We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website. cookie policy and privacy policy.
 
+0  
 
0
447
1
avatar+4330 

A bag contains ten b***s, some of which are red and the rest of which are yellow. When two b***s are drawn at random at the same time, the probability that both b***s are red is 1/15 . How many b***s in the bag are red?
 

 Sep 10, 2016
 #1
avatar+105411 
0

Let the number of yellow b***s = Y

 

Then the number of red b***s =  10 - Y

 

And we know that   the chance of drawing a red ball on the first draw = [10 - Y] / 10

And the chance of drawing a red ball on the second draw [without rplacement]  = [9 -Y] / 9

 

So.....the chance of drawing a red ball on both draws is given as :

 

[ 10 - Y] / 10 * [9 - Y] / 9  = 1/15     simplify

 

 

[90 - 19Y + Y^2] / 90  = 1/15     multiply both sides by 90

 

Y^2 - 19Y + 90 = 90/15

 

Y^2 - 19Y + 90  = 6     subtract 6 from both sides

 

Y^2 - 19Y + 84 = 0    factor

 

(Y - 12)(Y - 7)  = 0     ans setting each factor to 0, we have that  Y = 12  [impossible]   or Y  = 7

 

So.....the number of red ones = 10 - Y   = 10 - 7   = 3

 

Proof :    (3/10) (2/9)  =  6/90   = 1/15

 

 

 

cool cool cool

 Sep 10, 2016

11 Online Users

avatar
avatar