This **question aims**Â to find the **joint and conditional**Â **probabilities. **Probability is a measure of the likelihood that an event will occur. Many events are not able to predicted with **absolute certainty.**Â We can only expect the probability of an event, i.e., how likely it is to occur, using it. **Probability ranges from**Â 0 to 1, where 0 means the event is **impossible**Â and 1Â **indicates a particular event.**

**Conditional Probability**

**Conditional probability**Â is the **probability o**f an event\outcome Â occurring based on the **occurrence of a previous event.**Â **Conditional probability**Â is calculated by **multiplying**Â probability of the last event by the updated probability of the **subsequent or conditional event.**

**For example:**

**Event**Â**A**is that an**individual applying to college will be accepted.**Â There is an**80%**chance that the individual will be accpeted into college.**Event B**Â is that this**Â person**Â will be**allocated accommodation**Â in the dormitory**. Accommodation in the dormitories**Â will be provided to only**60%**of all admitted students.- P (
**Accepted and Dorm Accommodation**) = P (Dorm Accommodation | Accepted) P (Accepted) =$ (0.60)*(0.80) = 0.48$.

**Expert Answer**

**Part(1)**

**Events:**

$A-$ **choose balls are white.**

$E_{i}-$ **result of the die rolls**Â $1,2,3,4,5,6$

**Probabilities**

Since the**Â die is fair,**Â all the outcomes have an **equal probability**Â to appear.

\[P(E_{i})=\dfrac{1}{6} \:where\: i=1,2,3,4,5,6\]

if the die is rolled, choose a combination of $i$ balls, among black and white balls, therefore:

\[P(A|E_{1})=\dfrac{\binom {5} {1}}{\binom {15} {1}}=\dfrac{5}{15}=\dfrac{1}{3}\]

\[P(A|E_{2})=\dfrac{\binom {5} {2}}{\binom {15} {2}}=\dfrac{10}{105}=\dfrac{2}{21}\]

\[P(A|E_{3})=\dfrac{\binom {5} {3}}{\binom {15} {3}}=\dfrac{10}{455}=\dfrac{2}{91}\]

\[P(A|E_{4})=\dfrac{\binom {5} {4}}{\binom {15} {4}}=\dfrac{1}{273}\]

\[P(A|E_{5})=\dfrac{\binom {5} {5}}{\binom {15} {5}}=\dfrac{1}{3003}\]

\[P(A|E_{6})=\dfrac{\binom {5} {6}}{\binom {15} {6}}=0\]

Calculate $P(A),P(A_{3}|A)$.

$E_{1},E_{2},E_{3},E_{4},E_{5},E_{6}$ are competing hypotheses, i.e. mutually exclusive events, the connection of which is the entire resulting space, so the conditional is a roll of the dice:

\[P(A)=\sum_{i=1}^{6} P(A|E_{i})P(E_{i})\]

**Plug values**Â of $P(E_{i})$ and $P(E|A_{i})$.

\[P(A)=\dfrac{1}{6}(\dfrac{1}{3}+\dfrac{2}{21}+\dfrac{2}{91}+\dfrac{1}{273}+\dfrac{1}{3003})=\dfrac{5}{66}\]

$P(E_{3}|A)$ can be **calculated**Â from the $P(E_{3})$ and $P(A|E_{3})$.

\[P(E_{3}|A)=P(A|E_{3})P(E_{3})\]

\[P(E_{3}|A)=\dfrac{2}{91}\dfrac{1}{6}=\dfrac{1}{273}\]

**Numerical Result**

- The probability that all the balls selected are white is $P(A)=\dfrac{5}{66}$.
- The conditional probability of $P(E_{3}|A)$ is $\dfrac{1}{273}$.

**Example**

A jar contains $4$ white and $10$ black balls. A fair die is rolled, and this number of marbles is randomly drawn from the jar. What is the probability that all the balls selected are white? What is the conditional probability that the die rolls a $2$ if all the chosen balls are white?

**Solution**

**Part(1)**

**Events:**

$A-$ **choose balls are white.**

$E_{i}-$ **result of the die rolls**Â $1,2,3,4,5,6$

**Probabilities**

Since the**Â die is fair,**Â all the outcomes have an **equal probability**Â to appear.

\[P(E_{i})=\dfrac{1}{6} \:where\: i=1,2,3,4,5,6\]

if theÂ **d****ie is rolled**,**Â choose a combination**Â of $i$ balls among **black and white balls**, therefore:

\[P(A|E_{1})=\dfrac{\binom {4} {1}}{\binom {14} {1}}=\dfrac{2}{7}\]

\[P(A|E_{2})=\dfrac{\binom {4} {2}}{\binom {14 {2}}=\dfrac{6}{91}\]

\[P(A|E_{3})=\dfrac{\binom {4} {3}}{\binom {14} {3}}=\dfrac{1}{91}\]

\[P(A|E_{4})=\dfrac{\binom {4} {4}}{\binom {14} {4}}=\dfrac{1}{1001}\]

\[P(A|E_{5})=\dfrac{\binom {4} {5}}{\binom {14} {5}}=0\]

\[P(A|E_{6})=\dfrac{\binom {4} {6}}{\binom {14} {6}}=0\]

Calculate $P(A),P(A_{3}|A)$.

$E_{1},E_{2},E_{3},E_{4},E_{5},E_{6}$ are **competing hypotheses**, i.e. **mutually exclusive events**, the connection of which is the entire resulting space, so the conditional is a roll of the dice:

\[P(A)=\sum_{i=1}^{6} P(A|E_{i})P(E_{i})\]

**Plug values**Â of $P(E_{i})$ and $P(E|A_{i})$.

\[P(A)=\dfrac{1}{6}(\dfrac{2}{7}+\dfrac{6}{91}+\dfrac{1}{91}+\dfrac{1}{1001})=\dfrac{2}{33}\]

$P(E_{2}|A)$ can be **calculated**Â from the $P(E_{2})$ and $P(A|E_{2})$.

\[P(E_{2}|A)=P(A|E_{2})P(E_{2})\]

\[P(E_{2}|A)=\dfrac{6}{91}\dfrac{1}{6}=\dfrac{1}{91}\]

**The probability** that all the balls selected are white are $P(A)=\dfrac{2}{33}$.

**The conditional probability** of $P(E_{3}|A)$ is $\dfrac{1}{91}$.