Seven balls are randomly withdrawn from an urn that contains 12 red, 1

Joanna Benson 2021-12-20 Answered
Seven balls are randomly withdrawn from an urn that contains 12 red, 16 blue, and 18 green balls. Find the probability that (a) 3 red, 2 blue, and 2 green balls are withdrawn; (b) at least 2 red balls are withdrawn; (c) all withdrawn balls are the same color; (d) either exactly 3 red balls or exactly 3 blue balls are withdrawn.

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Expert Answer

Barbara Meeker
Answered 2021-12-21 Author has 5533 answers

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
\(P(3R, 2B, 2G)=\frac{\binom{12}{3}\cdot\binom{16}{2}\cdot \binom{18}{2}}{\binom{46}{7}}=\frac{3060}{40549}=0.075464\)

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zesponderyd
Answered 2021-12-22 Author has 5143 answers

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.
\[\begin{matrix} P(At\ least\ two\ R) &=1-P(0R or 1R)\\ &= 1- P(0R) - P(1R)\\ &= 1-\frac{\binom{34}{7}}{\binom{46}{7}}-\frac{12\binom{34}{6}}{\binom{46}{7}}\\ &=0.59797... \end{matrix}\]

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nick1337
Answered 2021-12-27 Author has 9672 answers

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.
\(\begin{matrix} P(precisely\ 3\ R\ or\ precisely\ 3\ B)& = P(A\cup B)\\ &= P(A)+P(B)-P(A\cap B)\\ &=\frac{\binom{12}{3}\binom{34}{4}}{\binom{46}{7}}+\frac{\binom{16}{3}\binom{30}{4}}{\binom{46}{7}}-\frac{\binom{12}{3}\binom{16}{3}18}{\binom{46}{7}} \\ &=0.4359... \end{matrix}\)

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