Examples (Revised) - Chapter 13 - Probability - Ncert Solutions class 12 - Maths
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NCERT Class 12 Maths Solutions Chapter 13: Probability
Example 1
If $\mathrm{P}(\mathrm{A})=\frac{7}{13}, \mathrm{P}(\mathrm{B})=\frac{9}{13}$ and $\mathrm{P}(\mathrm{A} \cap \mathrm{B})=\frac{4}{13}$, evaluate $\mathrm{P}(\mathrm{A} \mid \mathrm{B})$.
Solution
We have $P(A \mid B)=\frac{P(A \cap B)}{P(B)}=\frac{\frac{4}{13}}{\frac{9}{13}}=\frac{4}{9}$
Example 2
A family has two children. What is the probability that both the children are boys given that at least one of them is a boy ?
Solution
Let $b$ stand for boy and $g$ for girl. The sample space of the experiment is
$
\mathrm{S}=\{(b, b),(g, b),(b, g),(g, g)\}
$
Let $\mathrm{E}$ and $\mathrm{F}$ denote the following events :
$\mathrm{E}$ : 'both the children are boys'
$\mathrm{F}$ : 'at least one of the child is a boy'
Then
$
\mathrm{E}=\{(b, b)\} \text { and } \mathrm{F}=\{(b, b),(g, b),(b, g)\}
$
Now
$
\mathrm{E} \cap \mathrm{F}=\{(b, b)\}
$
Thus
$
\mathrm{P}(\mathrm{F})=\frac{3}{4} \text { and } \mathrm{P}(\mathrm{E} \cap \mathrm{F})=\frac{1}{4}
$
Therefore
$
\mathrm{P}(\mathrm{E} \mid \mathrm{F})=\frac{\mathrm{P}(\mathrm{E} \cap \mathrm{F})}{\mathrm{P}(\mathrm{F})}=\frac{\frac{1}{4}}{\frac{3}{4}}=\frac{1}{3}
$
Example 3
Ten cards numbered 1 to 10 are placed in a box, mixed up thoroughly and then one card is drawn randomly. If it is known that the number on the drawn card is more than 3 , what is the probability that it is an even number?
Solution
Let A be the event 'the number on the card drawn is even' and B be the event 'the number on the card drawn is greater than 3'. We have to find $\mathrm{P}(\mathrm{AlB})$.
Now, the sample space of the experiment is $S=\{1,2,3,4,5,6,7,8,9,10\}$
Then
$
A=\{2,4,6,8,10\}, B=\{4,5,6,7,8,9,10\}
$
and
$
A \cap B=\{4,6,8,10\}
$
Also
$
\mathrm{P}(\mathrm{A})=\frac{5}{10}, \mathrm{P}(\mathrm{B})=\frac{7}{10} \text { and } \mathrm{P}(\mathrm{A} \cap \mathrm{B})=\frac{4}{10}
$
Then
$
P(A \mid B)=\frac{P(A \cap B)}{P(B)}=\frac{\frac{4}{10}}{\frac{7}{10}}=\frac{4}{7}
$
Example 4
In a school, there are 1000 students, out of which 430 are girls. It is known that out of $430,10 \%$ of the girls study in class XII. What is the probability that a student chosen randomly studies in Class XII given that the chosen student is a girl?
Solution Let E denote the event that a student chosen randomly studies in Class XII and $\mathrm{F}$ be the event that the randomly chosen student is a girl. We have to find $\mathrm{P}(\mathrm{E} \mid \mathrm{F})$.
Now
$
\mathrm{P}(\mathrm{F})=\frac{430}{1000}=0.43 \text { and } \mathrm{P}(\mathrm{E} \cap \mathrm{F})=\frac{43}{1000}=0.043 \quad(\text { Why?) }
$
Then
$
\mathrm{P}(\mathrm{E} \mid \mathrm{F})=\frac{\mathrm{P}(\mathrm{E} \cap \mathrm{F})}{\mathrm{P}(\mathrm{F})}=\frac{0.043}{0.43}=0.1
$
Example 5
A die is thrown three times. Events A and B are defined as below:
A : 4 on the third throw
B : 6 on the first and 5 on the second throw
Find the probability of A given that B has already occurred.
Solution
The sample space has 216 outcomes.
Now
$
\begin{aligned}
& B=\{(6,5,1),(6,5,2),(6,5,3),(6,5,4),(6,5,5),(6,5,6)\} \\
&
\end{aligned}
$
and
$
A \cap B=\{(6,5,4)\} .
$
Now
$
P(B)=\frac{6}{216} \text { and } P(A \cap B)=\frac{1}{216}
$
Then $\quad P(A \mid B)=\frac{P(A \cap B)}{P(B)}=\frac{\frac{1}{216}}{\frac{6}{216}}=\frac{1}{6}$
Example 6
A die is thrown twice and the sum of the numbers appearing is observed to be 6 . What is the conditional probability that the number 4 has appeared at least once?
Solution
Let $\mathrm{E}$ be the event that 'number 4 appears at least once' and $\mathrm{F}$ be the event that 'the sum of the numbers appearing is 6 '.
Then,
$
E=\{(4,1),(4,2),(4,3),(4,4),(4,5),(4,6),(1,4),(2,4),(3,4),(5,4),(6,4)\}
$
and
$
\mathrm{F}=\{(1,5),(2,4),(3,3),(4,2),(5,1)\}
$
We have $\quad \mathrm{P}(\mathrm{E})=\frac{11}{36}$ and $\mathrm{P}(\mathrm{F})=\frac{5}{36}$
Also
$
\mathrm{E} \cap \mathrm{F}=\{(2,4),(4,2)\}
$
Therefore
$
\mathrm{P}(\mathrm{E} \cap \mathrm{F})=\frac{2}{36}
$
Hence, the required probability
$
\mathrm{P}(\mathrm{E} \mid \mathrm{F})=\frac{\mathrm{P}(\mathrm{E} \cap \mathrm{F})}{\mathrm{P}(\mathrm{F})}=\frac{\frac{2}{36}}{\frac{5}{36}}=\frac{2}{5}
$
For the conditional probability discussed above, we have considered the elementary events of the experiment to be equally likely and the corresponding definition of the probability of an event was used. However, the same definition can also be used in the general case where the elementary events of the sample space are not equally likely, the probabilities $\mathrm{P}(\mathrm{E} \cap \mathrm{F})$ and $\mathrm{P}(\mathrm{F})$ being calculated accordingly. Let us take up the following example.
