NCERT Solutions: Class 12 Maths Chapter 13 - Probability

$P(E|F)=\dfrac{P(E\cap F)}{P(F)}=\dfrac{0.2}{0.3}=\dfrac23$. Also $P(F|E)=\dfrac{P(E\cap F)}{P(E)}=\dfrac{0.2}{0.6}=\dfrac13$.

$P(A|B)=\dfrac{P(A\cap B)}{P(B)}=\dfrac{0.32}{0.5}=0.64=\dfrac{16}{25}$.

$P(A\cap B)=P(A)P(B|A)=0.8\cdot0.4=0.32$. Thus $P(A|B)=0.32/0.5=0.64$. Also $P(A\cup B)=0.8+0.5-0.32=0.98$.

Here $P(A)=5/26$ and $P(B)=5/13$. Also $P(A\cap B)=P(A|B)P(B)=\dfrac25\cdot\dfrac5{13}=\dfrac2{13}$. Hence $P(A\cup B)=\dfrac5{26}+\dfrac5{13}-\dfrac2{13}=\dfrac{11}{26}$.

$P(A\cap B)=P(A)+P(B)-P(A\cup B)=\dfrac6{11}+\dfrac5{11}-\dfrac7{11}=\dfrac4{11}$. Therefore $P(A|B)=\dfrac{4/11}{5/11}=\dfrac45$ and $P(B|A)=\dfrac{4/11}{6/11}=\dfrac23$.

For three tosses there are $8$ equally likely outcomes. (i) Given $F=\{HHH,HHT\}$, only $HHH$ has head on the third toss, so $P(E|F)=1/2$. (ii) Given at most two heads, all outcomes except $HHH$ remain; exactly three of them have at least two heads, so $3/7$. (iii) Given at least one tail, all outcomes except $HHH$ remain; six of these have at most two tails, so $6/7$.

The sample space is $\{HH,HT,TH,TT\}$. (i) Given at least one head, the possible outcomes are $HH,HT,TH$; exactly $HT,TH$ have a tail on one coin, so $P(E|F)=2/3$. (ii) Given no head, the only outcome is $TT$, which has tails, so no tail appears is impossible; probability $0$.

Once $F$ has occurred, the first two throws are fixed as $6$ and $5$. The third throw is still independent and has probability $1/6$ of being $4$.

If the father is in the middle, mother and son occupy the two ends. Therefore the son is certainly on one end, so $P(E|F)=1$.

(a) With black die $5$, sum greater than $9$ needs red die $5$ or $6$, so probability $2/6=1/3$. (b) Given red die is less than $4$, there are $3\cdot6=18$ possible ordered outcomes. Sum $8$ occurs for red $2$ with black $6$, or red $3$ with black $5$, so probability $2/18=1/9$.

$E\cap F=\{3\}$, so $P(E|F)=1/2$ and $P(F|E)=1/3$. Also $E\cap G=\{3,5\}$, so $P(E|G)=2/4=1/2$ and $P(G|E)=2/3$. Further $(E\cup F)\cap G=\{2,3,5\}$ and $(E\cap F)\cap G=\{3\}$, giving $3/4$ and $1/4$.

The equally likely ordered outcomes are $BB,BG,GB,GG$. (i) If the youngest is a girl, the outcomes are $BG,GG$, so probability of both girls is $1/2$. (ii) If at least one is a girl, the outcomes are $BG,GB,GG$, so probability is $1/3$.

Among multiple choice questions there are $500+400=900$ questions. Of these, $500$ are easy. Hence the conditional probability is $500/900=5/9$.

Given the two numbers are different, there are $36-6=30$ ordered outcomes. Sum $4$ with different numbers occurs as $(1,3)$ and $(3,1)$, so the probability is $2/30=1/15$.

A coin is tossed only when the first die is not a multiple of $3$, so the first die is not $3$ and no second die is thrown. Thus the event the coin shows a tail is incompatible with at least one die showing $3$. Hence the conditional probability is $0$.

