cistG

2021-08-15

An urn contains 5 white and 10 black balls. A fair die is rolled and that number of balls is randomly chosen from the urn. What is the probability that all of the balls selected are white? What is the conditional probability that the die landed on 3 if all the balls selected are white?

insonsipthinye

1.Events:
A- ll of the choose balls are white
${E}_{i}$ - result of the die rill is i, $i=1,2,3,4,5,6$
Probabilities:
Since the die is fair:
$P\left({E}_{i}\right)=\frac{1}{6}f\phantom{\rule{1em}{0ex}}\text{or}\phantom{\rule{1em}{0ex}}i=\left\{1,2,3,4,5,6\right\}$
If the die rolls i we choose a combination of i balls, among black and five white balls, therefore $P\left(A\mid {E}_{1}\right)=\frac{\left(\begin{array}{c}5\\ 1\end{array}\right)}{\left(\begin{array}{c}15\\ 1\end{array}\right)}=\frac{5}{15}=\frac{1}{3}$
$P\left(A\mid {E}_{2}\right)=\frac{\left(\begin{array}{c}5\\ 2\end{array}\right)}{\left(\begin{array}{c}15\\ 2\end{array}\right)}=\frac{10}{105}=\frac{2}{21}$
$P\left(A\mid {E}_{3}\right)=\frac{\left(\begin{array}{c}5\\ 3\end{array}\right)}{\left(\begin{array}{c}15\\ 3\end{array}\right)}=\frac{10}{455}=\frac{2}{91}$
$P\left(A\mid {E}_{4}\right)=\frac{\left(\begin{array}{c}5\\ 4\end{array}\right)}{\left(\begin{array}{c}15\\ 4\end{array}\right)}=\frac{1}{273}$
$P\left(A\mid {E}_{5}\right)=\frac{\left(\begin{array}{c}5\\ 5\end{array}\right)}{\left(\begin{array}{c}15\\ 5\end{array}\right)}=\frac{1}{3003}$
$P\left(A\mid {E}_{6}\right)=\frac{\left(\begin{array}{c}5\\ 6\end{array}\right)}{\left(\begin{array}{c}15\\ 6\end{array}\right)}=0$
Calculate: $P\left(A\right),P\left({E}_{3}\mid A\right)$
2.${E}_{1},{E}_{2},{E}_{3},{E}_{4},{E}_{5},{E}_{6}$ are competing hypothesis, that is, mutually exclusive events, union of which is the whole outcome space, so conditioning on the roll of the die: $P\left(A\right)=\sum _{i=1}^{6}P\left(A\mid {E}_{i}\right)P\left({E}_{i}\right)$
Substituting $P\left({E}_{i}\right),P\left(A\mid {E}_{i}\right)$ yields: $P\left(A\right)=\frac{1}{6}\left(\frac{1}{3}+\frac{2}{21}+\frac{2}{91}+\frac{1}{273}+\frac{1}{3003}\right)=\frac{5}{66}$
$P\left({E}_{3}\mid A\right)$ can be calculated from the definition if we note the identity $P\left({}_{}$

Jeffrey Jordon

A-event that all balls drawn are white

${D}_{i}$ - outcome of roll of the die is $i,i=1,2,\dots ,6$

$P\left(A\right)=\sum _{i=1}^{6}P\left(A|{D}_{i}\right)P\left({D}_{i}\right)$

$=\frac{1}{6}\left(P\left(A|{D}_{i}\right)+\dots +P\left(A|{D}_{6}\right)\right)$

$=\frac{1}{6}\left(\frac{5}{15}+\frac{{5}_{{C}_{2}}}{{15}_{{C}_{2}}}+\frac{{5}_{{C}_{3}}}{{15}_{{C}_{3}}}+\frac{{5}_{{C}_{4}}}{{15}_{{C}_{4}}}+\frac{{5}_{{C}_{5}}}{{15}_{{C}_{5}}}+0\right)=\frac{5}{66}$

$P\left({D}_{3}|A\right)=\frac{P\left(A|{D}_{3}\right)P\left({D}_{3}\right)}{P\left(A\right)}$

$=\frac{\frac{{5}_{{C}_{3}}}{{15}_{{C}_{3}}}}{\sum _{i=1}^{5}\frac{{5}_{{C}_{i}}}{{15}_{{C}_{i}}}}=\frac{22}{455}$

