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 Simple probability

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Billl has some counters in a bag

 

3 of the counters are red.

7 of the counters are blue.

The rest are yellow.

 

Bill takes at random a counter from the bag.

The probability that he takes a tellow counter is 2/7

How many yellow counters are in the bag before bill takes a counter?

 

Please explain all answers.

 Nov 30, 2015

Best Answer 

 #2
avatar+129899 
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Let x be the number of yellow counters in the bag.  And we know that:

 

[   x ] /  [ 3  +  7 +  x ]   = 2/7    cross-multiply

 

7x =  2(10 + x)

 

7x  = 20  - 2x

 

5x = 20

 

x = 4   .......this is the number of yellow counters

 

 

 

cool cool cool

 Nov 30, 2015

4+0 Answers

 #1
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+5

here are the numbers: 3 red, 7 blue, X yellow

 

X yellow = 2 / 7

3 red + 7 blue must equl the other 5 /7

 

3 + 7 = 10, 5 * 2 = 10

 

3 red has 3 / 20 chance

7 blue has 7 /20 chance

X yellow has 10 / 20 chance

 

there are 10 yellow

 Nov 30, 2015
 #2
avatar+129899 
+5
Best Answer

Let x be the number of yellow counters in the bag.  And we know that:

 

[   x ] /  [ 3  +  7 +  x ]   = 2/7    cross-multiply

 

7x =  2(10 + x)

 

7x  = 20  - 2x

 

5x = 20

 

x = 4   .......this is the number of yellow counters

 

 

 

cool cool cool

CPhill Nov 30, 2015
 #3
avatar+10 
+5

The guest that posted is Konzon Of Kondraga.

 Nov 30, 2015
 #4
avatar+129899 
0

Welcome to the site, KOK......glad to have you aboard......!!!!!

 

 

 

cool cool cool

 Nov 30, 2015

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