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A bag contains red marbles, white marbles, green marbles, and blue marbles. There are an equal number of red marbles and white marbles, and five times as many green marbles as blue marbles. There is a $35\%$ chance of selecting a red marble first. What is the fewest possible number of green marbles in the bag?

 Apr 6, 2015

Best Answer 

 #3
avatar+129847 
+14

Consider this......20 total marbles

7 red  .....7/20 = 35%

7 white

1 blue

5 green

 

  

 Apr 6, 2015
 #1
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I need help on the exact same question :)

 Apr 6, 2015
 #2
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Just guessing:

There is a 70% chance of selecting either a red or white marble. Therefore there must be a 30% chance of either a green or a blue marble. Since green has a five time higher likelihood, then the 30% must be divided between 0.25 and 0.05.

There should be a minimum of 25 green marbles.

 Apr 6, 2015
 #3
avatar+129847 
+14
Best Answer

Consider this......20 total marbles

7 red  .....7/20 = 35%

7 white

1 blue

5 green

 

  

CPhill Apr 6, 2015
 #4
avatar+118667 
+14

A bag contains red marbles, white marbles, green marbles, and blue marbles. There are an equal number of red marbles and white marbles, and five times as many green marbles as blue marbles. There is a 35% chance of selecting a red marble first. What is the fewest possible number of green marbles in the bag?

 

I want to talk about this one a bit

There is a 35% chance of getting a red so there must be a 35% chance of getting a white.

35+35=70%

There is 100% chance that you will pick a marbles  and 70+30=100

SO  there must be a 30% chance that you will choose a blue or green marble.

and let the number of Blue marbles be B  the number of green marbles will be 5B     

B+5B=30%

6B=30%

Blue=5%

Green = 30-5=25%

 

so lets look at the ratios

Red: White: Green : blue

35  :35     :       25:    5

They are all divisable by 5

7    : 7      :      5   :   1

So the fewest number of green marbles is 5    

 Apr 7, 2015

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