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NCERT Question.8 : If each element of a second-order determinant is either 0 or 1, what is the probability that the value of the determinant is positive? (Assume that the individual entries of the determinant are chosen independently, each value being assumed with probability $1/2$).
Solution
Let the second-order determinant be
$$
\begin{vmatrix} a & b \ \\ c & d \end{vmatrix} = ad – bc
$$
where each of $a, b, c, d$ is independently $0$ or $1$ with probability $1/2$.
Step 1: Total possible determinants
Each element has 2 choices (0 or 1), so the total number of determinants is:
$$
2^4 = 16
$$
Step 2: Determinants with positive value
For the determinant to be positive, we need:
$$
ad – bc > 0 \implies ad = 1 \text{ and } bc = 0
$$
Now, count the cases:
- $a = 1, d = 1$ (so $ad = 1$)
- $bc = 0$ (so at least one of $b$ or $c$ is 0)
Possible combinations for $(b, c)$ satisfying $bc = 0$:
- $(0,0)$
- $(0,1)$
- $(1,0)$
Thus, there are $3$ favorable cases:
$$
\begin{bmatrix}1 & 0 \ \\ 0 & 1\end{bmatrix}, \quad
\begin{bmatrix}1 & 0 \\ \ 1 & 1\end{bmatrix}, \quad
\begin{bmatrix}1 & 1 \ \\ 0 & 1\end{bmatrix}
$$
Step 3: Probability
The probability that the determinant is positive:
$$
P(\text{determinant positive}) = \frac{\text{Number of favorable cases}}{\text{Total cases}} = \frac{3}{16}
$$
Final Result
$$
P(\text{determinant positive}) = \frac{3}{16}
$$
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NCERT Question.9 : An electronic assembly consists of two subsystems, A and B. From previous testing procedures, the following probabilities are known:
$P(\text{A fails}) = 0.2$
$P(\text{B fails alone}) = 0.15$
$P(\text{A and}\;\text{B fail}) = 0.15$
Evaluate the following:
(i) $P(\text{A fails}\mid \;\text{B has failed})$
(ii) $P(\text{A fails alone})$
Solution
Let event $E_A$ denote “A fails” and $E_B$ denote “B fails.”
Given:
$$P(E_A) = 0.2$$
$$P(E_A \cap E_B) = 0.15$$
$$P(B\;\text{fails alone}) = P(E_B) – P(E_A \cap E_B) = 0.15$$
So,
$$P(E_B) = 0.15 + 0.15 = 0.30$$
(i) Probability that A fails given that B has failed:
$$
P(E_A \mid E_B) = \frac{P(E_A \cap E_B)}{P(E_B)}
$$
Substitute values:
$$
P(E_A \mid E_B) = \frac{0.15}{0.30} = 0.5
$$
(ii) Probability that A fails alone:
$$
P(E_A\;\text{alone}) = P(E_A) – P(E_A \cap E_B)
$$
$$
P(E_A\;\text{alone}) = 0.2 – 0.15 = 0.05
$$
Final Result
$$
P(E_A \mid E_B) = 0.5,\quad P(E_A\;\text{alone}) = 0.05
$$
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NCERT Question 10 : Bag 1 contains 3 red and 4 black balls, and Bag II contains 4 red and 5 black balls. One ball is transferred from Bag I to Bag II, and then a ball is drawn from Bag II. The ball so drawn is found to be red in color. Find the probability that the transferred ball is black.
Solution
Let us first assume $A_{1}$ denote the event that a red ball is transferred from Bag I to Bag II.
And $A_{2}$ denote the event that a black ball is transferred from Bag I to Bag II.
$$P(A_{1})=\frac{3}{7}$$
$$P(A_{2})=\frac{4}{7}$$
Let $X$ be the event that the drawn ball is red.
When the red ball is transferred from Bag I to Bag II,
$$P(X \mid A_{1})=\frac{5}{10}=\frac{1}{2}$$
And when the black ball is transferred from Bag I to Bag II,
$$P(X \mid A_{2})=\frac{4}{10}=\frac{2}{5}$$
Hence,
$$P(A_{2} \mid X)=\frac{P(A_{2})P(X \mid A_{2})}{P(A_{1})P(X \mid A_{1}) + P(A_{2})P(X \mid A_{2})}$$
$$P(A_{2} \mid X)=\frac{\frac{4}{7} \times \frac{2}{5}}{\frac{3}{7} \times \frac{1}{2} + \frac{4}{7} \times \frac{2}{5}}$$
$$P(A_{2} \mid X)=\frac{16}{31}$$
NCERT Question 11 : Choose the correct answer: A and B are two events such that $P(A) \ne 0$ and $P(B \mid A) = 1$, then
A. $A \subset B$
B. $B \subset A$
C. $B = \varphi$
D. $A = \varphi$
Solution
By definition of conditional probability,
$$
P(B\mid A)=\frac{P(A\cap B)}{P(A)}.