Example 7
Consider the experiment of tossing a coin. If the coin shows head, toss it again but if it shows tail, then throw a die. Find the conditional probability of the event that 'the die shows a number greater than 4 ' given that 'there is at least one tail'.
Solution
The outcomes of the experiment can be represented in following diagrammatic manner called the 'tree diagram'.
The sample space of the experiment may be described as
.png)
$
\mathrm{S}=\{(\mathrm{H}, \mathrm{H}),(\mathrm{H}, \mathrm{T}),(\mathrm{T}, 1),(\mathrm{T}, 2),(\mathrm{T}, 3),(\mathrm{T}, 4),(\mathrm{T}, 5),(\mathrm{T}, 6)\}
$
where $(\mathrm{H}, \mathrm{H})$ denotes that both the tosses result into head and $(\mathrm{T}, i)$ denote the first toss result into a tail and the number $i$ appeared on the die for $i=1,2,3,4,5,6$. Thus, the probabilities assigned to the 8 elementary events
$
(\mathrm{H}, \mathrm{H}),(\mathrm{H}, \mathrm{T}),(\mathrm{T}, 1),(\mathrm{T}, 2),(\mathrm{T}, 3)(\mathrm{T}, 4),(\mathrm{T}, 5),(\mathrm{T}, 6)
$
are $\frac{1}{4}, \frac{1}{4}, \frac{1}{12}, \frac{1}{12}, \frac{1}{12}, \frac{1}{12}, \frac{1}{12}, \frac{1}{12}$ respectively which is clear from the Fig 13.2.
.png)
Let $\mathrm{F}$ be the event that 'there is at least one tail' and $\mathrm{E}$ be the event 'the die shows a number greater than 4 '. Then
$
\begin{aligned}
\mathrm{F}= & \{(\mathrm{H}, \mathrm{T}),(\mathrm{T}, 1),(\mathrm{T}, 2),(\mathrm{T}, 3),(\mathrm{T}, 4),(\mathrm{T}, 5),(\mathrm{T}, 6)\} \\
\mathrm{E}= & \{(\mathrm{T}, 5),(\mathrm{T}, 6)\} \text { and } \mathrm{E} \cap \mathrm{F}=\{(\mathrm{T}, 5),(\mathrm{T}, 6)\} \\
\mathrm{P}(\mathrm{F})= & \mathrm{P}(\{(\mathrm{H}, \mathrm{T})\})+\mathrm{P}(\{(\mathrm{T}, 1)\})+\mathrm{P}(\{(\mathrm{T}, 2)\})+\mathrm{P}(\{(\mathrm{T}, 3)\}) \\
& +\mathrm{P}(\{(\mathrm{T}, 4)\})+\mathrm{P}(\{(\mathrm{T}, 5)\})+\mathrm{P}(\{(\mathrm{T}, 6)\}) \\
\text { Now } \quad & \frac{1}{4}+\frac{1}{12}+\frac{1}{12}+\frac{1}{12}+\frac{1}{12}+\frac{1}{12}+\frac{1}{12}=\frac{3}{4} \\
\text { and } \quad \mathrm{P}(\mathrm{E} \cap \mathrm{F})= & \mathrm{P}(\{(\mathrm{T}, 5)\})+\mathrm{P}(\{(\mathrm{T}, 6)\})=\frac{1}{12}+\frac{1}{12}=\frac{1}{6} \\
\text { Hence } \quad \mathrm{P}(\mathrm{E} \mid \mathrm{F})= & \frac{\mathrm{P}(\mathrm{E} \cap \mathrm{F})}{\mathrm{P}(\mathrm{F})}=\frac{\frac{1}{6}}{\frac{3}{4}}=\frac{2}{9}
\end{aligned}
$
Example 8
An urn contains 10 black and 5 white balls. Two balls are drawn from the urn one after the other without replacement. What is the probability that both drawn balls are black?
Solution
Let $\mathrm{E}$ and $\mathrm{F}$ denote respectively the events that first and second ball drawn are black. We have to find $\mathrm{P}(\mathrm{E} \cap \mathrm{F})$ or $\mathrm{P}(\mathrm{EF})$.
Now
$
\mathrm{P}(\mathrm{E})=\mathrm{P}(\text { black ball in first draw })=\frac{10}{15}
$
Also given that the first ball drawn is black, i.e., event $\mathrm{E}$ has occurred, now there are 9 black balls and five white balls left in the urn. Therefore, the probability that the second ball drawn is black, given that the ball in the first draw is black, is nothing but the conditional probability of $\mathrm{F}$ given that $\mathrm{E}$ has occurred.
i.e.
$
\mathrm{P}(\mathrm{F} \mid \mathrm{E})=\frac{9}{14}
$
By multiplication rule of probability, we have
$
\begin{aligned}
\mathrm{P}(\mathrm{E} \cap \mathrm{F}) & =\mathrm{P}(\mathrm{E}) \mathrm{P}(\mathrm{F} \mid \mathrm{E}) \\
& =\frac{10}{15} \times \frac{9}{14}=\frac{3}{7}
\end{aligned}
$
Multiplication rule of probability for more than two events If $\mathrm{E}, \mathrm{F}$ and $\mathrm{G}$ are three events of sample space, we have
$
\mathrm{P}(\mathrm{E} \cap \mathrm{F} \cap \mathrm{G})=\mathrm{P}(\mathrm{E}) \mathrm{P}(\mathrm{F} \mid \mathrm{E}) \mathrm{P}(\mathrm{G} \mid(\mathrm{E} \cap \mathrm{F}))=\mathrm{P}(\mathrm{E}) \mathrm{P}(\mathrm{F} \mid \mathrm{E}) \mathrm{P}(\mathrm{G} \mid \mathrm{EF})
$
Similarly, the multiplication rule of probability can be extended for four or more events.
