Alejandra is designing a game of chance. For one part of the game,

Question

Alejandra is designing a game of chance. For one part of the game, a player is to randomly choose 3 marbles, without replacement, from a box containing B blue marbles, R red marbles, and no other marbles. Alejandra correctly determined the positive integers B and R so that the number of possible selections in which 1 blue marble and 2 red marbles are chosen is twice the number of possible selections in which 2 blue marbles and 1 red marble are chosen.

The positive integers B and R that Alejandra determined must be such that B is the number that is __ 1 __ and R is the number that is __ 2 __.

Based on the information provided, select for 1 and for 2 the options that create the most accurate statement. Make only two selections, one in each column.

nisen20 · Accepted Answer

\frac{B}{B+R} × \frac{R}{B+R-1} × \frac{R-1}{B+R-2} = \frac{B}{B+R} × \frac{B-1}{B+R-1} × \frac{R}{B+R-2} × 2

R-1 = 2B-2

B = \frac{R+1}{2}

R = 2B-1

Kinshook · Answer

jackg0101

Given: Alejandra is designing a game of chance. For one part of the game, a player is to randomly choose 3 marbles, without replacement, from a box containing B blue marbles, R red marbles, and no other marbles. Alejandra correctly determined the positive integers B and R so that the number of possible selections in which 1 blue marble and 2 red marbles are chosen is twice the number of possible selections in which 2 blue marbles and 1 red marble are chosen.

Asked: The positive integers B and R that Alejandra determined must be such that B is the number that is __ 1 __ the number that is __ 2 __R.

Number of possible selections in which 1 blue marble and 2 red marbles are chosen = BC1*RC2 = BR(R-1)/2 
Number of possible selections in which 2 blue marbles and 1 red marble are chosen = BC2*RC1 = B(B-1)/2 * R= BR(B-1)/2

BR(R-1)/2 = 2 * BR(B-1)/2 
R-1 = 2(B-1) = 2B - 2
R = 2B -1 
B = (R+1)/2

tusharrgarg93 · Answer

Let B = X and R = Y,

Then it could be solved by creating an algebraic equation. 
x/x+y*x-1/x+y-1*y/x+y-2 = 2*( x/x+y*y/x+y-1*y-1/x+y-2)
-> x-1 = 2(y-1) 
-> x= 2y-1

nazii · Answer

Dear  Bunuel 
This test is a math test, however, the wording is odd, would you pls help translation? It seems a hard question since only 11% got it right!
I would appreciate in advance for your kind explanation

Kinshook · Answer

jackg0101 
Please recheck the correct answer.

nazii · Answer

Dear  Kinshook 
Thanks for you approach shared 
Would you pls just tell me why you divide each formula by 2?
I would appreciate if you explain the concept used in this question

jawnjon · Answer

kennaanna  can you explain why Y=(B/2) and not R x (B/2)?

rmahe11 · Answer

Yeah this is nice question , i technically made an error by forgetting the twice part in last sentence . That's wasnt getting the other part right

Perni94 · Answer

Hi guys,

this question just popped up in a mock exam and I got it wrong from probably misunderstanding the text.

Specifically, why do you not consider the scenarios:
Scenario 1: first pick red, then blue
Scenario 2: first pick blue, then red

Is it because of the prompt saying that you first choose the blue?

tdew · Answer

I just tried some simple numbers, 
but it's not the best and quickest way as equation approach, though.

Goal: Getting 1B and 2R marbles should happen twice as often as getting 2B and 1R.

first, assume B = 2, R = 2 -> so that there are enough marbles to be chosen for either 1B2R or 2B1R
2 x 2 x 1 = 2 x 1 x 2 -> 4:4 = 1:1, it's not twice.
(B-R-R)-----(B-B-R)

we know that the LHS must be higher, then change R to 3
2 x 3 x 2 = 2 x 1 x 3 -> 12:6 = 2:1, it's what we are looking for.
(B-R-R)-----(B-B-R)

so, B = 2, R =3.  
align with choice, 
B = (R+1)/2 -> B = (3+1)/2 = 2 
R = 2B-1 = 2(2)-1 = 3

1	2
		R+2
		2B-1
		B-2
		\(\frac{B+1}{2}\)
		\(\frac{R+1}{2}\)
		\(\frac{R+2}{2}\)

Alejandra is designing a game of chance. For one part of the game,

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