Question

There is a bag filled with 3 blue,2 red and 4 green marbles.A marble is taken at random from the bag,the colour is noted and then replaced.Another marble is taken at random.What is the probability of getting 2 different colours?

Answer 1

Let A - "two different"

$P(A)=1-P(\neg A)$

$\neg A$ - "both same"

Let B - "both blue", C - "both red", D - "both green", then

$P(\neg A)=P(B)+P(C)+P(D)=({\frac 3 {3+2+4}})^2+({\frac 2 {3+2+4}})^2+({\frac 4 {3+2+4}})^2={\frac{29} {81}}$

So, $P(A)=1-{\frac {29} {81}}={\frac {52} {81}}$