The following example illustrates the extension of multiplication rule of probability for three events.
Example 9
Three cards are drawn successively, without replacement from a pack of 52 well shuffled cards. What is the probability that first two cards are kings and the third card drawn is an ace?
Solution
Let $\mathrm{K}$ denote the event that the card drawn is king and $\mathrm{A}$ be the event that the card drawn is an ace. Clearly, we have to find $\mathrm{P}(\mathrm{KKA})$
Now
$
\mathrm{P}(\mathrm{K})=\frac{4}{52}
$
Also, $\mathrm{P}(\mathrm{K} \mid \mathrm{K})$ is the probability of second king with the condition that one king has already been drawn. Now there are three kings in $(52-1)=51$ cards.
Therefore
$
\mathrm{P}(\mathrm{K} \mid \mathrm{K})=\frac{3}{51}
$
Lastly, $\mathrm{P}(\mathrm{A} \mid \mathrm{KK})$ is the probability of third drawn card to be an ace, with the condition that two kings have already been drawn. Now there are four aces in left 50 cards.
Therefore
$
\mathrm{P}(\mathrm{A} \mid \mathrm{KK})=\frac{4}{50}
$
By multiplication law of probability, we have
$
\begin{aligned}
\mathrm{P}(\mathrm{KKA}) & =\mathrm{P}(\mathrm{K}) \quad \mathrm{P}(\mathrm{K} \mid \mathrm{K}) \quad \mathrm{P}(\mathrm{A} \mid \mathrm{KK}) \\
& =\frac{4}{52} \times \frac{3}{51} \times \frac{4}{50}=\frac{2}{5525}
\end{aligned}
$
Example 10
$\mathrm{~A}$ die is thrown. If $\mathrm{E}$ is the event 'the number appearing is a multiple of 3' and $F$ be the event 'the number appearing is even' then find whether $E$ and $F$ are independent?
Solution
We know that the sample space is $\mathrm{S}=\{1,2,3,4,5,6\}$
Now
$
\mathrm{E}=\{3,6\}, \mathrm{F}=\{2,4,6\} \text { and } \mathrm{E} \cap \mathrm{F}=\{6\}
$
Then
$
\mathrm{P}(\mathrm{E})=\frac{2}{6}=\frac{1}{3}, \mathrm{P}(\mathrm{F})=\frac{3}{6}=\frac{1}{2} \text { and } \mathrm{P}(\mathrm{E} \cap \mathrm{F})=\frac{1}{6}
$
Clearly $\quad \mathrm{P}(\mathrm{E} \cap \mathrm{F})=\mathrm{P}(\mathrm{E}) . \mathrm{P}(\mathrm{F})$
Hence $\quad E$ and $F$ are independent events.
Example 11
An unbiased die is thrown twice. Let the event A be 'odd number on the first throw' and B the event 'odd number on the second throw'. Check the independence of the events A and B.
Solution
If all the 36 elementary events of the experiment are considered to be equally likely, we have
$
\mathrm{P}(\mathrm{A})=\frac{18}{36}=\frac{1}{2} \text { and } \mathrm{P}(\mathrm{B})=\frac{18}{36}=\frac{1}{2}
$
Also
$
\begin{aligned}
\mathrm{P}(\mathrm{A} \cap \mathrm{B}) & =\mathrm{P} \text { (odd number on both throws) } \\
& =\frac{9}{36}=\frac{1}{4}
\end{aligned}
$
Now
$
\mathrm{P}(\mathrm{A}) \mathrm{P}(\mathrm{B})=\frac{1}{2} \times \frac{1}{2}=\frac{1}{4}
$
Clearly
$
\mathrm{P}(\mathrm{A} \cap \mathrm{B})=\mathrm{P}(\mathrm{A}) \times \mathrm{P}(\mathrm{B})
$
Thus,
$A$ and $B$ are independent events
Example 12
Three coins are tossed simultaneously. Consider the event $E$ 'three heads or three tails', F 'at least two heads' and G 'at most two heads'. Of the pairs (E,F), (E,G) and (F,G), which are independent? which are dependent?
