Maths Case study (4 Marks)

(ii) $\begin{aligned} & \mathrm{P}(\mathrm{A})=\frac{1}{3}, \mathrm{P}\left(\mathrm{A}^{\prime}\right)=1-\frac{1}{3}=\frac{2}{3} \\ & \mathrm{P}(\mathrm{B})=\frac{1}{2}, \mathrm{P}\left(\mathrm{b}^{\prime}\right)=1-\frac{1}{3}=\frac{1}{2} \\ & \mathrm{P}(\text { none of them selected })=P\left(A^{\prime} \cap B^{\prime}\right)=P\left(A^{\prime}\right) \cdot P\left(B^{\prime}\right)=\frac{2}{3} \cdot \frac{1}{2} \\ & \mathrm{P}(\text { Both are selected })=\frac{1}{3}\end{aligned}$

\[\begin{aligned}& \mathrm{P}\left(\mathrm{E}_1\right)=0.5, \mathrm{P}\left(\mathrm{E}_2\right)=0.2, \mathrm{P}\left(\mathrm{E}_3\right)=0.3 \\& P\left(\frac{A}{E_1}\right)=0.06, P\left(\frac{A}{E_2}\right)=0.04, P\left(\frac{A}{E_3}\right)=0.03\end{aligned}\]

Using Bayes' theorem, we have

$\begin{aligned} & P\left(\frac{E_1}{A}\right)=\frac{P\left(E_1\right) \cdot P\left(\frac{A}{E_1}\right)}{P\left(E_1\right) \cdot P\left(\frac{A}{E_1}\right)+P\left(E_2\right) \cdot P\left(\frac{A}{E_2}\right)+P\left(E_3\right) \cdot P\left(\frac{A}{E_3}\right)} \\ & =\frac{0.5 \times 0.06}{0.5 \times 0.06+0.2 \times 0.04+0.3 \times 0.03}=\frac{30}{47} \\ & \therefore \text { Required probability }=P\left(\frac{\bar{E}_1}{A}\right) \\ & =1-P\left(\frac{E_1}{A}\right)=1-\frac{30}{47}=\frac{17}{47}\end{aligned}$

(ii) Let $\mathrm{A}$ be the event of committing an error and $\mathrm{E}_1, \mathrm{E}_2$ and $\mathrm{E}_3$ be the events that Govind, Priyanka and Tahseen processed the form.

$\begin{aligned} & \mathrm{P}\left(\mathrm{E}_1\right)=0.5, \mathrm{P}\left(\mathrm{E}_2\right)=0.2, \mathrm{P}\left(\mathrm{E}_3\right)=0.3 \\ & P\left(\frac{A}{E_1}\right)=0.06, P\left(\frac{A}{E_2}\right)=0.04, P\left(\frac{A}{E_3}\right)=0.03 \\ & P\left(A \cap E_2\right)=P\left(\frac{A}{E_2}\right) \cdot P\left(E_2\right) \\ & \Rightarrow 0.04 \times 0.2=0.008\end{aligned}$

(ii) $\begin{aligned} & \mathrm{P}(\text { Grade } \mathrm{A} \text { in Maths) }=\mathrm{P}(\mathrm{M})=0.2 \\ & \mathrm{P}(\text { Grade } \mathrm{A} \text { in Physics })=\mathrm{P}(\mathrm{P})=0.3 \\ & \mathrm{P} \text { (Grade } \mathrm{A} \text { in Chemistry) }=\mathrm{P}(\mathrm{C})=0.5 \\ & \mathrm{P}(\text { not } \mathrm{A} \text { garde in Maths) }=\mathrm{P}(\bar{M})=1-0.2=0.8 \\ & \mathrm{P}(\text { not } \mathrm{A} \text { garde in Physics })=\mathrm{P}(\bar{P})=1-0.3=0.7 \\ & \mathrm{P}(\text { not } \mathrm{A} \text { garde in Chemistry) }=\mathrm{P}(\bar{C})=1-0.5=0.5 \\ & \mathrm{P}(\text { getting grade } \mathrm{A} \text { in on subjects })=P(\bar{M} \cap \bar{P} \cap \bar{C}) \\ & =P(\bar{M}) \times P(\bar{P}) \times P(\bar{C}) \\ & =0.8 \times 0.7 \times 0.5=0.280\end{aligned}$

Let $\mathrm{A}$ denotes the event that daughter is at one end $n(A)=12$ and $B$ denotes the event that father, and mother are in the middle $n(B)=4$

Also, $\mathrm{n}(\mathrm{A} \cap \mathrm{B})=4$

\[P(A / B)=\frac{P(A \cap B)}{P(B)}=\frac{\frac{4}{24}}{\frac{4}{24}}=1\]

(ii) Sample space is given by \{MFSD, MFDS, MSFD, MSDF, MDFS, MDSF, FMSD, FMDS, FSMD, FSDM, FDMS, FDSM, SFMD, SFDM, SMFD, SMDF, SDMF, SDFM DFMS, DFSM, DMSF, DMFS, DSMF, DSFM\}, where F, M, D and $S$ represent father, mother, daughter and son respectively. $n(S)=24$

Let $\mathrm{A}$ denotes the event that mother is at right end. $n(A)=6$ and $B$ denotes the event that son and daughter are together.

\[n(B)=12\]

Also, $\mathrm{n}(\mathrm{A} \cap \mathrm{B})=4$

\[P(A / B)=\frac{P(A \cap B)}{P(B)}=\frac{\frac{4}{24}}{\frac{12}{24}}=\frac{1}{3}\]