$$
We are given $P(B\mid A)=1$ and $P(A)\ne 0$. Substitute into the definition:
$$
1=\frac{P(A\cap B)}{P(A)}.
$$
Multiply both sides by $P(A)$ (permitted because $P(A)\ne 0$):
$$
P(A\cap B)=P(A).
$$
The equality $P(A\cap B)=P(A)$ means that every outcome that lies in $A$ also lies in $A\cap B$. But $A\cap B$ consists exactly of those outcomes that are in both $A$ and $B$. So every outcome of $A$ is an outcome of $B$. That is the precise set definition of “$A$ is a subset of $B$”:
$$
A\subset B.
$$
Why other options are wrong
(B) $B\subset A$ would mean every outcome of $B$ is in $A$. The condition $P(B\mid A)=1$ says nothing about outcomes of $B$ that are outside $A$; $B$ may well contain extra outcomes. For example, take a sample space with outcomes $\{1,2,3\}$, let $A=\{1\}$ and $B=\{1,2\}$. Then $P(B\mid A)=1$ but $B\not\subset A$.
(C) $B=\varnothing$ is impossible here because if $B=\varnothing$ then $P(B\mid A)=0$, not $1$ (provided $P(A)\ne0$).
(D) $A=\varnothing$ is excluded by the hypothesis $P(A)\ne0$; moreover, $P(B\mid A)$ would be undefined if $P(A)=0$.
Intuition and a concrete example
Think of $A$ as “students who did all homework” and $B$ as “students who passed the course.” If $P(\text{passed}\mid\text{did all homework})=1$, then every student who did all homework passed — so the set of students who did all homework is contained in the set of students who passed. But there can be students who passed without doing all homework (so $B$ can be bigger than $A$).
Formal conclusion
$$
\boxed{A\subset B}
$$
so option (A) is correct.
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NCERT 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. $P(B \mid A) = P(B)$
Solution
We are given:
$$P(A \mid B) > P(A)$$
Using the definition of conditional probability:
$$P(A \mid B) = \frac{P(A \cap B)}{P(B)}$$
So the condition becomes:
$$\frac{P(A \cap B)}{P(B)} > P(A)$$
Multiply both sides by $P(B)$ (which is positive):
$$P(A \cap B) > P(A)P(B)$$
This inequality means:
- The joint probability $P(A \cap B)$ is greater than the product $P(A)P(B)$.
- So, A and B are positively associated (they increase each other’s chances).
Now check the options.
Check Option C
Use:
$$P(B \mid A) = \frac{P(A \cap B)}{P(A)}$$
From above we have:
$$P(A \cap B) > P(A),P(B)$$
Divide both sides by $P(A)$:
$$\frac{P(A \cap B)}{P(A)} > P(B)$$
But the left side is exactly $P(B \mid A)$.
Thus:
$$P(B \mid A) > P(B)$$
This matches option (C).
Therefore, the correct option is:
Option (C): $P(B \mid A) > P(B)$
NCERT Question 13: If $A$ and $B$ are any two events such that
$$P(A)+P(B)-P(A \;and \;B)=P(A),$$
then which of the following is true?
(A) $P(B\mid A)=1$
(B) $P(A\mid B)=1$
(C) $P(B\mid A)=0$
(D) $P(A\mid B)=0$
Solution
From the given equality
$$P(A)+P(B)-P(A\cap B)=P(A)$$
subtract $P(A)$ from both sides to obtain
$$P(B)-P(A\cap B)=0.$$
Hence
$$P(B)=P(A\cap B). \tag{1}$$
Interpretation of (1): the probability of $B$ equals the probability that both $A$ and $B$ occur. Two cases arise.
Case 1: $P(B)>0$.
Then (1) implies $P(A\cap B)=P(B)$. By the definition of conditional probability,
$$P(A\mid B)=\frac{P(A\cap B)}{P(B)}=\frac{P(B)}{P(B)}=1.$$
So when $P(B)>0$ we get $P(A\mid B)=1$, which is option (B).
Case 2: $P(B)=0$.
If $P(B)=0$ then (1) gives $P(A\cap B)=0$ as well. In that situation $P(A\mid B)$ is undefined (division by zero), so none of the conditional statements involving $P(A\mid B)$ have a meaningful numerical value. However, most textbook/contest interpretations of the given identity implicitly consider the nontrivial case $P(B)>0$, and the intended conclusion is (B).
Why the other options are not correct (for $P(B)>0$): from (1) we have $P(A\cap B)=P(B)$, so $P(B\mid A)=P(A\cap B)/P(A)=P(B)/P(A)$ which need not equal $1$ (unless $P(B)=P(A)$). Option (C) ($P(B\mid A)=0$) is false in general. Option (D) ($P(A\mid B)=0$) contradicts our derivation when $P(B)>0$.
Final answer: (B)
$P(A\mid B)=1$ (with the usual proviso that $P(B)>0$; if $P(B)=0$ the conditional probability $P(A\mid B)$ is undefined).
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