1. i. $0$
2. ii. $\dfrac12$
3. iii. not defined
4. iv. $1$

$P(A|B)=\dfrac{P(A\cap B)}{P(B)}$ is defined only when $P(B)\ne0$. Since $P(B)=0$, it is not defined.

1. i. $A\subset B$ but $A\ne B$
2. ii. $A=B$
3. iii. $A\cap B=\phi$
4. iv. $P(A)=P(B)$

When $P(A\cap B)\ne0$, the equality gives $\dfrac{P(A\cap B)}{P(B)}=\dfrac{P(A\cap B)}{P(A)}$, hence $P(A)=P(B)$. This matches option (iv).

For independent events, $P(A\cap B)=P(A)P(B)=\dfrac35\cdot\dfrac15=\dfrac3{25}$.

There are $26$ black cards. The probability is $\dfrac{26}{52}\cdot\dfrac{25}{51}=\dfrac12\cdot\dfrac{25}{51}=\dfrac{25}{102}$.

The box is approved if all three selected oranges are good. Thus probability $=\dfrac{12}{15}\cdot\dfrac{11}{14}\cdot\dfrac{10}{13}=\dfrac{44}{91}$.

$P(A)=1/2$, $P(B)=1/6$, and $P(A\cap B)=1/12$. Since $P(A)P(B)=\dfrac12\cdot\dfrac16=\dfrac1{12}=P(A\cap B)$, the events are independent.

$P(A)=3/6=1/2$, $P(B)=3/6=1/2$, and $A\cap B=\{2\}$ so $P(A\cap B)=1/6$. Since $1/6\ne1/4=P(A)P(B)$, the events are not independent.

$P(E)P(F)=\dfrac35\cdot\dfrac3{10}=\dfrac9{50}$, but $P(E\cap F)=\dfrac15=\dfrac{10}{50}$. These are unequal, so the events are not independent.

If mutually exclusive, $P(A\cup B)=P(A)+P(B)$, so $3/5=1/2+p$ and $p=1/10$. If independent, $3/5=1/2+p-(1/2)p=1/2+p/2$, so $p=1/5$.

Independence gives $P(A\cap B)=0.3\cdot0.4=0.12$. Then $P(A\cup B)=0.3+0.4-0.12=0.58$. Also $P(A|B)=P(A)=0.3$ and $P(B|A)=P(B)=0.4$.

Not $A$ and not $B$ is $(A\cup B)^prime$. Now $P(A\cup B)=\dfrac14+\dfrac12-\dfrac18=\dfrac58$. Hence the required probability is $1-\dfrac58=\dfrac38$.

Using De Morgan, $P(A\cap B)=1-P(A^prime\cup B^prime)=1-\dfrac14=\dfrac34$. But $P(A)P(B)=\dfrac12\cdot\dfrac7{12}=\dfrac7{24}$. Since these are not equal, $A$ and $B$ are not independent.

$P(A\cap B)=0.3\cdot0.6=0.18$. $P(A\cap B^prime)=P(A)-P(A\cap B)=0.12$. $P(A\cup B)=0.3+0.6-0.18=0.72$. Hence $P(A^prime\cap B^prime)=1-0.72=0.28$.

The complement is getting no odd number, i.e. all three throws are even. Its probability is $(1/2)^3=1/8$. Hence the required probability is $1-1/8=7/8$.

With replacement, $P(R)=8/18=4/9$ and $P(B)=10/18=5/9$. Thus $P(RR)=16/81$, $P(BR)=20/81$, and one black and one red can occur as $BR$ or $RB$, giving $2\cdot\dfrac59\cdot\dfrac49=\dfrac{40}{81}$.

The problem is not solved only if both fail, with probability $\dfrac12\cdot\dfrac23=\dfrac13$. Hence it is solved with probability $2/3$. Exactly one solves it with probability $\dfrac12\cdot\dfrac23+\dfrac12\cdot\dfrac13=\dfrac12$.

(i) $P(E\cap F)=1/52=P(E)P(F)=\dfrac14\cdot\dfrac1{13}$. (ii) $P(E\cap F)=2/52=1/26=P(E)P(F)=\dfrac12\cdot\dfrac1{13}$. (iii) $P(E\cap F)=P($queen$)=4/52=1/13$, but $P(E)P(F)=\dfrac2{13}\cdot\dfrac2{13}=\dfrac4{169}$, so not independent.

