Probability
5.0 Conditional Probability
5.0 Conditional Probability
$P(A/B)$ denotes the probability of event $A$ happening, given that event $B$ has happened. It is the conditional probability of $A$, given $B$. In other words, $P(A/B)$ implies that the event $B$ has already occurred, after which $A$ occurs. It implies that the outcomes favourable to $B$ become the total outcomes, and $P(A/B)$ states the outcomes common to $A$ and $B$.
$$P(A/B) = P(A\;given\;B)$$ $$ = {{Total\;no.\;\;of\;favourable\;cases} \over {Total\;no.\;of\;cases}} = {{P(A \cap B)} \over {P(B)}}$$
The probability of part of $A$ that belongs to $B$ is calculated.
Illustration 21. Let there be a containing $5$ white and $4$ red balls. Two balls are drawn from the bag one after the other without replacement. Consider the following events
$A$, drawing a white ball in the first draw and $B$, drawing a red ball in the second draw. What are the possible ways conditional probability can be applied here?
Solution: $A$ = drawing a white ball in the first draw.
$B$ = drawing a red ball in the second draw.
Let us consider, happening of $B$ after $A$. Then, the conditional probability is represented as $P(B/A)$.
$P(B/A)$ = Probability of drawing a red ball in second draw given that a white ball has already been drawn in the first draw.
After event $A$ occurs, there are $4$ white and $4$ red balls left.
Thus
$P(B/A)$ = Probability of drawing a red ball from a bag of $4$ white and $4$ red balls.
$$P(B/A) = {4 \over 8} = {1 \over 2}$$
This shows that $P(B/A)$ is a valid conditional probability.
For this random experiment, $P(A/B)$ is not meaningful, as the expression $P(A/B)$ means, drawing of white in the first draw, given red is drawn in the second. But $A$ cannot occur after $B$ according to the event description.
Note: The events $A$ and $B$ are subsets of two different sample spaces as they are outcomes of two different trials which are performed one after the other.
Illustration 22. Consider the random experiment of throwing a pair of dice and two events associated with it given by
$A$ - The sum of the numbers on the two dice is $8$.
$B$ - There is an even number on the first die.
What are the possible ways conditional probability can be applied here?
Solution: The sample space for the throwing of two dice is
$S =(1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (2,1), (2,2), (2,3), (2,4), (2,5), (2,6), (3,1), (3,2), (3,3), (3,4), (3,5), (3,6),$$(4,1), (4,2), (4,3), (4,4), (4,5), (4,6), (5,1), (5,2), (5,3), (5,4), (5,5), (5,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6)$
Both the events mentioned have the same sample space.
$A$ - The sum of the numbers on the two dice is $8$ = $\{ (2,6), (3,5), (4,4), (5,3), (6,2) \} $
$B$ - There is an even number on the first die = $ \{ (2,1), (2,2), (2,3), (2,4), (2,5), (2,6), (4,1), (4,2), (4,3), (4,4), (4,5), (4,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6)\} $
Let us now consider that the sum of the numbers on the dice is $8$, given that the first die shows an even number.
This conditional probability is the occurrence of $A$, after the occurrence of $B$.
Thus is represented as $P(A/B)$.
$P(A/B)$ = Probability of occurrence of $A$, when $B$ has occurred
$P(A/B)$ = Probability of occurrence of $A$ when $B$ is taken as the sample space.
$$\begin{equation} \begin{aligned} A \cap B = \{ (2,6),(4,4),(6,2)\} \\ P(A/B) = {{n(A \cap B)} \over {n(B)}} = {3 \over {18}} \\\end{aligned} \end{equation} $$
Let us now consider that the first die shows an even number given that the sum of the numbers on the dice is $8$.
This conditional probability is the occurrence of $B$, after the occurrence of $A$.
Thus is represented as $P(B/A)$.
$P(B/A)$ = Probability of occurrence of $B$, when $A$ has occurred
$P(B/A)$ = Probability of occurrence of $B$ when $A$ is taken as the sample space.
$$\begin{equation} \begin{aligned} B \cap A = \{ (2,6),(4,4),(6,2)\} \\ P(B/A) = {{n(A \cap B)} \over {n(A)}} = {3 \over 5} \\\end{aligned} \end{equation} $$
Illustration 23. A pair of dice is thrown. If the two numbers appearing on them are different, find the probability that
i. The sum of the numbers is $6$
ii. The sum of the numbers is $4$ or less.
iii. The sum of the numbers is $4$.