Solution
The sample space of the experiment is given by
$\mathrm{S}=\{\mathrm{HHH}$, HHT, HTH, THH, HTT, THT, TTH, TTT $\}$
Clearly
$\mathrm{E}=\{\mathrm{HHH}$, TTT $\}, \mathrm{F}=\{\mathrm{HHH}, \mathrm{HHT}, \mathrm{HTH}, \mathrm{THH}\}$
and $\mathrm{G}=\{\mathrm{HHT}, \mathrm{HTH}, \mathrm{THH}, \mathrm{HTT}$, THT, TTH, TTT $\}$
Also $\mathrm{E} \cap \mathrm{F}=\{\mathrm{HHH}\}, \mathrm{E} \cap \mathrm{G}=\{\mathrm{TTT}\}, \mathrm{F} \cap \mathrm{G}=\{\mathrm{HHT}, \mathrm{HTH}, \mathrm{THH}\}$
Therefore $\mathrm{P}(\mathrm{E})=\frac{2}{8}=\frac{1}{4}, \mathrm{P}(\mathrm{F})=\frac{4}{8}=\frac{1}{2}, \mathrm{P}(\mathrm{G})=\frac{7}{8}$
and
$
\mathrm{P}(\mathrm{E} \cap \mathrm{F})=\frac{1}{8}, \mathrm{P}(\mathrm{E} \cap \mathrm{G})=\frac{1}{8}, \mathrm{P}(\mathrm{F} \cap \mathrm{G})=\frac{3}{8}
$
Also
$
\mathrm{P}(\mathrm{E}) . \mathrm{P}(\mathrm{F})=\frac{1}{4} \times \frac{1}{2}=\frac{1}{8}, \mathrm{P}(\mathrm{E}) \cdot \mathrm{P}(\mathrm{G})=\frac{1}{4} \times \frac{7}{8}=\frac{7}{32}
$
and
$
\mathrm{P}(\mathrm{F}) \cdot \mathrm{P}(\mathrm{G})=\frac{1}{2} \times \frac{7}{8}=\frac{7}{16}
$
Thus
$
\begin{aligned}
\mathrm{P}(\mathrm{E} \cap \mathrm{F}) & =\mathrm{P}(\mathrm{E}) \cdot \mathrm{P}(\mathrm{F}) \\
\mathrm{P}(\mathrm{E} \cap \mathrm{G}) & \neq \mathrm{P}(\mathrm{E}) \cdot \mathrm{P}(\mathrm{G}) \\
\mathrm{P}(\mathrm{F} \cap \mathrm{G}) & \neq \mathrm{P}(\mathrm{F}) \cdot \mathrm{P}(\mathrm{G})
\end{aligned}
$
and
Hence, the events (E and F) are independent, and the events (E and G) and (F and G) are dependent.
Example 13
Prove that if $\mathrm{E}$ and $\mathrm{F}$ are independent events, then so are the events $\mathrm{E}$ and $\mathrm{F}^{\prime}$.
Solution
Since $\mathrm{E}$ and $\mathrm{F}$ are independent, we have
$
\mathrm{P}(\mathrm{E} \cap \mathrm{F})=\mathrm{P}(\mathrm{E}) . \mathrm{P}(\mathrm{F})
$
From the venn diagram in Fig 13.3, it is clear
that $\mathrm{E} \cap \mathrm{F}$ and $\mathrm{E} \cap \mathrm{F}^{\prime}$ are mutually exclusive events and also $\mathrm{E}=(\mathrm{E} \cap \mathrm{F}) \cup\left(\mathrm{E} \cap \mathrm{F}^{\prime}\right)$.
.png)
Therefore
$
\begin{aligned}
\mathrm{P}(\mathrm{E})= & \mathrm{P}(\mathrm{E} \cap \mathrm{F})+\mathrm{P}\left(\mathrm{E} \cap \mathrm{F}^{\prime}\right) \\
\mathrm{P}\left(\mathrm{E} \cap \mathrm{F}^{\prime}\right)= & \mathrm{P}(\mathrm{E})-\mathrm{P}(\mathrm{E} \cap \mathrm{F}) \\
= & \mathrm{P}(\mathrm{E})-\mathrm{P}(\mathrm{E}) \cdot \mathrm{P}(\mathrm{F}) \\
& (\text { by }(1)) \\
= & \mathrm{P}(\mathrm{E})(1-\mathrm{P}(\mathrm{F})) \\
= & \mathrm{P}(\mathrm{E}) . \quad \mathrm{P}\left(\mathrm{F}^{\prime}\right)
\end{aligned}
$
or
(by (1))
Hence, $\mathrm{E}$ and $\mathrm{F}^{\prime}$ are independent
Example 14
If A and B are two independent events, then the probability of occurrence of at least one of $A$ and $B$ is given by $1-\mathrm{P}\left(\mathrm{A}^{\prime}\right) \mathrm{P}\left(\mathrm{B}^{\prime}\right)$
Solution
We have
$
\begin{aligned}
\mathrm{P}(\text { at least one of } \mathrm{A} \text { and } \mathrm{B}) & =\mathrm{P}(\mathrm{A} \cup \mathrm{B}) \\
& =\mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B})-\mathrm{P}(\mathrm{A} \cap \mathrm{B}) \\
& =\mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B})-\mathrm{P}(\mathrm{A}) \mathrm{P}(\mathrm{B}) \\
& =\mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B})[1-\mathrm{P}(\mathrm{A})] \\
& =\mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B}) . \mathrm{P}\left(\mathrm{A}^{\prime}\right) \\
& =1-\mathrm{P}\left(\mathrm{A}^{\prime}\right)+\mathrm{P}(\mathrm{B}) \mathrm{P}\left(\mathrm{A}^{\prime}\right) \\
& =1-\mathrm{P}\left(\mathrm{A}^{\prime}\right)[1-\mathrm{P}(\mathrm{B})] \\
& =1-\mathrm{P}\left(\mathrm{A}^{\prime}\right) \mathrm{P}\left(\mathrm{B}^{\prime}\right)
\end{aligned}
$
Example 15
A person has undertaken a construction job. The probabilities are 0.65 that there will be strike, 0.80 that the construction job will be completed on time if there is no strike, and 0.32 that the construction job will be completed on time if there is a strike. Determine the probability that the construction job will be completed on time.
Solution
Let A be the event that the construction job will be completed on time, and B be the event that there will be a strike. We have to find $\mathrm{P}(\mathrm{A})$.
We have
$
\begin{aligned}
\mathrm{P}(\mathrm{B}) & =0.65, \mathrm{P}(\text { no strike })=\mathrm{P}\left(\mathrm{B}^{\prime}\right)=1-\mathrm{P}(\mathrm{B})=1-0.65=0.35 \\
\mathrm{P}(\mathrm{A} \mid \mathrm{B}) & =0.32, \mathrm{P}\left(\mathrm{A} \mid \mathrm{B}^{\prime}\right)=0.80
\end{aligned}
$
Since events B and B' form a partition of the sample space S, therefore, by theorem on total probability, we have
$
\begin{aligned}
\mathrm{P}(\mathrm{A}) & =\mathrm{P}(\mathrm{B}) \mathrm{P}(\mathrm{A} \mid \mathrm{B})+\mathrm{P}\left(\mathrm{B}^{\prime}\right) \mathrm{P}\left(\mathrm{AlB}^{\prime}\right) \\
& =0.65 \times 0.32+0.35 \times 0.8 \\
& =0.208+0.28=0.488
\end{aligned}
$
Thus, the probability that the construction job will be completed in time is 0.488 . We shall now state and prove the Bayes' theorem.