Let $H$ and $E$ denote reading Hindi and English newspapers. $P(H\cup E)=0.6+0.4-0.2=0.8$, so neither has probability $0.2$. Also $P(E|H)=0.2/0.6=1/3$, and $P(H|E)=0.2/0.4=1/2$.

1. i. $0$
2. ii. $\dfrac13$
3. iii. $\dfrac1{12}$
4. iv. $\dfrac1{36}$

The only even prime number on a die is $2$. Thus both dice must show $2$, whose probability is $\dfrac16\cdot\dfrac16=\dfrac1{36}$.

1. i. $A$ and $B$ are mutually exclusive
2. ii. $P(A^prime B^prime)=[1-P(A)][1-P(B)]$
3. iii. $P(A)=P(B)$
4. iv. $P(A)+P(B)=1$

If $A$ and $B$ are independent, then their complements are also independent. Hence $P(A^prime\cap B^prime)=P(A^prime)P(B^prime)=[1-P(A)][1-P(B)]$. Conversely this condition is equivalent to independence.

If the first ball is red, probability $1/2$, the second red probability is $7/12$. If the first ball is black, probability $1/2$, the second red probability is $5/12$. Hence total probability $=\dfrac12\cdot\dfrac7{12}+\dfrac12\cdot\dfrac5{12}=\dfrac12$.

Let $B_1,B_2$ be the choices of the first and second bag, and $R$ the event of drawing red. Then $P(B_1)=P(B_2)=1/2$, $P(R|B_1)=4/8=1/2$, $P(R|B_2)=2/8=1/4$. By Bayes theorem, $P(B_1|R)=\dfrac{(1/2)(1/2)}{(1/2)(1/2)+(1/2)(1/4)}=\dfrac23$.

Let $H$ be hostlier and $A$ be A grade. Then $P(H)=0.6$, $P(H^prime)=0.4$, $P(A|H)=0.3$, $P(A|H^prime)=0.2$. Thus $P(H|A)=\dfrac{0.6\cdot0.3}{0.6\cdot0.3+0.4\cdot0.2}=\dfrac{0.18}{0.26}=\dfrac9{13}$.

Let $K$ be knows and $C$ be correct. Then $P(K)=3/4$, $P(G)=1/4$, $P(C|K)=1$, $P(C|G)=1/4$. Hence $P(K|C)=\dfrac{(3/4)(1)}{(3/4)(1)+(1/4)(1/4)}=\dfrac{3/4}{13/16}=\dfrac{12}{13}$.

Let $D$ be the disease and $+$ be a positive test. $P(D)=0.001$, $P(D^prime)=0.999$, $P(+|D)=0.99$, $P(+|D^prime)=0.005$. Thus $P(D|+)=\dfrac{0.001\cdot0.99}{0.001\cdot0.99+0.999\cdot0.005}=\dfrac{0.00099}{0.005985}=\dfrac{22}{133}$.

Let the coins be $C_1,C_2,C_3$. Priors are $1/3$ each, and $P(H|C_1)=1$, $P(H|C_2)=3/4$, $P(H|C_3)=1/2$. Hence $P(C_1|H)=\dfrac{1}{1+3/4+1/2}=\dfrac49$.

The prior probabilities are $1/6$, $1/3$ and $1/2$. Thus $P(S|A)=\dfrac{(1/6)(0.01)}{(1/6)(0.01)+(1/3)(0.03)+(1/2)(0.15)}=\dfrac{1/600}{52/600}=\dfrac1{52}$.

Let $D$ be defective. Then $P(B|D)=\dfrac{0.4\cdot0.01}{0.6\cdot0.02+0.4\cdot0.01}=\dfrac{0.004}{0.016}=\dfrac14$.

Let $G_1,G_2$ be the winning groups and $N$ be new product introduced. Then $P(G_2|N)=\dfrac{0.4\cdot0.3}{0.6\cdot0.7+0.4\cdot0.3}=\dfrac{0.12}{0.54}=\dfrac29$.

