# A bag contains 12 sweets, of which 5 are red, 4 are green and 3 are yellow. 3 sweets are chosen without replacement. What is the probability that he chooses no red sweets?

Jun 30, 2018

$= \frac{7}{44}$

#### Explanation:

For:

• $12 = \left\{\begin{matrix}5 R \\ 4 G \\ 3 Y\end{matrix}\right.$

There are $\left(\begin{matrix}12 \\ 3\end{matrix}\right)$ ways of choosing 3 sweets from 12, regardless of color

There are $\left(\begin{matrix}7 \\ 3\end{matrix}\right)$ ways of choosing 3 sweets from the non-red sweets.

So:

P("all 3 are non-Red") = (((7),(3)))/(((12),(3))) = (7!9!3!)/(3!4!12!)

$= \frac{7 \cdot 6 \cdot 5}{12 \cdot 11 \cdot 10} = \frac{7}{44}$

Alternatively, using the Hypergeometric distribution:

• $P \left(X = k\right) = \frac{\left(\begin{matrix}K \\ k\end{matrix}\right) \left(\begin{matrix}N - K \\ n - k\end{matrix}\right)}{\left(\begin{matrix}N \\ n\end{matrix}\right)}$

• $N = 12$, is the population size (ie,  balls):

• $K = 5$, is the number of success states (ie, Red balls) in the population,
• $n = 3$, is the number of draws
• $k = 0$, is the number of observed successes

$P \left(X = 0\right) = \frac{\left(\begin{matrix}12 \\ 0\end{matrix}\right) \left(\begin{matrix}7 \\ 3\end{matrix}\right)}{\left(\begin{matrix}12 \\ 3\end{matrix}\right)}$

= (1* ((7!)/(3!4!)))/(((12!)/(9!3!))) = 7/44