Bayes' Theorem If $\mathrm{E}_1, \mathrm{E}_2, \ldots, \mathrm{E}_n$ are $n$ non empty events which constitute a partition of sample space $S$, i.e. $\mathrm{E}_1, \mathrm{E}_2, \ldots, \mathrm{E}_n$ are pairwise disjoint and $\mathrm{E}_1 \cup \mathrm{E}_2 \cup \ldots \cup \mathrm{E}_n=\mathrm{S}$ and $\mathrm{A}$ is any event of nonzero probability, then
Proof By formula of conditional probability, we know that
$
\begin{aligned}
\mathrm{P}\left(\mathrm{E}_i \mid \mathrm{A}\right) & =\frac{\mathrm{P}\left(\mathrm{A} \cap \mathrm{E}_i\right)}{\mathrm{P}(\mathrm{A})} \\
& =\frac{\mathrm{P}\left(\mathrm{E}_i\right) \mathrm{P}\left(\mathrm{AlE}_i\right)}{\mathrm{P}(\mathrm{A})} \text { (by multiplication rule of probability) } \\
& =\frac{\mathrm{P}\left(\mathrm{E}_i\right) \mathrm{P}\left(\mathrm{AlE}_i\right)}{\sum_{j=1}^n \mathrm{P}\left(\mathrm{E}_j\right) \mathrm{P}\left(\mathrm{AlE}_j\right)} \text { (by the result of theorem of total probability) }
\end{aligned}
$
Remark The following terminology is generally used when Bayes' theorem is applied.
The events $\mathrm{E}_1, \mathrm{E}_2, \ldots, \mathrm{E}_n$ are called hypotheses.
The probability $\mathrm{P}\left(\mathrm{E}_i\right)$ is called the priori probability of the hypothesis $\mathrm{E}_i$
The conditional probability $\mathrm{P}\left(\mathrm{E}_i \mid \mathrm{A}\right)$ is called a posteriori probability of the hypothesis $\mathrm{E}_i$.
Bayes' theorem is also called the formula for the probability of "causes". Since the $\mathrm{E}_i$ 's are a partition of the sample space $\mathrm{S}$, one and only one of the events $\mathrm{E}_i$ occurs (i.e. one of the events $\mathrm{E}_i$ must occur and only one can occur). Hence, the above formula gives us the probability of a particular $\mathrm{E}_i$ (i.e. a "Cause"), given that the event $\mathrm{A}$ has occurred.
The Bayes' theorem has its applications in variety of situations, few of which are illustrated in following examples.
Example 16
Bag I contains 3 red and 4 black balls while another Bag II contains 5 red and 6 black balls. One ball is drawn at random from one of the bags and it is found to be red. Find the probability that it was drawn from Bag II.
Solution
Let $\mathrm{E}_1$ be the event of choosing the bag I, $\mathrm{E}_2$ the event of choosing the bag II and $\mathrm{A}$ be the event of drawing a red ball.
Then
$
\mathrm{P}\left(\mathrm{E}_1\right)=\mathrm{P}\left(\mathrm{E}_2\right)=\frac{1}{2}
$
Also
$
\mathrm{P}\left(\mathrm{AIE}_1\right)=\mathrm{P}(\text { drawing a red ball from Bag } \mathrm{I})=\frac{3}{7}
$
and
$
\mathrm{P}\left(\mathrm{AlE}_2\right)=\mathrm{P}(\text { drawing a red ball from Bag II })=\frac{5}{11}
$
Now, the probability of drawing a ball from Bag II, being given that it is red, is $\mathrm{P}\left(\mathrm{E}_2 / \mathrm{A}\right)$
By using Bayes' theorem, we have
$
\mathrm{P}\left(\mathrm{E}_2 \mid \mathrm{A}\right)=\frac{\mathrm{P}\left(\mathrm{E}_2\right) \mathrm{P}\left(\mathrm{AlE}_2\right)}{\mathrm{P}\left(\mathrm{E}_1\right) \mathrm{P}\left(\mathrm{AlE}_1\right)+\mathrm{P}\left(\mathrm{E}_2\right) \mathrm{P}\left(\mathrm{AlE}_2\right)}=\frac{\frac{1}{2} \times \frac{5}{11}}{\frac{1}{2} \times \frac{3}{7}+\frac{1}{2} \times \frac{5}{11}}=\frac{35}{68}
$
Example 17
Given three identical boxes I, II and III, each containing two coins. In box I, both coins are gold coins, in box II, both are silver coins and in the box III, there is one gold and one silver coin. A person chooses a box at random and takes out a coin. If the coin is of gold, what is the probability that the other coin in the box is also of gold?
Solution
Let $\mathrm{E}_1, \mathrm{E}_2$ and $\mathrm{E}_3$ be the events that boxes I, II and III are chosen, respectively.