Vasquez

Result:
$P\left(A\right)=\frac{22}{455}$
$P\left(C|A\right)=\frac{1}{22}$
Solution:
1. Probability of selecting all white balls:
Let's denote the event of selecting all white balls as $A$, and the event of rolling a die and choosing that number of balls from the urn as $B$. We need to find the probability of event $A$ given that event $B$ has occurred.
The total number of balls in the urn is 15 (5 white + 10 black). The probability of selecting all white balls from the urn depends on the number rolled on the die. Let $n$ be the number rolled on the die.
The probability of selecting all white balls given that $n$ is rolled is the ratio of the number of ways to select all white balls from the urn to the total number of possible selections when rolling $n$ on the die.
The number of ways to select all white balls from the urn is $\left(\genfrac{}{}{0}{}{5}{n}\right)$ (choosing $n$ white balls out of 5), and the total number of possible selections when rolling $n$ is $\left(\genfrac{}{}{0}{}{15}{n}\right)$ (choosing $n$ balls out of 15).
Therefore, the probability of selecting all white balls given that $n$ is rolled can be expressed as:
$P\left(A|B=n\right)=\frac{\left(\genfrac{}{}{0}{}{5}{n}\right)}{\left(\genfrac{}{}{0}{}{15}{n}\right)}$
To find the overall probability of selecting all white balls, we need to consider the probabilities for each possible value of $n$ and weigh them by their respective probabilities.
The die is fair, so each number from 1 to 6 has an equal probability of $\frac{1}{6}$.
Hence, the probability of selecting all white balls can be calculated as:
$P\left(A\right)={\sum }_{n=1}^{6}P\left(A|B=n\right)·P\left(B=n\right)={\sum }_{n=1}^{6}\left(\frac{\left(\genfrac{}{}{0}{}{5}{n}\right)}{\left(\genfrac{}{}{0}{}{15}{n}\right)}\right)·\left(\frac{1}{6}\right)$
2. Conditional probability of the die landing on 3 given that all selected balls are white:
Let's denote the event of the die landing on 3 as $C$. We need to find the conditional probability of event $C$ given event $A$.
Using Bayes' theorem, the conditional probability can be calculated as:
$P\left(C|A\right)=\frac{P\left(A|C\right)·P\left(C\right)}{P\left(A\right)}$
We already have the expression for $P\left(A\right)$ from the previous part. The probability of the die landing on 3 is $\frac{1}{6}$, since it is a fair die. To find $P\left(A|C\right)$, we need to substitute $n=3$ in the expression we derived earlier for $P\left(A|B=n\right)$.
Substituting all the values, the conditional probability can be calculated as:
$P\left(C|A\right)=\frac{\left(\frac{\left(\genfrac{}{}{0}{}{5}{3}\right)}{\left(\genfrac{}{}{0}{}{15}{3}\right)}\right)·\left(\frac{1}{6}\right)}{P\left(A\right)}$
To get the final answer, we substitute the expression for $P\left(A\right)$ and calculate the conditional probability.
Calculating these values yields the following result:
$P\left(A\right)=\frac{22}{455}$
$P\left(C|A\right)=\frac{1}{22}$
Therefore, the probability that all the balls selected are white is $\frac{22}{455}$, and the conditional probability that the die landed on 3 given that all the balls selected are white is $\frac{1}{22}$.

