# Choosing one type of ball without replacement.

Suppose I have $9$ balls, among which $3$ are green and $6$ are red. What is the probability that a ball randomly chosen is green?

It is $\dfrac{3}{9}=\dfrac{1}{3}$.

If three balls are randomly chosen without replacement, then what is the probability that the three balls are green?

Is it $\dfrac{3}{9}\times\left\{\dfrac{2}{8}+\dfrac{3}{8}\right\}\times\left\{\dfrac{1}{7}+\dfrac{2}{7}+\dfrac{3}{7}\right\}=\dfrac{90}{504}$?

But in hypergeometric distribution formula, it is

$$f(X=3)=\dfrac{\binom{3}{3}\binom{9-3}{3-3}}{\binom{9}{3}}=\dfrac{1}{84}.$$

• I've no idea whatsoever where you got that calculation that yields $90/504$. – Wildcard Dec 29 '16 at 5:12

b) All 3 are green without replacement

= $\dfrac39 \cdot \dfrac28 \cdot \dfrac17$ = $\dfrac1{84}$

c) 2 balls are green out of 3 without replacement.

= $\left(\dfrac{3}{9} \cdot \dfrac{2}{8} \cdot \dfrac{6}{7} \right ) + \left(\dfrac{3}{9} \cdot \dfrac{6}{8} \cdot \dfrac{2}{7} \right ) + \left(\dfrac{6}{9} \cdot \dfrac{3}{8} \cdot \dfrac{2}{7} \right )$

= $3 \cdot \left(\dfrac{3 \cdot 2 \cdot 6}{9 \cdot 8 \cdot 7}\right)$ = $\dfrac{3}{14}$

Or

= $\binom{3}{2} \cdot \left(\frac{3}{9} \cdot \dfrac{2}{8} \cdot \dfrac{6}{7} \right )$ = $\frac{3}{14}$

Here $\binom{3}{2}$ because 2 greens can come to any place out of 3.

• Could you please explain how does $\left[\frac{3}{9} \cdot \frac{2}{8} \cdot \frac{6}{7} \right ]$ in $\binom{3}{2} \cdot \left[\frac{3}{9} \cdot \frac{2}{8} \cdot \frac{6}{7} \right ]$ come? – user81411 Dec 29 '16 at 5:38
• I already updated my answer. Like we have 3 options. (1st green * 2nd green * 3rd red) but it is not necessary 3rd is red. May be 1st or 2nd ball is red. So we have 3 cases. We can write as $\binom{3}{2}$ or C(3,2) for number of possible cases i.e $\frac{3!}{2!*1*!} = 3$ – Kanwaljit Singh Dec 29 '16 at 5:47
• If still any doubt you can ask. – Kanwaljit Singh Dec 29 '16 at 5:48

The probability that the first ball selected is green is $P_1 = \frac {3}{9}$. After selecting a green balls, there are now $2$ green and a total of $8$ balls. Now the probability of selecting a green balls is $P_2 = \frac {2}{8}$. Now there are a total of $7$ balls and a green ball. Now the probability is $P_3=\frac {1}{7}$. Thus, $$P_{\text {req}} = P_1 \times P_2 \times P_3 = \frac {3}{9} \times \frac {2}{8} \times \frac {1}{7} = \frac {1}{84}$$ which is the same as that got using the hypergeometric distribution formula. Hope it helps.

• but if the question is: If three balls are randomly chosen without replacement, what is the probability that two balls are green? Will it be: $(3/9)\times\{(2/8)+(3/8)\}$? – user81411 Dec 29 '16 at 5:11

$$P = \frac{3}{9} \cdot \frac{2}{8}\cdot \frac{1}{7} = \frac{1}{3} \cdot \frac{1}{4} \cdot \frac{1}{7} = \frac{1}{84}$$