1
$\begingroup$

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

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

$\endgroup$
  • $\begingroup$ I've no idea whatsoever where you got that calculation that yields $90/504$. $\endgroup$ – Wildcard Dec 29 '16 at 5:12
1
$\begingroup$

a) Your first answer is correct.

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.

$\endgroup$
  • $\begingroup$ 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? $\endgroup$ – user81411 Dec 29 '16 at 5:38
  • $\begingroup$ 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$ $\endgroup$ – Kanwaljit Singh Dec 29 '16 at 5:47
  • $\begingroup$ If still any doubt you can ask. $\endgroup$ – Kanwaljit Singh Dec 29 '16 at 5:48
2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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: $(3/9)\times\{(2/8)+(3/8)\}$? $\endgroup$ – user81411 Dec 29 '16 at 5:11
1
$\begingroup$

The answer is given by:

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.