Let $A$ be getting $1,2,3$ or $4$ and $B$ be getting $5$ or $6$. Then $P(A)=2/3$, $P(B)=1/3$. If $A$ occurs, exactly one head in one toss has probability $1/2$. If $B$ occurs, exactly one head in three tosses has probability $3/8$. Hence $P(A|E)=\dfrac{(2/3)(1/2)}{(2/3)(1/2)+(1/3)(3/8)}=\dfrac8{11}$.

$P(A|D)=\dfrac{0.5\cdot0.01}{0.5\cdot0.01+0.3\cdot0.05+0.2\cdot0.07}=\dfrac{0.005}{0.034}=\dfrac5{34}$.

Let $D$ be the event the lost card is diamond and $A$ the event the two drawn cards are diamonds. $P(D)=1/4$, $P(D^prime)=3/4$. If $D$ occurred, $P(A|D)=\binom{12}{2}/\binom{51}{2}$; otherwise $P(A|D^prime)=\binom{13}{2}/\binom{51}{2}$. Therefore $P(D|A)=\dfrac{(1/4)66}{(1/4)66+(3/4)78}=\dfrac{66}{300}=\dfrac{11}{50}$.

1. i. $\dfrac45$
2. ii. $\dfrac12$
3. iii. $\dfrac15$
4. iv. $\dfrac25$

Let $H$ be actual head and $R$ be reported head. Then $P(H)=P(T)=1/2$, $P(R|H)=4/5$, and $P(R|T)=1/5$. Thus $P(H|R)=\dfrac{(1/2)(4/5)}{(1/2)(4/5)+(1/2)(1/5)}=\dfrac45$.

1. i. $P(A|B)=\dfrac{P(B)}{P(A)}$
2. ii. $P(A|B)\lt P(A)$
3. iii. $P(A|B)\ge P(A)$
4. iv. None of these

Since $A\subset B$, $A\cap B=A$. Therefore $P(A|B)=P(A)/P(B)$. As $0\lt P(B)\le1$, $P(A)/P(B)\ge P(A)$. Hence option (iii) is correct.

(i) If $A \subset B$, then $A \cap B=A$. Hence $P(B|A)=\dfrac{P(A \cap B)}{P(A)}=\dfrac{P(A)}{P(A)}=1$. (ii) If $A \cap B=\emptyset$, then $P(A \cap B)=0$, so $P(B|A)=0$.

The equally likely outcomes are $MM,MF,FM,FF$. (i) Given at least one child is male, the possible outcomes are $MM,MF,FM$. Only $MM$ has both males, so the probability is $\dfrac13$. (ii) Given the elder child is female, the possible outcomes are $FM,FF$. Only $FF$ has both females, so the probability is $\dfrac12$.

Let $M$ and $W$ denote male and female, and $G$ denote grey hair. Then $P(M)=P(W)=\dfrac12$, $P(G|M)=0.05$ and $P(G|W)=0.0025$. By Bayes' theorem, $P(M|G)=\dfrac{P(M)P(G|M)}{P(M)P(G|M)+P(W)P(G|W)}=\dfrac{0.5\cdot0.05}{0.5\cdot0.05+0.5\cdot0.0025}=\dfrac{0.05}{0.0525}=\dfrac{20}{21}$.

Let $X$ be the number of right-handed people in the sample. Then $X\sim B(10,0.9)$. Therefore $P(X\le 6)=\sum_{k=0}^{6}\binom{10}{k}(0.9)^k(0.1)^{10-k}=0.0127951984$.

A leap year has $366=52\times7+2$ days, so two consecutive weekdays occur $53$ times. Tuesday will occur $53$ times if the two extra days are Monday-Tuesday or Tuesday-Wednesday. There are $7$ equally likely pairs of extra days, and $2$ favourable pairs. Hence the required probability is $\dfrac27$.

