Elitmus
Exam
Numerical Ability
Probability
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 ?
Read Solution (Total 16)
-
- Least Chance.
P= 1/2 * 5/16 + 1/2 * 0 = 5/32
greatest chance.
1/2*1 +1/2*4/16=5/8 - 9 years agoHelpfull: Yes(47) No(3)
- Least chance
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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
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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 - 9 years agoHelpfull: Yes(19) No(27)
- 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 - 9 years agoHelpfull: Yes(7) No(7)
- UTSAV GUPTA "" THE BEST SOLUTION AWARDEE "" is actually WRONG !! admin plzz review this !!!!
- 7 years agoHelpfull: Yes(5) No(0)
- [please give the answer properly]
- 9 years agoHelpfull: Yes(4) No(1)
- 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 - 8 years agoHelpfull: Yes(4) No(5)
- Greatest chance when second bag contains only red ball. And all other balls must be in the second beg.
- 9 years agoHelpfull: Yes(1) No(0)
- 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.
- 8 years agoHelpfull: Yes(1) No(2)
- (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 - 6 years agoHelpfull: Yes(1) No(0)
- (ii) greatest chance of drawing red ball is
1/2*1/5=1/10
- 9 years agoHelpfull: Yes(0) No(3)
- i) (1/2)*(3/9) + (1/2)*(2/7)
ii) (1/2)*(1/5) - 9 years agoHelpfull: Yes(0) No(3)
- 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 - 9 years agoHelpfull: Yes(0) No(2)
- 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.
- 9 years agoHelpfull: Yes(0) No(2)
- 5/321/2 so 5/8 can be the greater chance.....
ADMIN PLZ LOOK AT THIS....
- 9 years agoHelpfull: Yes(0) No(1)
- (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
- 8 years agoHelpfull: Yes(0) No(4)
- least chance=
1/2 *5/16 +1/2 * 0 =5/32
greatest chance=
1/2 * 4/16 + 1/2 * 1/1 =5/8 - 7 years agoHelpfull: Yes(0) No(0)
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