Anand Classes provides comprehensive and well-structured NCERT Solutions for Probability Exercise 13.1 of Class 12 Chapter 13, helping students build strong conceptual understanding for board exams and competitive entrance tests. These expertly crafted solutions simplify complex probability concepts, offer step-by-step explanations, and ensure better clarity for solving exam-oriented questions with confidence. Click the print button to download study material and notes.
Access NCERT Solutions for Probability Exercise 13.1 of Class 12 Math Chapter 13
NCERT Question.1 : Given that $E$ and $F$ are events such that $P(E)=0.6$, $P(F)=0.3$ and $P(E \cap F)=0.2$, find $P(E\mid F)$ and $P(F\mid E)$.
Solution
Given
$P(E)=0.6$
$P(F)=0.3$
$P(E \cap F)=0.2$
So,
$$P(F \cap E)=P(E \cap F)=0.2$$
Now,
$$P(E \mid F)=\frac{P(E \cap F)}{P(F)}=\frac{0.2}{0.3}=\frac{2}{3}$$
and
$$P(F \mid E)=\frac{P(F \cap E)}{P(E)}=\frac{0.2}{0.6}=\frac{2}{6}=\frac{1}{3}$$
Final Result
$$\boxed{P(E\mid F)=\frac{2}{3}, P(F\mid E)=\frac{1}{3}}$$
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NCERT Question.2 : Compute $P(A \mid B)$ if $P(B)=0.5$ and $P(A \cap B)=0.32$
Solution
Given
$P(B)=0.5$
$P(A \cap B)=0.32$
Then
$$P(A \mid B)=\frac{P(A \cap B)}{P(B)}=\frac{0.32}{0.5}=\frac{32}{50}=\frac{16}{25}$$
Final Result
$$\boxed{P(A \mid B)=\frac{16}{25}}$$
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NCERT Question.3 : If $P(A)=0.8$, $P(B)=0.5$ and $P(B \mid A)=0.4$, find
(i) $P(A \cap B)$
(ii) $P(A \mid B)$
(iii) $P(A \cup B)$
Solution
Given
$P(A)=0.8$
$P(B)=0.5$
$P(B \mid A)=0.4$
(i)
Using
$$P(B \mid A)=\frac{P(A \cap B)}{P(A)}$$
So,
$$0.4=\frac{P(A \cap B)}{0.8}$$
Hence,
$$P(A \cap B)=0.4 \times 0.8=0.32$$
(ii)
Now,
$$P(A \mid B)=\frac{P(A \cap B)}{P(B)}=\frac{0.32}{0.5}=0.64$$
(iii)
Using union formula,
$$P(A \cup B)=P(A)+P(B)-P(A \cap B)$$
So,
$$P(A \cup B)=0.8+0.5-0.32=0.98$$
Final Result
$$\boxed{P(A \cap B)=0.32 ; P(A \mid B)=0.64 ; P(A \cup B)=0.98}$$
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NCERT Question.4 : Evaluate $P(A \cup B)$ if $2P(A)=P(B)=\dfrac{5}{13}$ and $P(A \mid B)=\dfrac{2}{5}$.
Solution
Given
$$2P(A)=\frac{5}{13}$$
So,
$$P(A)=\frac{1}{2} \times \frac{5}{13}=\frac{5}{26}$$
Also,
$$P(B)=\frac{5}{13}$$
Now,
$$P(A \mid B)=\frac{P(A \cap B)}{P(B)}$$
Thus,
$$\frac{2}{5}=\frac{P(A \cap B)}{\dfrac{5}{13}}$$
So,
$$P(A \cap B)=\frac{2}{5} \times \frac{5}{13}=\frac{2}{13}$$
Now evaluate
$$P(A \cup B)=P(A)+P(B)-P(A \cap B)$$
Hence,
$$P(A \cup B)=\frac{5}{26}+\frac{5}{13}-\frac{2}{13}$$
Convert $\dfrac{5}{13}$ and $\dfrac{2}{13}$ to denominator $26$:
$$P(A \cup B)=\frac{5}{26}+\frac{10}{26}-\frac{4}{26}=\frac{11}{26}$$
Final Result
$$\boxed{P(A \cup B)=\frac{11}{26}}$$
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NCERT Question.5 : If $P(A)=\dfrac{6}{11}$, $P(B)=\dfrac{5}{11}$ and $P(A \cup B)=\dfrac{7}{11}$ find
(i) $P(A \cap B)$
(ii) $P(A \mid B)$
(iii) $P(B \mid A)$
Solution
Given
$P(A)=\dfrac{6}{11}$
$P(B)=\dfrac{5}{11}$
$P(A \cup B)=\dfrac{7}{11}$
(i)
Using
$$P(A \cup B)=P(A)+P(B)-P(A \cap B)$$
So,
$$\frac{7}{11}=\frac{6}{11}+\frac{5}{11}-P(A \cap B)$$
Hence,
$$P(A \cap B)=\frac{11}{11}-\frac{7}{11}=\frac{4}{11}$$
(ii)
Now,
$$P(A \mid B)=\frac{P(A \cap B)}{P(B)}=\frac{\dfrac{4}{11}}{\dfrac{5}{11}}=\frac{4}{5}$$
(iii)
And,
$$P(B \mid A)=\frac{P(B \cap A)}{P(A)}=\frac{\dfrac{4}{11}}{\dfrac{6}{11}}=\frac{2}{3}$$
Final Result
$$\boxed{P(A \cap B)=\frac{4}{11} ; P(A \mid B)=\frac{4}{5} ; P(B \mid A)=\frac{2}{3}}$$
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NCERT Question.6 : A coin is tossed three times.