Then
$
\mathrm{P}\left(\mathrm{E}_1\right)=\mathrm{P}\left(\mathrm{E}_2\right)=\mathrm{P}\left(\mathrm{E}_3\right)=\frac{1}{3}
$
Also, let A be the event that 'the coin drawn is of gold'
Then
$
\begin{aligned}
& \mathrm{P}\left(\mathrm{AIE}_1\right)=\mathrm{P}(\text { a gold coin from bag } \mathrm{I})=\frac{2}{2}=1 \\
& \mathrm{P}\left(\mathrm{AIE}_2\right)=\mathrm{P}(\text { a gold coin from bag II })=0 \\
& \mathrm{P}\left(\mathrm{AIE}_3\right)=\mathrm{P}(\text { a gold coin from bag III })=\frac{1}{2}
\end{aligned}
$
Now, the probability that the other coin in the box is of gold
$
\begin{aligned}
& =\text { the probability that gold coin is drawn from the box I. } \\
& =\mathrm{P}\left(\mathrm{E}_1 \mid \mathrm{A}\right)
\end{aligned}
$
By Bayes' theorem, we know that
$
\begin{aligned}
\mathrm{P}\left(\mathrm{E}_1 \mid \mathrm{A}\right) & =\frac{\mathrm{P}\left(\mathrm{E}_1\right) \mathrm{P}\left(\mathrm{AlE}_1\right)}{\mathrm{P}\left(\mathrm{E}_1\right) \mathrm{P}\left(\mathrm{AlE}_1\right)+\mathrm{P}\left(\mathrm{E}_2\right) \mathrm{P}\left(\mathrm{AlE}_2\right)+\mathrm{P}\left(\mathrm{E}_3\right) \mathrm{P}\left(\mathrm{AlE}_3\right)} \\
& =\frac{\frac{1}{3} \times 1}{\frac{1}{3} \times 1+\frac{1}{3} \times 0+\frac{1}{3} \times \frac{1}{2}}=\frac{2}{3}
\end{aligned}
$
Example 18
Suppose that the reliability of a HIV test is specified as follows: Of people having HIV, $90 \%$ of the test detect the disease but $10 \%$ go undetected. Of people free of HIV, $99 \%$ of the test are judged HIV-ive but $1 \%$ are diagnosed as showing HIV+ive. From a large population of which only $0.1 \%$ have HIV, one person is selected at random, given the HIV test, and the pathologist reports him/her as HIV+ive. What is the probability that the person actually has HIV?
Solution
Let E denote the event that the person selected is actually having HIV and A the event that the person's HIV test is diagnosed as +ive. We need to find P(EIA).
Also $\mathrm{E}^{\prime}$ denotes the event that the person selected is actually not having HIV.
Clearly, $\left\{E, E^{\prime}\right\}$ is a partition of the sample space of all people in the population. We are given that
$
\mathrm{P}(\mathrm{E})=0.1 \%=\frac{0.1}{100}=0.001
$
$
\begin{aligned}
\mathrm{P}\left(\mathrm{E}^{\prime}\right)= & 1-\mathrm{P}(\mathrm{E})=0.999 \\
\mathrm{P}(\mathrm{AIE})= & \mathrm{P}(\text { Person tested as HIV+ive given that he/she } \\
& \text { is actually having HIV }) \\
= & 90 \%=\frac{90}{100}=0.9
\end{aligned}
$
and
$\mathrm{P}\left(\mathrm{AIE}^{\prime}\right)=\mathrm{P}($ Person tested as HIV +ive given that he/she is actually not having HIV)
$
=1 \%=\frac{1}{100}=0.01
$
Now, by Bayes' theorem
$
\begin{aligned}
\mathrm{P}(\mathrm{E} \mid \mathrm{A}) & =\frac{\mathrm{P}(\mathrm{E}) \mathrm{P}(\mathrm{A} \mid \mathrm{E})}{\mathrm{P}(\mathrm{E}) \mathrm{P}(\mathrm{A} \mid \mathrm{E})+\mathrm{P}\left(\mathrm{E}^{\prime}\right) \mathrm{P}\left(\mathrm{A}^{\prime} E^{\prime}\right)} \\
& =\frac{0.001 \times 0.9}{0.001 \times 0.9+0.999 \times 0.01}=\frac{90}{1089} \\
& =0.083 \text { approx. }
\end{aligned}
$
Thus, the probability that a person selected at random is actually having HIV given that he/she is tested HIV+ive is 0.083 .
Example 19
In a factory which manufactures bolts, machines A, B and C manufacture respectively $25 \%, 35 \%$ and $40 \%$ of the bolts. Of their outputs, 5,4 and 2 percent are respectively defective bolts. A bolt is drawn at random from the product and is found to be defective. What is the probability that it is manufactured by the machine B?
Solution
Let events $B_1, B_2, B_3$ be the following :
$\mathrm{B}_1$ : the bolt is manufactured by machine $A$
$\mathrm{B}_2$ : the bolt is manufactured by machine $\mathrm{B}$
$\mathrm{B}_3$ : the bolt is manufactured by machine $\mathrm{C}$
Clearly, $\mathrm{B}_1, \mathrm{~B}_2, \mathrm{~B}_3$ are mutually exclusive and exhaustive events and hence, they represent a partition of the sample space.
Let the event $\mathrm{E}$ be 'the bolt is defective'.
The event $E$ occurs with $B_1$ or with $B_2$ or with $B_3$. Given that,
$
\mathrm{P}\left(\mathrm{B}_1\right)=25 \%=0.25, \mathrm{P}\left(\mathrm{B}_2\right)=0.35 \text { and } \mathrm{P}\left(\mathrm{B}_3\right)=0.40
$
factured by machine $\mathrm{A}=5 \%=0.05$
$\text { Similarly, } \quad \mathrm{P}\left(\mathrm{ElB}_2\right)=0.04, \mathrm{P}\left(\mathrm{ElB}_3\right)=0.02 \text {. }$
Hence, by Bayes' Theorem, we have
$
\begin{aligned}
\mathrm{P}\left(\mathrm{B}_2 \mid \mathrm{E}\right) & =\frac{\mathrm{P}\left(\mathrm{B}_2\right) \mathrm{P}\left(\mathrm{ElB}_2\right)}{\mathrm{P}\left(\mathrm{B}_1\right) \mathrm{P}\left({\mathrm{E} \mid B_1}_1\right)+\mathrm{P}\left(\mathrm{B}_2\right) \mathrm{P}\left(\mathrm{ElB}_2\right)+\mathrm{P}\left(\mathrm{B}_3\right) \mathrm{P}\left(\mathrm{ElB}_3\right)} \\
& =\frac{0.35 \times 0.04}{0.25 \times 0.05+0.35 \times 0.04+0.40 \times 0.02} \\
& =\frac{0.0140}{0.0345}=\frac{28}{69}
\end{aligned}
$
Example 20
A doctor is to visit a patient. From the past experience, it is known that the probabilities that he will come by train, bus, scooter or by other means of transport are respectively $\frac{3}{10}, \frac{1}{5}, \frac{1}{10}$ and $\frac{2}{5}$. The probabilities that he will be late are $\frac{1}{4}, \frac{1}{3}$, and $\frac{1}{12}$, if he comes by train, bus and scooter respectively, but if he comes by other means of transport, then he will not be late. When he arrives, he is late. What is the probability that he comes by train?
