 Joanna Benson

2021-12-20

Seven balls are randomly withdrawn from an urn that contains 12 red, 16 blue, and 18 green balls. Find the probability that (a) 3 red, 2 blue, and 2 green balls are withdrawn; (b) at least 2 red balls are withdrawn; (c) all withdrawn balls are the same color; (d) either exactly 3 red balls or exactly 3 blue balls are withdrawn.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

Step 1
Given:
An urn that contains 12R, 16B, 18G balls
7 balls are randomly chosen
The result space S exhibits an equal distribution of probabilities.
$S=\left\{\left\{{x}_{1},{x}_{2},{x}_{3},{x}_{4},{x}_{5},{x}_{6},{x}_{7}\right\};{x}_{i}$ are different balls from the urn
S having 7 different valued combinations $|S|=$
$\left(\genfrac{}{}{0}{}{46}{7}\right)$
($|X|$ is the number of elements in X)
For an event $A\subseteq S$$|A|=n$
$P\left(A\right)=\frac{n}{\left(\genfrac{}{}{0}{}{46}{7}\right)}$
a) $P\left(3R,2B,2G\right)=?$
Find the number of ball combinations with 7 different values where 3 are red, 2 are blue, and 2 are green.
Choose the three red (R) from 12 of them in $\left(\genfrac{}{}{0}{}{12}{3}\right)$ ways, then choose 2 more blue to generate $\left(\genfrac{}{}{0}{}{16}{2}\right)$ times more possibilities, and choose the 2 green ones - $\left(\genfrac{}{}{0}{}{18}{2}\right)$ possibilities
$P\left(3R,2B,2G\right)=\frac{\left(\genfrac{}{}{0}{}{12}{3}\right)\cdot \left(\genfrac{}{}{0}{}{16}{2}\right)\cdot \left(\genfrac{}{}{0}{}{18}{2}\right)}{\left(\genfrac{}{}{0}{}{46}{7}\right)}=\frac{3060}{40549}=0.075464$ zesponderyd

Step 2
b) P(At least two R)=?
Remember Proposition 4.1
$P\left(A\right)=1-P\left({A}^{c}\right)$
A draw of 0 or 1 red ball is a complement to an event in which at least 2 red balls were drawn.
mutually exclusive events
There are $\left(\genfrac{}{}{0}{}{34}{7}\right)$a selection of 7 balls without any reds (that is choosing between 34 non red balls)
And if there is only one red ball, it can be chosen from the red balls in 12 different ways, but for every red ball selection, the other six balls can be selected in $\left(\genfrac{}{}{0}{}{34}{6}\right)$ different ways. nick1337

Step 3
c) PP(7R or 7B or 7G)=?
P(7R or 7B  or 7G)= P(7R)+P(7B)+P(7G) $⇒$ mutually exclusive events
The reds can be chosen in $\left(\genfrac{}{}{0}{}{12}{7}\right)$ ways because there are 12 red balls to chose from, similarly the blue balls can be any of the $\left(\genfrac{}{}{0}{}{16}{7}\right)$ possible choices, and 7G has $\left(\genfrac{}{}{0}{}{18}{7}\right)$ 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 sum of the selections for the three red balls produces the number of options where  - $\left(\genfrac{}{}{0}{}{12}{3}\right)$ and the number of choices for the remaining non red balls - $\left(\genfrac{}{}{0}{}{34}{4}\right)$
Same for the blue, the number of draws of precisely three blue balls is $\left(\genfrac{}{}{0}{}{16}{3}\right)\left(\genfrac{}{}{0}{}{30}{4}\right)$
Additionally, if there are three red balls and three blue balls among the seven balls available $\left(A\cap BA\cap B\right)$ it can be any of $\left(\genfrac{}{}{0}{}{12}{3}\right)\left(\genfrac{}{}{0}{}{16}{3}\right)\left(\genfrac{}{}{0}{}{18}{1}\right)$
Due to the fact that the initial calculation $A\cup B$  is the Proposition 4.4.

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