Question 945409
  
Given:
Two bottles containing red and blue balls.
Bottle 1: 4R+6B
Bottle 2: xR+12B
One ball is drawn from each bottle.
P(RR∪BB)=0.56 (i.e. probability of drawing both red or both blue)
Find x (number of red balls in bottle 2).
  
Solution:
Let 
R=event of drawing a red ball
B=event of drawing a blue ball
Since events of drawing from different bottles are independent, we can multiply the individual probabilities, i.e.
P(RR)=P(R) from bottle 1  *  P(R) from bottle 2, and similarly for P(BB)
  
P(RR)={{{(4/(4+6))*(x/(x+12))=0.4x/(x+12)}}}
P(BB)={{{(6/(4+6))*(12/(x+12))=7.2/(x+12)}}}
Since the events RR and BB are mutually exclusive, probability of the union is the sum of the individual probabilities.
P(RR∪BB)={{{0.4x/(x+12)+7.2/(x+12)}}}
which is given to be 0.56
So we have an equation in x from which we can solve for x:
{{{0.4x/(x+12)+7.2/(x+12) = 0.56}}}
{{{0.4x+7.2=0.56(x+12)}}}
{{{7.2-6.72=0.16x}}}
{{{x=0.48/0.16=3}}}
  
Answer:
There are 3 red balls in the second bottle.