Solution
Let $\mathrm{E}$ be the event that the doctor visits the patient late and let $\mathrm{T}_1, \mathrm{~T}_2, \mathrm{~T}_3, \mathrm{~T}_4$ be the events that the doctor comes by train, bus, scooter, and other means of transport respectively.
Then
$
\begin{aligned}
\mathrm{P}\left(\mathrm{T}_1\right) & =\frac{3}{10}, \mathrm{P}\left(\mathrm{T}_2\right)=\frac{1}{5}, \mathrm{P}\left(\mathrm{T}_3\right)=\frac{1}{10} \text { and } \mathrm{P}\left(\mathrm{T}_4\right)=\frac{2}{5} \quad \text { (given) } \\
\mathrm{P}\left(\mathrm{E}_1\right) & =\text { Probability that the doctor arriving late comes by train }=\frac{1}{4}
\end{aligned}
$
(given)
Similarly, $\mathrm{P}\left(\mathrm{EIT}_2\right)=\frac{1}{3}, \mathrm{P}\left(\mathrm{EIT}_3\right)=\frac{1}{12}$ and $\mathrm{P}\left(\mathrm{EIT}_4\right)=0$, since he is not late if he comes by other means of transport.
Therefore, by Bayes' Theorem, we have
$
\mathrm{P}\left(\mathrm{T}_1 \mid \mathrm{E}\right)=\text { Probability that the doctor arriving late comes by train }
$
$
\begin{aligned}
& =\frac{\mathrm{P}\left(\mathrm{T}_1\right) \mathrm{P}\left(\mathrm{ElT}_1\right)}{\mathrm{P}\left(\mathrm{T}_1\right) \mathrm{P}\left(\mathrm{EIT}_1\right)+\mathrm{P}\left(\mathrm{T}_2\right) \mathrm{P}\left(\mathrm{ElT}_2\right)+\mathrm{P}\left(\mathrm{T}_3\right) \mathrm{P}\left(\mathrm{ElT}_3\right)+\mathrm{P}\left(\mathrm{T}_4\right) \mathrm{P}\left(\mathrm{ElT}_4\right)} \\
& =\frac{\frac{3}{10} \times \frac{1}{4}}{\frac{3}{10} \times \frac{1}{4}+\frac{1}{5} \times \frac{1}{3}+\frac{1}{10} \times \frac{1}{12}+\frac{2}{5} \times 0}=\frac{3}{40} \times \frac{120}{18}=\frac{1}{2}
\end{aligned}
$
Hence, the required probability is $\frac{1}{2}$.
Example 21
A man is known to speak truth 3 out of 4 times. He throws a die and reports that it is a six. Find the probability that it is actually a six.
Solution
Let $\mathrm{E}$ be the event that the man reports that six occurs in the throwing of the die and let $S_1$ be the event that six occurs and $S_2$ be the event that six does not occur.
Then
$
\begin{aligned}
\mathrm{P}\left(\mathrm{S}_1\right)= & \text { Probability that six occurs }=\frac{1}{6} \\
\mathrm{P}\left(\mathrm{S}_2\right)= & \text { Probability that six does not occur }=\frac{5}{6} \\
\mathrm{P}\left(\mathrm{EIS}_1\right)= & \text { Probability that the man reports that six occurs when six has } \\
& \text { actually occurred on the die } \\
= & \text { Probability that the man speaks the truth }=\frac{3}{4} \\
\mathrm{P}\left(\mathrm{EIS}_2\right)= & \text { Probability that the man reports that six occurs when } \text { six } \text { has } \\
& \text { not actually occurred on the die } \\
= & \text { Probability that the man does not speak the truth }=1-\frac{3}{4}=\frac{1}{4}
\end{aligned}
$
Thus, by Bayes' theorem, we get
$\mathrm{P}(\mathrm{S}, \mathrm{E})=$ Probability that the report of the man that six has occurred is actually a six
$
\begin{aligned}
& =\frac{\mathrm{P}\left(\mathrm{S}_1\right) \mathrm{P}\left(\mathrm{EIS}_1\right)}{\mathrm{P}\left(\mathrm{S}_1\right) \mathrm{P}\left(\mathrm{EIS}_1\right)+\mathrm{P}\left(\mathrm{S}_2\right) \mathrm{P}\left(\mathrm{EIS}_2\right)} \\
& =\frac{\frac{1}{6} \times \frac{3}{4}}{\frac{1}{6} \times \frac{3}{4}+\frac{5}{6} \times \frac{1}{4}}=\frac{1}{8} \times \frac{24}{8}=\frac{3}{8}
\end{aligned}
$
Hence, the required probability is $\frac{3}{8}$.
Example 22
Coloured balls are distributed in four boxes as shown in the following table:
.png)
A box is selected at random and then a ball is randomly drawn from the selected box. The colour of the ball is black, what is the probability that ball drawn is from the box III?
