Question

A bag contains red, green, and yellow balls of equal sizes. If one ball is chosen out of the box at random, then the probability of a red ball being taken out is $\frac{1}{2}$ and that of a yellow ball being taken out is $\frac{1}{3}$. If 3 yellow balls and 1 green ball are added to the bag, then the probability of a randomly chosen ball being yellow will become $\frac{3}{8}$. Find the total number of balls.

Answer 1

Toal no of sell are R+Y +G
$P\left(R\right)=\frac{R}{R+Y+G}=\frac{1}{2}$
2R=R +Y +G .....(i)
$P\left(Y\right)=\frac{4}{R+Y+G}=\frac{1}{3}$
3Y =R + Y + G ....(ii)
P(Y) after adding 3 yellow 1 green balls
$\frac{8}{8}=\frac{4}{R+4+Y+4}$
So, 3 (R + G + Y ) +12 =8Y .....(iii)
Subsblites the value of (R+ Y + G form cg (ii) to (iii)
3(3y +12 =8y
9y -8y =12
y =12
Subtract (i) & (ii)
2R -3Y =0
2R =3Y
$R=\left(\frac{3}{2}\right)y=\frac{3}{2}x12=18$
Now, G =6
So, total balls =12 +18 +6 =36