Latest IBM Aptitude Question SOLUTION: There are 9 popsicles: 3 orange, 3 cherry, 3 grape. There are 4 children. What is the probability that all 4 children will get the flavor of their choice?

• Placement Papers
• /
• IBM
• /
• Numerical Ability
• /
• Probability

IBM Question Papers: IBM Interview Experience(49) IBM Company Profile

(#M40005585) IBM Numerical Ability question Probability Keep an EYE

There are 9 popsicles: 3 orange, 3 cherry, 3 grape. There are 4 children. What is the probability that all 4 children will get the flavor of their choice?

Asked In IBM (12 years ago)

Unsolved Read Solution (3)

Is this Puzzle helpful?   (8)   (0) Submit Your Solution Probability

Read Solution (3) :

gyayak  (11 years ago)

Consider the 4 children. Each of them has a flavor preference, and so
there are 3x3x3x3 = 81 possibilities for what the kids want (we assume
that each kid does in fact want one of the three flavors: orange,
cherry, or grape).

Of these 81 possibilities, 3 of them mean that one child will be
unhappy: OOOO, CCCC, and GGGG (i.e., when all four of them want the
same flavor as the others). For all of the other possibilities (e.g.
OOCG, OOOC, OOCC,...) there are going to be enough popsicles of the
required flavors.

Assuming that each of the 81 possibilities is equally likely,

Pr(4 happy kids) = 1 - Pr(at least one unhappy kid)
= 1 - 3/81
= 1 - 1/27
= 26/27

A supply of four of each flavor would have guaranteed four happy kids,
but a supply of three of each flavor works in almost all (96%) of the
cases.

Is this solution Helpfull? Yes (6) | No   

ritwick  (11 years ago)

questm iz wrong...IF all the four have same choices..then nly 3 popsciles arre there of the same group..so one of the children will sit empty..which iz not posible.

Is this solution Helpfull? Yes (3) | No (2)  

Aditya sahu  (11 years ago)

Consider the 4 children. Each of them has a flavor preference, and so there are 3x3x3x3 = 81 possibilities for what the kids want Of these 81 Possibilities, 3 of them mean that one child will be unhappy
Pr(4 happy kids) = 1 - Pr(at least one unhappy kid)
= 1 - 3/81
= 1 - 1/27
= 26/27

Is this solution Helpfull? Yes (2) | No   

Please Login to Read Solution.