# Seven balls are randomly withdrawn from an urn that contains 12 red, 1

Seven balls are randomly withdrawn from an urn that contains 12 red, 16 blue, and 18 green balls. Find the probability that (a) 3 red, 2 blue, and 2 green balls are withdrawn; (b) at least 2 red balls are withdrawn; (c) all withdrawn balls are the same color; (d) either exactly 3 red balls or exactly 3 blue balls are withdrawn.

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Barbara Meeker

Step 1
Given:
An urn that contains 12R, 16B, 18G balls
7 balls are randomly chosen
Probability is equally distributed over the outcome space S:
$$​​​​​​​S=\{\{x_{1}, x_{2}, x_{3}, x_{4}, x_{5}, x_{6}, x_{7}\} ; x_{i}$$ are different balls from the urn
Since S contains 7 valued combinations $$\displaystyle{\left|{S}\right|}=$$
$$\binom{46}{7}$$
($$\displaystyle{\left|{X}\right|}$$ is the number of elements in X)
For an event $$A\subseteq S$$, $$\displaystyle{\left|{A}\right|}={n}$$
$$P(A)=\frac{n}{\binom{46}{7}}$$
a) $$\displaystyle{P}{\left({3}{R},{2}{B},{2}{G}\right)}=?$$
Find the number of 7 valued combinations of balls where three are red, two blue and 2 green.
Choose the three red (R) from 12 of them in $$\binom{12}{3}$$ ways, then choose 2 more blue to generate $$\binom{16}{2}$$ times more possibilities, and choose the 2 green ones - $$\binom{18}{2}$$ possibilities
$$P(3R, 2B, 2G)=\frac{\binom{12}{3}\cdot\binom{16}{2}\cdot \binom{18}{2}}{\binom{46}{7}}=\frac{3060}{40549}=0.075464$$

###### Not exactly what you’re looking for?
zesponderyd

Step 2
b) P(At least two R)=?
Remember Proposition 4.1
$$\displaystyle{P}{\left({A}\right)}={1}-{P}{\left({A}^{{{c}}}\right)}$$
And complement from an event where at least two red balls were drawn is an event where 0 or 1 red ball is drawn.
$$P(0\ R\ \textit{or}\ R)=P(0R)+P(1R) \Rightarrow$$ mutually exclusive events
There are $$\binom{34}{7}$$ choices of 7 balls that contain no reds (that is choosing between 34 non red balls)
And if there is only 1 red ball, it can be chosen among red balls in 12 ways, but for every choice of the red ball the remaining six can be chosen in $$\binom{34}{6}$$ ways.
$\begin{matrix} P(At\ least\ two\ R) &=1-P(0R or 1R)\\ &= 1- P(0R) - P(1R)\\ &= 1-\frac{\binom{34}{7}}{\binom{46}{7}}-\frac{12\binom{34}{6}}{\binom{46}{7}}\\ &=0.59797... \end{matrix}$

nick1337

Step 3
c) PP(7R or 7B or 7G)=?
P(7R or 7B  or 7G)= P(7R)+P(7B)+P(7G) $$\Rightarrow$$ mutually exclusive events
The reds can be chosen in $$\binom{12}{7}$$ ways because there are 12 red balls to chose from, similarly the blue balls can be any of the $$\binom{16}{7}$$ possible choices, and 7G has $$\binom{18}{7}$$ possibilities.
Step 4
d) P(precisely 3 R or precisely 3 B)=?
A - event that excatly 3 red balls are drawn
B - event that excatly 3 blue balls are drawn
The number of choices where three balls are red is the product of number of choices for the three red balls - $$\binom{12}{3}$$ and the number of choices for the remaining non red balls - $$\binom{34}{4}$$
Same for the blue, the number of draws of precisely three blue balls is $$\binom{16}{3}\binom{30}{4}$$
Also, if the choice of seven balls consists of three red balls and three blue balls $$(A\cap BA∩B)$$ it can be any of $$\binom{12}{3}\binom{16}{3}\binom{18}{1}$$
This is needed because the first step in calculating $$A\cup B$$  is the Proposition 4.4.
$$\begin{matrix} P(precisely\ 3\ R\ or\ precisely\ 3\ B)& = P(A\cup B)\\ &= P(A)+P(B)-P(A\cap B)\\ &=\frac{\binom{12}{3}\binom{34}{4}}{\binom{46}{7}}+\frac{\binom{16}{3}\binom{30}{4}}{\binom{46}{7}}-\frac{\binom{12}{3}\binom{16}{3}18}{\binom{46}{7}} \\ &=0.4359... \end{matrix}$$