Bunuel wrote:
malikshilpa wrote:
Is ' picking of 3 marbles at random' same as 'picking 3 marbles one by on without replacement' ? Although looking at the solution these two scenarios appear to be similar. However, when 3 marbles are picked up together (in one shot) then should the problem not be treated differently. Please clarify .
Mathematically the probability of picking 3 marbles simultaneously, or picking them one at a time (without replacement) is the same.
A basket contains 3 blue, 3 red and 3 yellow marbles. If 3 marbles are extracted from the basket at random, what is the probability that a marble of each color is among the extracted?A. 2/21
B. 3/25
C. 1/6
D. 9/28
E. 11/24
We need to find the probability of BRY (a marble of each color).
Probability approach:\(P(BRY)=\frac{3}{9}*\frac{3}{8}*\frac{3}{7}*3!=\frac{9}{28}\), we are multiplying by 3! since BRY scenario can occur in several different ways: BRY, RYB, RBY (# of permutations of 3 distinct letters BRY is 3!).
Combination approach:\(P(BRY)=\frac{C^1_3*C^1_3*C^1_3}{C^3_9}=\frac{9}{28}\).
Answer: D.
Hope it's clear.
Hi Bunuel,
I have a question with the methodology here -- I can follow the work you're doing but i'm having a hard time understanding why. Can you please clarify the reasoning?
a) In your statement below, can you please elaborate WHY we multiply by 3!? --
[\(P=\frac{3}{9}*\frac{3}{8}*\frac{3}{7}*3!=\frac{9}{28}\)If I reference the GMAT club book, I don't see it referencing the multiplication by a factorial(Page 109 of the book).
b) What is the difference between the method you used vs. doing the following? (9/9)(6/8)(3/7)
c) lastly, regarding your combinatorial approach, isn't that technically a reverse approach(pg 109 of the book)? Sorry if i'm being trivial here but i'm having a hard time connecting the dots hence the trivial questions. How can you use the 1C3 approach because don't you have 9 to begin with? Sorry - i'm going nuts here.