Solution
Let $\mathrm{A}, \mathrm{E}_1, \mathrm{E}_2, \mathrm{E}_3$ and $\mathrm{E}_4$ be the events as defined below :
A : a black ball is selected $\quad \mathrm{E}_1:$ box $\mathrm{I}$ is selected
$\mathrm{E}_2:$ box II is selected $\quad \mathrm{E}_3:$ box III is selected
$\mathrm{E}_4$ : box IV is selected
Since the boxes are chosen at random,
Therefore
$
\mathrm{P}\left(\mathrm{E}_1\right)=\mathrm{P}\left(\mathrm{E}_2\right)=\mathrm{P}\left(\mathrm{E}_3\right)=\mathrm{P}\left(\mathrm{E}_4\right)=\frac{1}{4}
$
Also
$\mathrm{P}($ box III is selected, given that the drawn ball is black $)=\mathrm{P}\left(\mathrm{E}_3 \mathrm{I}\right)$. By Bayes' theorem,
$\begin{aligned}
\mathrm{P}\left(\mathrm{E}_3 \mathrm{I}\right) & =\frac{\mathrm{P}\left(\mathrm{E}_3\right) \cdot \mathrm{P}\left(\mathrm{AlE}_3\right)}{\mathrm{P}\left(\mathrm{E}_1\right) \mathrm{P}\left(\mathrm{AlE}_1\right)+\mathrm{P}\left(\mathrm{E}_2\right) \mathrm{P}\left(\mathrm{AlE}_2\right)+\mathrm{P}\left(\mathrm{E}_3\right) \mathrm{P}\left(\mathrm{AlE}_3\right)+\mathrm{P}\left(\mathrm{E}_4\right) \mathrm{P}\left(\mathrm{AlE}_4\right)} \\
& =\frac{\frac{1}{4} \times \frac{1}{7}}{\frac{1}{4} \times \frac{3}{18}+\frac{1}{4} \times \frac{1}{4}+\frac{1}{4} \times \frac{1}{7}+\frac{1}{4} \times \frac{4}{13}}=0.165
\end{aligned}$
Example 23
A and B throw a die alternatively till one of them gets a ' 6 ' and wins the game. Find their respective probabilities of winning, if A starts first.
Solution
Let $\mathrm{S}$ denote the success (getting a ' 6 ') and $\mathrm{F}$ denote the failure (not getting a ' 6 ').
Thus,
$
\begin{aligned}
\mathrm{P}(\mathrm{S}) & =\frac{1}{6}, \mathrm{P}(\mathrm{F})=\frac{5}{6} \\
\mathrm{P}(\text { A wins in the first throw }) & =\mathrm{P}(\mathrm{S})=\frac{1}{6}
\end{aligned}
$
A gets the third throw, when the first throw by A and second throw by B result into failures.
Therefore, $\quad \mathrm{P}(\mathrm{A}$ wins in the 3rd throw $)=\mathrm{P}(\mathrm{FFS})=\mathrm{P}(\mathrm{F}) \mathrm{P}(\mathrm{F}) \mathrm{P}(\mathrm{S})=\frac{5}{6} \times \frac{5}{6} \times \frac{1}{6}$
$
=\left(\frac{5}{6}\right)^2 \times \frac{1}{6}
$
$
\mathrm{P}(\mathrm{A} \text { wins in the 5th throw })=\mathrm{P}(\text { FFFFS })=\left(\frac{5}{6}\right)^4\left(\frac{1}{6}\right) \text { and so on. }
$
Hence,
$
\begin{aligned}
\mathrm{P}(\text { A wins }) & =\frac{1}{6}+\left(\frac{5}{6}\right)^2\left(\frac{1}{6}\right)+\left(\frac{5}{6}\right)^4\left(\frac{1}{6}\right)+\ldots \\
& =\frac{\frac{1}{6}}{1-\frac{25}{36}}=\frac{6}{11} \\
\mathrm{P}(\mathrm{B} \text { wins }) & =1-\mathrm{P}(\text { A wins })=1-\frac{6}{11}=\frac{5}{11}
\end{aligned}
$
Remark If $a+a r+a r^2+\ldots+a r^{n-1}+\ldots$, where $|r|<1$, then sum of this infinite G.P. is given by $\frac{a}{1-r}$. (Refer A.1.3 of Class XI Text book).
Example 24
If a machine is correctly set up, it produces $90 \%$ acceptable items. If it is incorrectly set up, it produces only $40 \%$ acceptable items. Past experience shows that $80 \%$ of the set ups are correctly done. If after a certain set up, the machine produces 2 acceptable items, find the probability that the machine is correctly setup.
Solution
Let A be the event that the machine produces 2 acceptable items.
Also let $B_1$ represent the event of correct set up and $B_2$ represent the event of incorrect setup.
Now
$
\begin{aligned}
\mathrm{P}\left(\mathrm{B}_1\right) & =0.8, \mathrm{P}\left(\mathrm{B}_2\right)=0.2 \\
\mathrm{P}\left(\mathrm{A}_1 \mid \mathrm{B}_1\right) & =0.9 \times 0.9 \text { and } \mathrm{P}\left(\mathrm{AlB}_2\right)=0.4 \times 0.4
\end{aligned}
$
Therefore
$
\begin{aligned}
\mathrm{P}\left(\mathrm{B}_1 \mid \mathrm{A}\right) & =\frac{\mathrm{P}\left(\mathrm{B}_1\right) \mathrm{P}\left(\mathrm{AlB}_1\right)}{\mathrm{P}\left(\mathrm{B}_1\right) \mathrm{P}\left(\mathrm{AlB}_1\right)+\mathrm{P}\left(\mathrm{B}_2\right) \mathrm{P}\left(\mathrm{AlB}_2\right)} \\
& =\frac{0.8 \times 0.9 \times 0.9}{0.8 \times 0.9 \times 0.9+0.2 \times 0.4 \times 0.4}=\frac{648}{680}=0.95
\end{aligned}
$