Solution: Consider the following events,
$A$ - Numbers appearing on the two dice are different.
$B$ -The sum of the numbers is $6$.
$C$ -The sum of the numbers is $4$ or less.
$D$ -The sum of the numbers is $4$.
Clearly,
$$A = (1,2),(1,3),(1,4),(1,5),(1,6),(2,1),(2,3),(2,4),(2,5),(2,6),(3,1),(3,2),(3,4),(3,5),(3,6),(4,1),(4,2),(4,3),(4,5),(4,6),$$$$(5,1),(5,2),(5,3),(5,4),(5,6),(6,1),(6,2),(6,3),(6,4),(6,5)$$$$B = { (1,5),(2,4),(4,2),(5,1),(3,3)}$$ $$C = { (1,2),(1,3),(2,1),(3,1),(1,1),(2,2)}$$ $$D = { (1,3),(2,2),(3,1)} $$
$$\begin{equation} \begin{aligned} A \cap B = \{ (1,5),(2,4),(4,2),(5,1)\} \\ A \cap C = \{ (1,2),(1,3),(2,1),(3,1)\} \\ A \cap D = \{ (1,3),(3,1)\} \\\end{aligned} \end{equation} $$
$$\begin{equation} \begin{aligned} n(A \cap B) = 4 \\ n(A \cap C) = 4 \\ n(A \cap D) = 2 \\ n(A) = 30 \\\end{aligned} \end{equation} $$
$P(B/A)$ = Probability of occurrence of $B$ given that $A$ has already occurred
i.e. $P(B/A)$ = Probability that we get the sum of the digits to be $6$, given that the dice show two different numbers
$$P(B/A) = {{n(A \cap B)} \over {n(A)}} = {4 \over {30}} = {2 \over {15}}$$
$P(C/A)$ = Probability of occurrence of $C$ given that $A$ has already occurred
i.e. $P(C/A)$ = Probability that we get the sum of the digits to be $4$ or less, given that the dice show two different numbers
$$P(C/A) = {{n(A \cap C)} \over {n(A)}} = {4 \over {30}} = {2 \over {15}}$$
$P(D/A)$ = Probability of occurrence of $D$ given that $A$ has already occurred
i.e. $P(D/A)$ = Probability that we get the sum of the digits to be $4$, given that the dice show two different numbers
$$P(D/A) = {{n(A \cap D)} \over {n(A)}} = {2 \over {30}} = {1 \over {15}}$$
Illustration 24. A fair dice is rolled. Consider the following events:
$A$ - Getting an odd number, $B$ - Getting a prime number less than $5$ and $C$ - Getting a number between, $1$ and $6$. Find
i. $P(A/B)$ and $P(B/A)$
ii. $P(A/C)$ and $P(C/A)$
iii. $P(A \cup B/C)$ iv. $P(A \cap B/C)$
Solution:
$A$ - Getting an odd number - $ \{1,3,5\}$
$B$ - Getting a prime number less than $5$ - $ \{2,3\}$
$C$ - Getting a number between, $1$ and $6$ - $ \{2,3,4,5\}$
$$\begin{equation} \begin{aligned} n(A) = 3 \\ n(B) = 2 \\ n(C) = 4 \\ \\ A \cap B = \{ 3\} \\ A \cap C = \{ 3,5\} \\ A \cap B \cap C = \{ 3\} \\ A \cup B \cap C = \{ 2,3,5\} \\ \\ n(A \cap B) = 1 \\ n(A \cap C) = 2 \\ n(A \cap B \cap C) = 1 \\ n(A \cup B \cap C) = 3 \\\end{aligned} \end{equation} $$
i. $$\begin{equation} \begin{aligned} P(A/B) = {{n(A \cap B)} \over {n(B)}} = {1 \over 2} \\ P(B/A) = {{n(A \cap B)} \over {n(A)}} = {1 \over 3} \\\end{aligned} \end{equation} $$
ii. $$\begin{equation} \begin{aligned} P(A/C) = {{n(A \cap C)} \over {n(C)}} = {1 \over 2} \\ P(C/A) = {{n(A \cap C)} \over {n(A)}} = {2 \over 3} \\\end{aligned} \end{equation} $$
iii. $$P(A \cup B/C) = {{n(A \cup B \cap C)} \over {n(C)}} = {3 \over 4}$$
iv. $$P(A \cap B/C) = {{n(A \cap B \cap C)} \over {n(C)}} = {1 \over 4}$$
Illustration 25. Ten cards numbered one to ten are placed in a box and mixed thoroughly. One card is drawn at random. If it is known that the number on the card is more than $3$, then find the probability that it is an even number.
Solution: The cards are numbered one to ten.
Hence, the sample space is $$S = \{ 1,2,3,4,5,6,7,8,9,10\} $$
Let the events be,
$A$ - A number greater than $3$.
$B$ - A number being even.
Then,
$$A = \{ 4,5,6,7,8,9,10\} $$
$$B = \{ 2,4,6,8,10\} $$
$$A \cap B = \{ 4,6,8,10\} $$
$$\begin{equation} \begin{aligned} n(A) = 7 \\ n(B) = 5 \\ n(A \cap B) = 4 \\\end{aligned} \end{equation} $$
Required probability is,
$$P(B/A) = {{n(A \cap B)} \over {n(A)}} = {4 \over 7}$$
Question 12. If a die is thrown, what is the probability of occurrence of a number lesser than $5$, if it is known that only odd numbers can come up.
Solution: Sample space $$S = \{ 1,2,3,4,5,6\} $$
Let $A$ = occurrence of an odd number = $\{ 1,3,5\} $
$B$ = occurrence of number less than five = $\{ 1,2,3,4\} $
Here
$$A \cap B = \{ 1,3\} $$
So,
$$P(B/A) = {{P(A \cap B)} \over {P(A)}} = {{2/6} \over {3/6}} = {2 \over 3}$$
Question 13. In an exam $30\% $ failed in Paper-I and $26\% $ failed in Paper-II. $11\% $ have failed in both the papers. A candidate is chosen at random:
i. if she has failed in paper-II, then find the chance of her failure in paper-I.
ii. if she has failed in paper-I, then find the chance of her failure in paper-II.
iii. find the chance of her failure in paper-I or paper-II.
Solution: Let $A$ and $B$ be events of failing in paper-I and paper-II respectively.
Since $30\% $ failed in paper-I,
$P(A) = 0.3$.
Similarly, since $26\% $ failed in paper-II,
$P(B) = 0.26$.
And since $11\% $ failed in both paper-I and paper-II,
$$P(A \cap B) = 0.11$$
i. The chance of failure in paper-I, given she has failed in paper-II is
$$P(A/B) = {{P(A \cap B)} \over {P(B)}} = {{0.11} \over {0.3}} = {{11} \over {30}}$$
ii. The chance of failure in paper-II, given she has failed in paper-I is
$$P(B/A) = {{P(A \cap B)} \over {P(A)}} = {{0.11} \over {0.26}} = {{11} \over {26}}$$
iii. The chance of failure in paper-I or paper-II is
$$P(A \cup B) = P(A) + P(B) - P(A \cap B) = 0.3 + 0.26 - 0.11 = 0.45 = 45\% $$
Question 14. A die is thrown twice and the sum of the numbers appearing is found to be $7$. What is the probability that the number $3$ has occurred at least once?
Solution: Let $A$ be the event that number $3$ appears at least once and $B$ be the event that the sum is $7$.
$$\begin{equation} \begin{aligned} B = \{ (6,1),(5,2),(3,4),(4,3),(2,5),(1,6)\} \\ A = \{ (1,3),(2,3),(3,3),(4,3),(5,3),(6,3),(3,6),(3,5),(3,4),(3,2),(3,1)\} \\ A \cap B = \{ (3,4),(4,3)\} \\\end{aligned} \end{equation} $$
$$P(A/B) = {{P(A \cap B)} \over {P(B)}} = {{2/36} \over {6/36}} = {2 \over 6} = {1 \over 3}$$
Question 15. Two integers are chosen at random from $1$ to $11$ . If the sum is even find the probability that both the numbers are odd.