Each box is chosen with probability $\dfrac14$. The probabilities of drawing a red marble from boxes $A,B,C,D$ are $\dfrac1{10},\dfrac6{10},\dfrac8{10},0$. Thus $P(R)=\dfrac14\left(\dfrac1{10}+\dfrac6{10}+\dfrac8{10}+0\right)=\dfrac38$. By Bayes' theorem, $P(A|R)=\dfrac{\frac14\cdot\frac1{10}}{\frac38}=\dfrac1{15}$, $P(B|R)=\dfrac{\frac14\cdot\frac6{10}}{\frac38}=\dfrac25$, and $P(C|R)=\dfrac{\frac14\cdot\frac8{10}}{\frac38}=\dfrac8{15}$.

Let $M$ be the meditation-yoga option, $D$ be the drug option and $H$ be heart attack. Since the original risk is $0.40$, $P(H|M)=0.40(1-0.30)=0.28$ and $P(H|D)=0.40(1-0.25)=0.30$. Also $P(M)=P(D)=\dfrac12$. Hence $P(M|H)=\dfrac{\frac12\cdot0.28}{\frac12\cdot0.28+\frac12\cdot0.30}=\dfrac{0.28}{0.58}=\dfrac{14}{29}$.

For determinant $\begin{vmatrix}a&b\\c&d\end{vmatrix}=ad-bc$, each of $a,b,c,d$ is independently $0$ or $1$, so there are $2^4=16$ equally likely determinants. The determinant is positive exactly when $ad=1$ and $bc=0$. This requires $a=d=1$, while $(b,c)$ is not $(1,1)$, giving $3$ favourable cases. Hence the probability is $\dfrac3{16}$.

The event that $B$ has failed consists of $B$ failing alone or both $A$ and $B$ failing. Thus $P(B\text{ fails})=0.15+0.15=0.30$. Therefore $P(A\text{ fails}|B\text{ has failed})=\dfrac{P(A\text{ and }B\text{ fail})}{P(B\text{ fails})}=\dfrac{0.15}{0.30}=\dfrac12$. Also $P(A\text{ fails alone})=P(A\text{ fails})-P(A\text{ and }B\text{ fail})=0.20-0.15=0.05$.

Let $B$ be the event that the transferred ball is black and $R$ be the event that the ball drawn from Bag $II$ is red. Then $P(B)=\dfrac47$ and $P(B')=\dfrac37$. If a black ball is transferred, Bag $II$ has $4$ red and $6$ black, so $P(R|B)=\dfrac4{10}$. If a red ball is transferred, Bag $II$ has $5$ red and $5$ black, so $P(R|B')=\dfrac5{10}$. Hence $P(B|R)=\dfrac{\frac47\cdot\frac4{10}}{\frac47\cdot\frac4{10}+\frac37\cdot\frac5{10}}=\dfrac{16}{31}$.

1. i. $A \subset B$
2. ii. $B \subset A$
3. iii. $B=\emptyset$
4. iv. $A=\emptyset$

Since $P(B|A)=\dfrac{P(A\cap B)}{P(A)}=1$, we get $P(A\cap B)=P(A)$. Thus whenever $A$ occurs, $B$ occurs; hence $A \subset B$.

1. i. $P(B|A) \lt P(B)$
2. ii. $P(A\cap B) \lt P(A)P(B)$
3. iii. $P(B|A) \gt P(B)$
4. iv. $P(B|A)=P(B)$

From $P(A|B) \gt P(A)$, we have $\dfrac{P(A\cap B)}{P(B)} \gt P(A)$, so $P(A\cap B) \gt P(A)P(B)$. Dividing by $P(A)$ gives $P(B|A)=\dfrac{P(A\cap B)}{P(A)} \gt P(B)$. Hence option (iii) is correct.

1. i. $P(B|A)=1$
2. ii. $P(A|B)=1$
3. iii. $P(B|A)=0$
4. iv. $P(A|B)=0$

The given condition gives $P(B)-P(A\cap B)=0$, so $P(B)=P(A\cap B)$. Therefore $B$ is contained in $A$ in probability, and $P(A|B)=\dfrac{P(A\cap B)}{P(B)}=1$. Hence option (ii) is correct.