Question

we have 3 blue and 5 red balls lay in a box three balls are chosen randomly out of the box find the probability distribution for number of blue balls without replacement.

Answer 1

Let B denotes Blue ball

and R denotes Red ball

Let X denotes the event of number of blue balls are chosen out of three balls drawn from the box

So,

Case 1:

When X = 0 means all three balls drawn are Red

$P(X_0)=\dfrac{5}{8}\times \dfrac{4}{7}\times\dfrac{3}{6}=\dfrac{60}{336}$

Case2:

When X=1 means only one Blue balls are chosen out of three balls

and it can be BRR + RBR +RRB

$So, P(X_1)=(\dfrac{3}{8}\times\dfrac{5}{7}\times\dfrac{4}{6})+(\dfrac{5}{8}\times\dfrac{3}{7}\times\dfrac{4}{6})+(\dfrac{5}{8}\times\dfrac{4}{7}\times\dfrac{3}{6})=\dfrac{180}{336}$

Case3:

When X=2 means two blue balls are there on choosing 3 balls from the box

and it can be BBR +BRB+ RBB

So, $P(X_2)=(\dfrac{3}{8}\times\dfrac{2}{7}\times\dfrac{5}{6})+(\dfrac{3}{8}\times\dfrac{5}{7}\times\dfrac{2}{6})+(\dfrac{5}{8}\times\dfrac{3}{7}\times\dfrac{2}{6})=\dfrac{90}{336}$

CAse4:

When X=3 means all blue balls are chosen from the box

So, $P(X_3)=\dfrac{3}{8}\times\dfrac{2}{7}\times\dfrac{1}{6}=\dfrac{6}{336}$

So, Probability distribution table will be: