You need two essential factors to plan your road trip since the driving conditions must have a working vehicle plus safe weather conditions. The likelihood of your planned journey becomes minimal once you know your car works properly, since rainfall conditions could prevent it. The NCERT solutions of Miscellaneous Exercise in Class 12 Maths Chapter 13 explores events which occur simultaneously or sequentially, thus making this a core decision-making technique.
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This NCERT exercise contains problems about basic probability and conditional probability, and includes independent and dependent events alongside the multiplication and addition theorems. This exercise combines all concepts to check comprehension while aiding students during board examinations. Here are the detailed solutions.
Question 1(i): A and B are two events such that$P(A)\neq 0.$ Find $P(B\mid A),$ if
$A$ is a subset of $B$
Answer:
A and B are two events such that$P(A)\neq 0.$
$A\subset B$
$\Rightarrow \, \, \, \, A\cap B=A$
$P(A\cap B)=P(B\cap A)=P(A)$
$P(B|A)=\frac{P(B\cap A)}{P(A)}$
$P(B|A)=\frac{P( A)}{P(A)}$
$P(B|A)=1$
Question 1(ii): $A$ and $B$ are two events such that $P(A)\neq 0.$ Find $P(B\mid A),$ if
Answer:
A and B are two events such that$P(A)\neq 0.$
$P(A\cap B)=P(B\cap A)=0$
$P(B|A)=\frac{P(B\cap A)}{P(A)}$
$P(B|A)=\frac{0}{P(A)}$
$P(B|A)=0$
Question 2 (i): A couple has two children,
Find the probability that both children are males, if it is known that at least one of the children is male.
Answer:
A couple has two children,
sample space $=\left \{ (b,b),(g,g),(b,g),(g,b) \right \}$
Let A be both children are males and B is at least one of the children is male.
$(A\cap B)=\left \{ (b,b) \right \}$
$P(A\cap B)=\frac{1}{4}$
$P(A)=\frac{1}{4}$
$P(B)=\frac{3}{4}$
$P(A|B)=\frac{P(A\cap B)}{P(B)}$
$P(A|B)=\frac{\frac{1}{4}}{\frac{3}{4}}=\frac{1}{3}$
Question 2 (ii): A couple has two children,
Find the probability that both children are females, if it is known that the elder child is a female.
Answer:
A couple has two children,
sample space $=\left \{ (b,b),(g,g),(b,g),(g,b) \right \}$
Let A be both children are females and B be the elder child is a female.
$(A\cap B)=\left \{ (g,g) \right \}$
$P(A\cap B)=\frac{1}{4}$
$P(A)=\frac{1}{4}$
$P(B)=\frac{2}{4}$
$P(A|B)=\frac{P(A\cap B)}{P(B)}$
$P(A|B)=\frac{\frac{1}{4}}{\frac{2}{4}}=\frac{1}{2}$
Answer:
We have $5^{o}/_{o}$ of men and $0.25^{o}/_{o}$ of women have grey hair.
Percentage of people with grey hairs $=(5+0.25)\%=5.25\%$
The probability that the selected haired person is male :
$=\frac{5}{5.25}=\frac{20}{21}$
Answer:
$90^{\circ} \%$ of people are right-handed.
$
\begin{aligned}
& P(\text { right }- \text { handed })=\frac{9}{10} \\
& P(\text { left }- \text { handed })=q=1-\frac{9}{10}=\frac{1}{10}
\end{aligned}
$
at most 6 of a random sample of 10 people are right-handed.
the probability that more than 6 of a random sample of 10 people are right-handed is given by,
$
\begin{aligned}
& { }^{10} C_r P^r q^{10-r} \\
& =\sum_T^{1010} C_r \frac{9}{10}^r \cdot\left(\frac{1}{10}\right)^{10-r}
\end{aligned}
$
the probability that at most 6 of a random sample of 10 people are right-handed is given by
$
=1-\sum_T^{1010} C_r \cdot \frac{9}{10}^r \cdot\left(\frac{1}{10}\right)^{10-r}
$
Answer:
In a leap year, there are 366 days.
In 52 weeks, there are 52 Tuesdays.
The probability that a leap year will have 53 Tuesday is equal to the probability that the remaining 2 days are Tuesday.
The remaining 2 days can be :
1. Monday and Tuesday
2. Tuesday and Wednesday
3. Wednesday and Thursday
4. Thursday and Friday
5.friday and Saturday
6.saturday and Sunday
7.sunday and Monday
Total cases = 7.
Favorable cases = 2
Probability of having 53 Tuesday in a leap year = P.
$P=\frac{2}{7}$
Question 6 (i): Suppose we have four boxes A,B,C and D containing coloured marbles as given below:
One of the boxes has been selected at random and a single marble is drawn from it. If the marble is red, what is the probability that it was drawn from box A ?
Answer:
'
Let R be the event of drawing red marble.
Let $E_A,E_B,E_C$ respectively denote the event of selecting box A, B, C.
Total marbles = 40
Red marbles =15
$P(R)=\frac{15}{40}=\frac{3}{8}$
Probability of drawing red marble from box A is $P(E_A|R)$
$P(E_A|R)=\frac{P(E_A\cap R)}{P(R)}$
$=\frac{\frac{1}{40}}{\frac{3}{8}}$
$=\frac{1}{15}$
Question 6 (ii): Suppose we have four boxes A,B,C and D containing coloured marbles as given below:
One of the boxes has been selected at random and a single marble is drawn from it. If the marble is red, what is the probability that it was drawn from box B?
Answer:
Let R be event of drawing red marble.
Let $E_A,E_B,E_C$ respectivly denote event of selecting box A,B,C.
Total marbles = 40
Red marbles =15
$P(R)=\frac{15}{40}=\frac{3}{8}$
Probability of drawing red marble from box B is $P(E_B|R)$
$P(E_B|R)=\frac{P(E_B\cap R)}{P(R)}$
$=\frac{\frac{6}{40}}{\frac{3}{8}}$
$=\frac{2}{5}$
Question 6 (iii): Suppose we have four boxes A,B,C and D containing coloured marbles as given below:
One of the boxes has been selected at random and a single marble is drawn from it. If the marble is red, what is the probability that it was drawn from box C?
Answer:
Let R be event of drawing red marble.
Let $E_A,E_B,E_C$ respectivly denote event of selecting box A,B,C.
Total marbles = 40
Red marbles =15
$P(R)=\frac{15}{40}=\frac{3}{8}$
Probability of drawing red marble from box C is $P(E_C|R)$
$P(E_C|R)=\frac{P(E_C\cap R)}{P(R)}$
$=\frac{\frac{8}{40}}{\frac{3}{8}}$
$=\frac{8}{15}$
Answer:
Let A,E1, E2 respectively denote the event that a person has a heart break, selected person followed the course of yoga and meditation , and the person adopted
the drug prescription.
$\therefore \, \, \, \, P(A)=0.40$
$\therefore \, \, \, \, P(E1)=P(E2)=\frac{1}{2}$
$P(A|E1)=0.40\times 0.70=0.28$
$P(A|E2)=0.40\times 0.75=0.30$
the probability that the patient followed a course of meditation and yoga is $P(E1,A)$
$P(E1,A)=\frac{P(E1).P(E1|A)}{P(E1).P(E1|A)+P(E2).P(E2|A)}$
$P(E1,A)=\frac{0.5\times 0.28}{0.5\times 0.28 + 0.5\times 0.30}$
$=\frac{14}{29}$
Answer:
Total number of determinant of second order with each element being 0 or 1 is $2^4=16$
The values of determinant is positive in the following cases $\begin{bmatrix} 1 & 0\\ 0 & 1\end{bmatrix}, \begin{bmatrix} 1 & 1\\ 0 & 1\end{bmatrix}, \begin{bmatrix} 1 & 0\\ 1 & 1\end{bmatrix}$
Probability is
$=\frac{3}{16}$
P(A fails) =$0.2$
P(B fails alone) = $0.15$
P(A and B fail) = $0.15$
Evaluate the following probabilities
$P(A \; fails\mid B\; has\; failed)$
Answer:
Let event in which A fails and B fails be $E_A,E_B$
$P(E_A)=0.2$
$P(E_A\, and \, E_B)=0.15$
$P(B\, fails\, alone)=P(E_B)-P(E_A\, and\, E_B)$
$\Rightarrow \, \, \, 0.15=P(E_B)-0.15$
$\Rightarrow \, \, \, P(E_B)=0.3$
$P(E_A|E_B)=\frac{P(E_A\cap E_B)}{P(E_B)}$
$=\frac{0.15}{0.3}=0.5$
Question 9 (ii): An electronic assembly consists of two subsystems, say, A and B. From previous testing procedures, the following probabilities are assumed to be known:
P(A fails) = $0.2$
P(B fails alone) = $0.15$
P(A and B fail) = $0.15$
Evaluate the following probabilities
Answer:
Let event in which A fails and B fails be $E_A,E_B$
$P(E_A)=0.2$
$P(E_A\, and \, E_B)=0.15$
$P(B\, fails\, alone)=P(E_B)-P(E_A\, and\, E_B)$
$\Rightarrow \, \, \, 0.15=P(E_B)-0.15$
$\Rightarrow \, \, \, P(E_B)=0.3$
$P(A\, fails\, \, alone)=P(E_A)-P(E_A\, and\, E_B)$
$=0.2-0.15=0.05$
Answer:
Let E1 and E2 respectively denote the event that red ball is transfered from bag 1 to bag 2 and a black ball is transfered from bag 1 to bag2.
