in the other. Probability of selecting a red ball would be 1/2

Is this solution Helpfull? Yes (19) | No (27)  

Abhishek Singh  (8 years ago)

Least Chance: One white ball in bag1 and all other balls in bag2
P= 1/2 * 5/16 = 5/32

Greatest chance: all red balls in bag1 and white balls in bag2
P= 1/2 * 5/5 = 1/2

Is this solution Helpfull? Yes (7) | No (7)  

sheetal singh  (7 years ago)

UTSAV GUPTA "" THE BEST SOLUTION AWARDEE "" is actually WRONG !! admin plzz review this !!!!

Is this solution Helpfull? Yes (5) | No   

priyanka  (8 years ago)

[please give the answer properly]

Is this solution Helpfull? Yes (4) | No (1)  

Akash Tiwari  (8 years ago)

for least chance (put everything in one bag and one bag empty)
p = 1/2 *5(no. of red balls)/17(total no. of balls) + 1/2 * 0
p=5/34

for best case( only 1 red ball in one bag and remaining 4 red balls with other 12 white balls in the 2nd bag)
p=1/2 * 1 + 1/2 * 4/16
p=5/8

Is this solution Helpfull? Yes (4) | No (5)  

manmohan  (8 years ago)

Greatest chance when second bag contains only red ball. And all other balls must be in the second beg.

Is this solution Helpfull? Yes (1) | No   

Nishant Chaturvedi  (7 years ago)

The least Chance is when one bag contains only one white ball, and the greatest chance is when one bag contains only one red ball, the chance being 1/11 and 6/11 respectively.

Is this solution Helpfull? Yes (1) | No (2)  

ANKITA BURNWAL  (5 years ago)

(1) Put 1 white in first bag, and all the remaining ball in second bag. Probability =
1/2*0+1/2*5/16= 5/32 (0 because no red ball in the first bag and we have to find the probability for the red ball)

(2) To maximize, we put 1 red in the first bag, and all the remaining in the second. Probability =
1/2*1+1/2*4/16=9/16

Is this solution Helpfull? Yes (1) | No   

abhishek  (8 years ago)

(ii) greatest chance of drawing red ball is
1/2*1/5=1/10

Is this solution Helpfull? Yes | No (3)  

priyanka  (8 years ago)

i) (1/2)*(3/9) + (1/2)*(2/7)

ii) (1/2)*(1/5)

Is this solution Helpfull? Yes | No (3)  

mukesh thakur  (8 years ago)

least chance is
1/2 *3/9 +1/2 *2/8=7/24
greatest chance is
1/2*1 +1/2*4/16=5/8

Is this solution Helpfull? Yes | No (2)  

Madhur  (8 years ago)

For the greatest chance, if I put 1 red ball in first bag and 12W and 4R balls in second bag, what would be the probability? Please explain the calculation in detail.

Is this solution Helpfull? Yes | No (2)  

lavanya  (8 years ago)

5/321/2 so 5/8 can be the greater chance.....
ADMIN PLZ LOOK AT THIS....

Is this solution Helpfull? Yes | No (1)  

ummehabiba  (7 years ago)

(1) Put 1 white in first bag, and all the remaining ball in second bag.  Probability = 12×0+12×51612×0+12×516 = 532532
(2) To maximize, we put 1 red in the first bag, and all the remaining in the second. Probability = 12×1+12×43212×1+12×432 = 916916

Is this solution Helpfull? Yes | No (4)  

Ravi kumar  (6 years ago)

least chance=
1/2 *5/16 +1/2 * 0 =5/32
greatest chance=
1/2 * 4/16 + 1/2 * 1/1 =5/8

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