RizerMix

To solve this problem, let's break it down into two parts.
Part 1: Probability of selecting all white balls
We have an urn containing 5 white balls and 10 black balls. A fair die is rolled, and that number of balls is randomly chosen from the urn. We want to find the probability that all of the balls selected are white.
Let's denote the event of selecting all white balls as $A$ and the event of rolling a particular number on the die as $B$. We need to find $P\left(A\right)$, the probability of event $A$ occurring.
The total number of balls in the urn is 15 (5 white + 10 black). Let's assume the die rolls a number $n$ (where $n$ can take values from 1 to 6). The number of ways we can choose $n$ balls from the urn is given by the binomial coefficient $\left(\genfrac{}{}{0}{}{15}{n}\right)$. The number of ways we can choose all white balls (5) from the 5 white balls in the urn is $\left(\genfrac{}{}{0}{}{5}{5}\right)=1$. Therefore, the probability of event $A$ given that the die rolls a number $n$ is:
$P\left(A|B=n\right)=\frac{\left(\genfrac{}{}{0}{}{5}{5}\right)}{\left(\genfrac{}{}{0}{}{15}{n}\right)}$
Since the die is fair, the probability of rolling any particular number $n$ is $\frac{1}{6}$. Therefore, the probability of event $A$ occurring is:
$P\left(A\right)=\frac{1}{6}\left(P\left(A|B=1\right)+P\left(A|B=2\right)+P\left(A|B=3\right)+P\left(A|B=4\right)+P\left(A|B=5\right)+P\left(A|B=6\right)\right)$
Now let's calculate the probabilities for each roll of the die:
$P\left(A|B=1\right)=\frac{\left(\genfrac{}{}{0}{}{5}{5}\right)}{\left(\genfrac{}{}{0}{}{15}{1}\right)}$
$P\left(A|B=2\right)=\frac{\left(\genfrac{}{}{0}{}{5}{5}\right)}{\left(\genfrac{}{}{0}{}{15}{2}\right)}$
$P\left(A|B=3\right)=\frac{\left(\genfrac{}{}{0}{}{5}{5}\right)}{\left(\genfrac{}{}{0}{}{15}{3}\right)}$
$P\left(A|B=4\right)=\frac{\left(\genfrac{}{}{0}{}{5}{5}\right)}{\left(\genfrac{}{}{0}{}{15}{4}\right)}$
$P\left(A|B=5\right)=\frac{\left(\genfrac{}{}{0}{}{5}{5}\right)}{\left(\genfrac{}{}{0}{}{15}{5}\right)}$
$P\left(A|B=6\right)=\frac{\left(\genfrac{}{}{0}{}{5}{5}\right)}{\left(\genfrac{}{}{0}{}{15}{6}\right)}$
Now we can substitute these values into the formula for $P\left(A\right)$:
$P\left(A\right)=\frac{1}{6}\left(\frac{\left(\genfrac{}{}{0}{}{5}{5}\right)}{\left(\genfrac{}{}{0}{}{15}{1}\right)}+\frac{\left(\genfrac{}{}{0}{}{5}{5}\right)}{\left(\genfrac{}{}{0}{}{15}{2}\right)}+\frac{\left(\genfrac{}{}{0}{}{5}{5}\right)}{\left(\genfrac{}{}{0}{}{15}{3}\right)}+\frac{\left(\genfrac{}{}{0}{}{5}{5}\right)}{\left(\genfrac{}{}{0}{}{15}{4}\right)}+\frac{\left(\genfrac{}{}{0}{}{5}{5}\right)}{\left(\genfrac{}{}{0}{}{15}{5}\right)}+\frac{\left(\genfrac{}{}{0}{}{5}{5}\right)}{\left(\genfrac{}{}{0}{}{15}{6}\right)}\right)$
Part 2: Conditional probability of die landing on 3 given all balls selected are white
We want to find the conditional probability of the die landing on 3 given that all the balls selected are white. Let's denote this event as $C$ (die lands on 3)
and event $A$ (all balls selected are white). We need to find $P\left(C|A\right)$.
Using Bayes' theorem, the conditional probability is given by:
$P\left(C|A\right)=\frac{P\left(A|C\right)·P\left(C\right)}{P\left(A\right)}$
We have already calculated $P\left(A\right)$ in Part 1. The probability of the die landing on 3 is $\frac{1}{6}$, since the die is fair. To calculate $P\left(A|C\right)$, we need to find the probability of selecting all white balls given that the die lands on 3.
$P\left(A|C\right)=\frac{\left(\genfrac{}{}{0}{}{5}{5}\right)}{\left(\genfrac{}{}{0}{}{15}{3}\right)}$
Now we can substitute these values into the formula for $P\left(C|A\right)$:
$P\left(C|A\right)=\frac{\frac{\left(\genfrac{}{}{0}{}{5}{5}\right)}{\left(\genfrac{}{}{0}{}{15}{3}\right)}·\frac{1}{6}}{P\left(A\right)}$
Therefore, the probability that all of the balls selected are white is $P\left(A\right)$, and the conditional probability that the die landed on 3 given all the balls selected are white is $P\left(C|A\right)$.

Do you have a similar question?