Three balls are drawn one by one without replacement from a bag containing 5 white and 4 green balls - Maths - Probability

Question

Three balls are drawn one by one without replacement from a bag
containing 5 white and 4 green balls FInd the probability
distribution of no of green balls drawn?

Shridhar Kaushik · Accepted Answer

Let X denote the total number of green balls drawn in three draws without replacement.

Clearly, there may be all green, 2 green, 1 green or no green at all.

Thus, X can assume values 0, 1, 2 and 3.

Let G₁ denote the event of getting a green ball in first draw.

G₂ denote the event of getting a green ball in second draw

and G₃ denote the event of getting a green ball in third draw.

Now,

P (X = 0) = Probability of getting no green ball in three draws

$= P ({\bar{G}}_{1} \cap {\bar{G}}_{2} \cap {\bar{G}}_{3}) = P ({\bar{G}}_{1}) P ({\bar{G}}_{2} / {\bar{G}}_{1}_{}) P ({\bar{G}}_{3} / {\bar{G}}_{1}_{} \cap {\bar{G}}_{2}) = \frac{5}{9} \times \frac{4}{8} \times \frac{3}{7} = \frac{5}{42}$

P (X = 1) = Probability of getting on green ball in three draws

$= P ((G_{1} \cap {\bar{G}}_{2} \cap {\bar{G}}_{3}) \cup ({\bar{G}}_{1} \cap G_{2} \cap {\bar{G}}_{3}) \cup ({\bar{G}}_{1} \cap {\bar{G}}_{2} \cap G_{3})) = P (G_{1} \cap {\bar{G}}_{2} \cap {\bar{G}}_{3}) + P ({\bar{G}}_{1} \cap G_{2} \cap {\bar{G}}_{3}) + P ({\bar{G}}_{1} \cap {\bar{G}}_{2} \cap G_{3}) = P (G_{1}) P ({\bar{G}}_{2} / G_{1}) P ({\bar{G}}_{3} / G_{1} \cap {\bar{G}}_{2}) + P ({\bar{G}}_{1}) P (G_{2} / {\bar{G}}_{1}) P ({\bar{G}}_{3} / {\bar{G}}_{1} \cap G_{2}) + P ({\bar{G}}_{1}) P ({\bar{G}}_{2} / {\bar{G}}_{1}) P (G_{3} / {\bar{G}}_{1} \cap {\bar{G}}_{2}) = \frac{4}{9} \times \frac{5}{8} \times \frac{4}{7} + \frac{5}{9} \times \frac{4}{8} \times \frac{4}{7} + \frac{5}{9} \times \frac{4}{8} \times \frac{4}{7} = \frac{80 + 80 + 80}{9 \times 8 \times 7} = \frac{240}{9 \times 8 \times 7} = \frac{10}{21}$

$P (X = 2) = P ((G_{1} \cap G_{2} \cap {\bar{G}}_{3}) \cap ({\bar{G}}_{1} \cap G_{2} \cap G_{3}) \cap (G_{1} \cap {\bar{G}}_{2} \cap G_{3})) = P (G_{1} \cap G_{2} \cap {\bar{G}}_{3}) + P ({\bar{G}}_{1} \cap G_{2} \cap G_{3}) + P (G_{1} \cap {\bar{G}}_{2} \cap G_{3}) = P (G_{1}) P (G_{2} / G_{1}) P ({\bar{G}}_{3} / G_{1} \cap G_{2}) + P ({\bar{G}}_{1}) P (G_{2} / {\bar{G}}_{1}) P (G_{3} / {\bar{G}}_{1} \cap G_{2}) + P (G_{1}) P ({\bar{G}}_{2} / G_{1}) P (G_{3} / G_{1} \cap {\bar{G}}_{2}) = \frac{4}{9} \times \frac{3}{8} \times \frac{5}{7} + \frac{5}{9} \times \frac{4}{8} \times \frac{3}{7} + \frac{4}{9} \times \frac{5}{8} \times \frac{3}{7} = 3 (\frac{5 \times 4 \times 3}{9 \times 8 \times 7}) = \frac{5}{14}$

$and = P (X = 3) = P (G_{1} \cap G_{2} \cap G_{3}) = P (G_{1}) P (G_{2} / G_{1}) P (G_{3} / G_{1} \cap G_{2}) = \frac{4}{9} \times \frac{3}{8} \times \frac{2}{7} = \frac{1}{21}$

Hence, the probability distribution of the number of green balls is given by –

$\begin{matrix} X & : & 0 & 1 & 2 & 3 \\ P (X) & : & \frac{5}{42} & \frac{10}{21} & \frac{5}{14} & \frac{1}{21} \end{matrix}$

Three balls are drawn one by one without replacement from a bag containing 5 white and 4 green balls. FInd the probability distribution of no. of green balls drawn?