This question aims to find the joint and conditional probabilities. Probability is a measure of the likelihood that an event will occur. Many events are not able to predicted with absolute certainty. We can only expect the probability of an event, i.e., how likely it is to occur, using it. Probability ranges from 0 to 1, where 0 means the event is impossible and 1 indicates a particular event.
Conditional Probability
Conditional probability is the probability of an event\outcome occurring based on the occurrence of a previous event. Conditional probability is calculated by multiplying probability of the last event by the updated probability of the subsequent or conditional event.
For example:
- Event A is that an individual applying to college will be accepted. There is an 80% chance that the individual will be accpeted into college.
- Event B is that this person will be allocated accommodation in the dormitory. Accommodation in the dormitories 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 $1,2,3,4,5,6$
Probabilities
Since the die is fair, all the outcomes have an equal probability 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 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 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 $1,2,3,4,5,6$
Probabilities
Since the die is fair, all the outcomes have an equal probability 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:
\[P(A|E_{1})=\dfrac{\binom {4} {1}}{\binom {14} {1}}=\dfrac{2}{7}\]
\[P(A|E_{2})=\dfrac{\binom {4} {2}}{\binom {14 {2}}=\dfrac{6}{91}\]
\[P(A|E_{3})=\dfrac{\binom {4} {3}}{\binom {14} {3}}=\dfrac{1}{91}\]
\[P(A|E_{4})=\dfrac{\binom {4} {4}}{\binom {14} {4}}=\dfrac{1}{1001}\]
\[P(A|E_{5})=\dfrac{\binom {4} {5}}{\binom {14} {5}}=0\]
\[P(A|E_{6})=\dfrac{\binom {4} {6}}{\binom {14} {6}}=0\]
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 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 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 $\dfrac{1}{91}$.