Determine $P(E\mid F)$ in the following cases.
(i) $E$ : head on third toss, $F$ : heads on first two tosses
(ii) $E$ : at least two heads, $F$ : at most two heads
(iii) $E$ : at most two tails, $F$ : at least one tail
Solution
Sample space
$$S=\{HHH, HHT, HTH, THH, TTH, THT, HTT, TTT\}$$
(i)
$E=\{HHH, HTH, THH, TTH\}$ so
$$P(E)=\frac{4}{8}=\frac{1}{2}$$
$F=\{HHH, HHT\}$ so
$$P(F)=\frac{2}{8}=\frac{1}{4}$$
$$E\cap F=\{HHH\}\quad\Rightarrow\quad P(E\cap F)=\frac{1}{8}$$
Therefore
$$P(E\mid F)=\frac{P(E\cap F)}{P(F)}=\frac{\dfrac{1}{8}}{\dfrac{1}{4}}=\frac{1}{2}$$
(ii)
$E=\{HHT, HTH, THH, HHH\}$ so
$$P(E)=\frac{4}{8}=\frac{1}{2}$$
$F=\{HHT, HTH, THH, TTH, THT, HTT, TTT\}$ so
$$P(F)=\frac{7}{8}$$
$$E\cap F=\{HHT, HTH, THH\}\quad\Rightarrow\quad P(E\cap F)=\frac{3}{8}$$
Therefore
$$P(E\mid F)=\frac{P(E\cap F)}{P(F)}=\frac{\dfrac{3}{8}}{\dfrac{7}{8}}=\frac{3}{7}$$
(iii)
$E=\{HHH, HHT, HTH, THH, TTH, THT, HTT\}$ so
$$P(E)=\frac{7}{8}$$
$F=\{HHT, HTH, THH, TTH, THT, HTT, TTT\}$ so
$$P(F)=\frac{7}{8}$$
$$E\cap F=\{HHT, HTH, THH, TTH, THT, HTT\}\quad\Rightarrow\quad P(E\cap F)=\frac{6}{8}$$
Therefore
$$P(E\mid F)=\frac{P(E\cap F)}{P(F)}=\frac{\dfrac{6}{8}}{\dfrac{7}{8}}=\frac{6}{7}$$
Final Result
$$\boxed{P(E\mid F)=\frac{1}{2}, \frac{3}{7}, \frac{6}{7}}$$
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NCERT Question.7 : Determine $P(E \mid F)$ in the following cases when two coins are tossed once:
(i) $E$ : tail appears on one coin, $F$ : one coin shows head
(ii) $E$ : no tail appears, $F$ : no head appears
Solution
Sample space
$$S=\{(H,H),(H,T),(T,H),(T,T)\}$$
(i)
$E$: tail appears on one coin
$$E=\{(H,T),(T,H)\}$$
$$P(E)=\frac{2}{4}=\frac{1}{2}$$
$F$: one coin shows head
$$F=\{(H,T),(T,H)\}$$
$$P(F)=\frac{2}{4}=\frac{1}{2}$$
Intersection
$$E\cap F=\{(H,T),(T,H)\}$$
$$P(E\cap F)=\frac{2}{4}=\frac{1}{2}$$
Thus,
$$P(E\mid F)=\frac{P(E\cap F)}{P(F)}=\frac{\dfrac{1}{2}}{\dfrac{1}{2}}=1$$
(ii)
$E$: no tail appears
$$E=\{(H,H)\}$$
$$P(E)=\frac{1}{4}$$
$F$: no head appears
$$F=\{(T,T)\}$$
$$P(F)=\frac{1}{4}$$
Intersection
$$E\cap F=\varnothing$$
$$P(E\cap F)=0$$
Thus,
$$P(E\mid F)=\frac{P(E\cap F)}{P(F)}=\frac{0}{\dfrac{1}{4}}=0$$
Final Result
$$\boxed{P(E\mid F)=1, 0}$$
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NCERT Question.8 : A die is thrown three times.
Determine $P(E \mid F)$ where
$E$ : 4 appears on the third toss
$F$ : 6 and 5 appear respectively on the first two tosses
Solution
A die is thrown 3 times.
S = {(1, 1, 1), …, (1, 6, 6), (2, 1, 1), …, (2, 6, 6), (3, 1, 1), …, (3, 6, 6), (4, 1, 1), …, (4, 6, 6), (5, 1, 1), …, (5, 6, 6), (6, 1, 1), …, (6, 6, 6)}
Sample space has
$$6 \times 6 \times 6 = 216$$
possible outcomes.
Event descriptions
$E$: 4 on the third toss
$$ E=\{(x,y,4) \mid x \in \{1,2,3,4,5,6\}; y \in \{1,2,3,4,5,6\}\} $$
$F$: 6 on the first toss and 5 on the second toss
$$F=\{(6,5,z) \mid z \in \{1,2,3,4,5,6\}\}$$
Probability of $F$
Six outcomes satisfy $F$
$$P(F)=\frac{6}{216}$$
Intersection $E \cap F$
Only one outcome satisfies both $E$ and $F$
$$E \cap F=\{(6,5,4)\}$$
$$P(E \cap F)=\frac{1}{216}$$
Conditional probability
$$P(E \mid F)=\frac{P(E \cap F)}{P(F)}=\frac{\frac{1}{216}}{\frac{6}{216}}=\frac{1}{6}$$
Final Result
$$\boxed{P(E \mid F)=\frac{1}{6}}$$
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