$P(E1)=\frac{3}{7}$ and $P(E2)=\frac{4}{7}$
Let A be the event that ball drawn is red.
When a red ball is transfered from bag 1 to bag 2.
$P(A|E1)=\frac{5}{10}=\frac{1}{2}$
When a black ball is transfered from bag 1 to bag 2.
$P(A|E2)=\frac{4}{10}=\frac{2}{5}$
$P(E2|A)=\frac{P(E2).P(A|E2)}{P(E2).P(A|E2)+P(E1).P(A|E1)}$
$=\frac{\frac{4}{7}\times \frac{2}{5}}{\frac{4}{7}\times \frac{2}{5}+\frac{3}{7}\times \frac{1}{2}}$
$=\frac{16}{31}$
Question 11: If A and B are two events such that $P(A\neq 0)$ and$P(B\mid A)=1,$ then
Choose the correct answer of the following:
(A) $A\subset B$
(B) $B\subset A$
(C) $B=\phi$
(D) $A=\phi$
Answer:
A and B are two events such that $P(A\neq 0)$ and$P(B\mid A)=1,$
$P(B|A)=\frac{P(B\cap A)}{P(A)}$
$1=\frac{P(B\cap A)}{P(A)}$
$P(B\cap A)=P(A)$
$\Rightarrow \, \, \, A\subset B$
Option A is correct.
Question 12: If $P(A\mid B)> P(A)$, then which of the following is correct :
(A) $P(B\mid A)< P(B)$
(B) $P(A\cap B)< P(A).P(B)$
(C) $P(B\mid A)> P(B)$
(D)$P(B\mid A)= P(B)$
Answer:
$P(A\mid B)> P(A)$
$\Rightarrow \, \, \frac{P(A\cap B)}{P(B)}> P(A)$
$\Rightarrow \, \, P(A\cap B)> P(A).P(B)$
$\Rightarrow \, \, \frac{P(A\cap B)}{P(A)}> P(B)$
$\Rightarrow \, \, P(B|A)> P(B)$
Option C is correct.
Question 13 If A and B are any two events such that $P(A)+P(B)-P(A \; and\; B)=P(A),$ then
(A) $P(B\mid A)=1$
(B) $P(A\mid B)=1$
(C) $P(B\mid A)=0$
(D) $P(A\mid B)=0$
Answer:
$P(A)+P(B)-P(A \; and\; B)=P(A),$
$P(A)+P(B)-P(A\cap B)=P(A)$
$\Rightarrow \, \, \, P(B)-P(A\cap B)=0$
$\Rightarrow \, \, \, P(B)=P(A\cap B)$
$P(A|B)=\frac{P(A\cap B)}{P(B)}=\frac{P(B)}{P(B)}=1$
Option B is correct.
1. Conditional Probability
Formula:
$
P(A \mid B)=\frac{P(A \cap B)}{P(B)}
$
2.
Multiplication Theorem
$P(A \cap B)=P(B) \cdot P(A \mid B)$
3. Independent Events
Events A and B are independent if:
$
P(A \cap B)=P(A) \cdot P(B)
$
4. Bayes' Theorem
Used to revise probability with new information:
$
P\left(E_i \mid A\right)=\frac{P\left(E_i\right) \cdot P\left(A \mid E_i\right)}{\sum P\left(E_j\right) \cdot P\left(A \mid E_j\right)}
$
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Frequently Asked Questions (FAQs)
More than 95% of questions in the board's exam are not asked from miscellaneous exercises but it is very important for competitive exams.
Miscellaneous exercise is consists of different questions from all the topics of probability Class 12 Maths.
The probability of an impossible event is 0.
The probability of getting 0 when we roll a die is zero.
The probability of a number less than 7 when we roll a die is one.
The event of getting a number less than 7 when we roll a die is an example of a certain event.
The event of getting number 0 we roll a die is an example of an impossible event.
Here you can get NCERT Solutions.
On Question asked by student community
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