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sanjoy18: Senior | Next Rank: 100 Posts; Posts: 81; Joined: Tue Jun 11, 2013 10:24 pm; Thanked: 7 times; Followed by:1 members

probability

by sanjoy18 » Thu Sep 12, 2013 1:20 pm

A cube is painted with red colour and then cut into 64 small identical cubes. If two cubes are picked
randomly from the heap of 64 cubes, what is the probability that both of them have exactly two faces
painted red?
(a)23/168
(b)47/84
(c)1/4
(d)31/63
(e)41/63

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Source: Beat The GMAT — Problem Solving | Thu Sep 12, 2013 1:20 pm

batwaraanirudh: Junior | Next Rank: 30 Posts; Posts: 10; Joined: Thu Mar 28, 2013 5:26 am; Thanked: 1 times

by batwaraanirudh » Thu Sep 12, 2013 4:06 pm

A cube is painted with red colour and then cut into 64 small identical cubes. If two cubes are picked
randomly from the heap of 64 cubes, what is the probability that both of them have exactly two faces
painted red?
(a)23/168
(b)47/84
(c)1/4
(d)31/63
(e)41/63

Total number of available cubes to choose from = 64
A cube has 12 sides and on each side we will find 2 identical cubes whose only 2 sides have red paint.
Hence total number of cubes with only 2 sides painted red = 12*2= 24

Therefore probability for picking the 2 cubes with only 2 sides painted red = (24C2)/(64C2)= 23/168

Hence choose A

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ganeshrkamath: Master | Next Rank: 500 Posts; Posts: 283; Joined: Sun Jun 23, 2013 11:56 pm; Location: Bangalore, India; Thanked: 97 times; Followed by:26 members; GMAT Score:750

by ganeshrkamath » Thu Sep 12, 2013 6:58 pm

sanjoy18 wrote:A cube is painted with red colour and then cut into 64 small identical cubes. If two cubes are picked
randomly from the heap of 64 cubes, what is the probability that both of them have exactly two faces
painted red?
(a)23/168
(b)47/84
(c)1/4
(d)31/63
(e)41/63

64 = 4^3
So each face is divided into 4^2 squares.
The cubes at the edges have exactly 2 faces painted red.
Number of cubes per edge with 2 faces painted red = 4 - 2 = 2 (removing the cubes at the corner)
Number of edges = 12
Total number of cubes with 2 faces painted red = 2 * 12 = 24

Favorable cases:
Select 2 out of 24 cubes : 24C2

Sample space:
Select 2 out of 64 cubes : 64C2

Probability = 24C2/64C2 = (24*23)/(64*63) = 23/168

Choose A

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vipulgoyal: Master | Next Rank: 500 Posts; Posts: 468; Joined: Mon Jul 25, 2011 10:20 pm; Thanked: 29 times; Followed by:4 members

by vipulgoyal » Thu Sep 12, 2013 9:00 pm

alt approach

let the side of big cube is A
cube divided in 64 inditical cubes
A^3/64 = I (Int)
the smallest value of A is 4 and the value of a(side of small cube is 1)
note:- the second large cube is of side 8 and small cube's side will be 2, the required probablity will be same is either cases
now consider cube with all faces painted, only cubes at edges will have two faces painted
these cube are 4*4 + 4*2 = 24
required probablity 24c2/64c2 = 23/168

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by [email protected] » Thu Sep 12, 2013 9:57 pm

Hi All,

In these types of situations, it's often a good idea to draw a picture for reference. To be painted on "two faces", a piece would have to be on the "edge" of the big cube but NOT be a corner piece (because corner pieces would have paint on 3 faces).

You'll find that 24 pieces fit the "2 face" description.

Then the probability math is straight-forward:

(24/64)(23/63 = 23/168

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