Solution: There are $5$ even and $6$ odd integers. Let event $A$ be both the numbers chosen are odd and $B$ be sum of the numbers is even.
$$P(A) = {{^6{C_2}} \over {^{11}{C_2}}}$$
$$P(B) = {{^6{C_2}{ + ^5}{C_2}} \over {^{11}{C_2}}}$$
$$P(A \cap B) = {{^6{C_2}} \over {^{11}{C_2}}}$$
Required probability
$$P(A/B) = {{^6{C_2}} \over {^5{C_2}{ + ^6}{C_2}}} = {3 \over 5}$$
Question 16. A green die and a red die are rolled,
i. Find the conditional probability of obtaining a sum greater than $9$, given that the green die resulted in $5$.
ii. Find the conditional probability of obtaining the sum of $8$, given that the red die resulted in a number less than $4$.
Solution: Consider the following events:
$A$ = Getting a sum greater than $9$
$$\begin{equation} \begin{aligned} A = \{ (4,6),(6,4),(5,5),(5,6),(6,5),(6,6)\} \\ \Rightarrow n(A) = 6 \\ \Rightarrow P(A) = {6 \over {36}} = {1 \over 6} \\\end{aligned} \end{equation} $$
$B$ = Getting $5$ on green die
$$\begin{equation} \begin{aligned} B = \{ (5,1),(5,2),(5,3),(5,4),(5,5),(5,6)\} \\ \Rightarrow n(B) = 6 \\ \Rightarrow P(B) = {6 \over {36}} = {1 \over 6} \\\end{aligned} \end{equation} $$
$C$ = Getting $8$ as the sum
$$\begin{equation} \begin{aligned} C = \{ (2,6),(6,2),(5,3),(4,4),(3,5)\} \\ \Rightarrow n(C) = 5 \\ \Rightarrow P(C) = {5 \over {36}} \\\end{aligned} \end{equation} $$
$D$ = Getting a number less than $4$ in red die
$$\begin{equation} \begin{aligned} D = \{ (1,1),(2,1),(3,1),(4,1),(5,1),(6,1),(1,2),(2,2),(3,2),(4,2),(5,2),(6,2),(1,3),(2,3),(3,3),(4,3),(5,3),(6,3)\} \\ \Rightarrow n(D) = 18 \\ \Rightarrow P(D) = {{18} \over {36}} = {1 \over 2} \\\end{aligned} \end{equation} $$
$n(A \cap B) = 2$
$n(C \cap D) = 2$.
Thus,
$$P(A \cap B) = \;{2 \over {36}} = {1 \over {18}}$$
$$P(C \cap D) = \;{2 \over {36}} = {1 \over {18}}$$
i. Required probability
$$ = P(A/B) = {{1/18} \over {1/6}} = {1 \over 3}$$
ii. Required probability
$$ = P(C/D) = {{1/18} \over {1/2}} = {1 \over 9}$$
Question 17. A quiz master has $30$ easy general knowledge questions, $40$ difficult general knowledge questions, $20$ easy science questions and $60$ difficult science questions. Find the probability that a question picked at random from difficult category is a science question.
Solution: Let $A$ be selecting easy question.
Therefore,
$$P(A) = \;{{30 + 20} \over {30 + 20 + 40 + 60}} = {{50} \over {150}} = {1 \over 3}$$
Let $B$ be selecting difficult question.
Therefore,
$$P(B) = \;{{40 + 60} \over {30 + 20 + 40 + 60}} = {{100} \over {150}} = {2 \over 3}$$
Let $C$ be selecting general knowledge question.
Therefore,
$$P(C) = \;{{30 + 40} \over {30 + 20 + 40 + 60}} = {{70} \over {150}} = {7 \over {15}}$$
Let $D$ be selecting science question.
Therefore,
$$P(D) = \;{{60 + 20} \over {30 + 20 + 40 + 60}} = {{80} \over {150}} = {8 \over {15}}$$
$$P(D \cap B) = {{60} \over {30 + 20 + 40 + 60}} = {{60} \over {150}} = {2 \over 5}$$
Required probability,
$$P(D/B) = {{P(B \cap D)} \over {P(B)}} = {{2/5} \over {2/3}} = {3 \over 5}$$