$\begin{aligned} & \text { (ii) } \mathrm{P}(\text { Grade A in Maths })=\mathrm{P}(\mathrm{M})=0.2 \\ & \mathrm{P}(\text { Grade A in Physics })=\mathrm{P}(\mathrm{P})=0.3 \\ & \mathrm{P}(\text { Grade A in Chemistry })=\mathrm{P}(\mathrm{C})=0.5 \\ & \mathrm{P}(\text { Grade A in no subjects })=P(\bar{M} \cap \bar{P} \cap \bar{C})=P(\bar{M}) \cdot P(\bar{P}) \cdot P(\bar{C}) \\ & \mathrm{P}(\text { Grade A in no subjects })=0.8 \times 0.7 \times 0.5=0.280\end{aligned}$

\[=2\left(\frac{15}{50} \times \frac{12}{49}\right)=\frac{36}{245}\]

(ii) $\mathrm{P}$ (First ticket shows an even number and second ticket shows an odd number)

\[=\frac{25}{50} \times \frac{25}{49}=\frac{25}{98}\]

Then $P\left(E_1\right)=\frac{\mathbf{5 0}}{\mathbf{1 0 0}}=\frac{\mathbf{5}}{\mathbf{1 0}}, P\left(E_2\right)=\frac{\mathbf{2 0}}{\mathbf{1 0 0}}=\frac{1}{5}$ and $P\left(E_3\right)=\frac{\mathbf{3 0}}{\mathbf{1 0 0}}=\frac{3}{10}$.

Also, let $\boldsymbol{E}$ be the event of committing an error.

We have, $P\left(E \mid E_1\right)=0.06, P\left(E \mid E_2\right)=0.04, P\left(E \mid E_3\right)=0.03$.

(i) The probability that Sophia processed the form and committed an error is given by

$P\left(E \cap E_2\right)=P\left(E_2\right) . P\left(E \mid E_2\right)=\frac{1}{5} \times 0.04=0.008$

(ii) The total probability of committing an error in processing the form is given by  

$\begin{array}{l}P(E)=P\left(E_1\right) \cdot P\left(E \mid E_1\right)+P\left(E_2\right) \cdot P\left(E \mid E_2\right)+P\left(E_3\right) . P\left(E \mid E_3\right) \\ P(E)=\frac{50}{100} \times 0.06+\frac{20}{100} \times 0.04+\frac{30}{100} \times 0.03=0.047 .\end{array}$

(iii) The probability that the form is processed by James given that form has an error is given by

$\begin{aligned} & P\left(E_1 \mid E\right)=\frac{P\left(E \mid E_1\right) \times P\left(E_1\right)}{P\left(E \mid E_1\right) . P\left(E_1\right)+P\left(E \mid E_2\right) . P\left(E_2\right)+P\left(E \mid E_3\right) . P\left(E_3\right)} \\ & =\frac{0.06 \times \frac{50}{100}}{0.06 \times \frac{50}{100}+0.04 \times \frac{20}{100}+0.03 \times \frac{30}{100}}=\frac{30}{47}\end{aligned}$

Therefore, the required probability that the form is not processed by James given that form has an error$=P\left(\overline{E_1} \mid E\right)=1-P\left(E_1 \mid E\right)=1-\frac{30}{47}=\frac{17}{47}$.

(iii) The probability that the form is processed by James given that form has an error is given by 

$\begin{aligned} & P\left(E_1 \mid E\right)=\frac{P\left(E \mid E_1\right) \times P\left(E_1\right)}{P\left(E \mid E_1\right) . P\left(E_1\right)+P\left(E \mid E_2\right) \cdot P\left(E_2\right)+P\left(E \mid E_3\right) \cdot P\left(E_3\right)} \\ & =\frac{0.06 \times \frac{50}{100}}{0.06 \times \frac{50}{100}+0.04 \times \frac{20}{100}+0.03 \times \frac{30}{100}}=\frac{30}{47}\end{aligned}$

Therefore, the required probability that the form is not processed by James given that form has an error$=P\left(\overline{E_1} \mid E\right)=1-P\left(E_1 \mid E\right)=1-\frac{30}{47}=\frac{17}{47}$.

(iii) OR $\sum_{i=1}^3 P\left(E_i \mid E\right)=P\left(E_1 \mid E\right)+P\left(E_2 \mid E\right)+P\left(E_3 \mid E\right)=1$

Since, sum of the posterior probabilities is 1. 

$\begin{aligned} & \text { (We have }, \sum_{i=1}^3 P\left(E_i \mid E\right)=P\left(E_1 \mid E\right)+P\left(E_2 \mid E\right)+P\left(E_3 \mid E\right) \\ & =\frac{P\left(E \cap E_1\right)+P\left(E \cap E_2\right)+P\left(E \cap E_3\right)}{P(E)} \\ & =\frac{P\left(\left(E \cap E_1\right) \cup\left(E \cap E_2\right) \cup\left(E \cap E_3\right)\right)}{P(E)} \text { as } E_i \& E_j ; i \neq j, \text { are mutually exclusive events } \\ & =\frac{P\left(E \cap\left(E_1 \cup E_2 \cup E_3\right)\right.}{P(E)}=\frac{P(E \cap S)}{P(E)}=\frac{P(E)}{P(E)}=1 ; \text {; } S \text { ' being the sample space ) }\end{aligned}$

1. (b) $\frac{5}{6^2}$

Solution:

P(X = 2) = (Probability of not getting six at first throw) × (Probability of getting six at second throw)

$=\frac{5}{6}\times\frac{1}{6}=\frac{5}{6^2}$

2. (c) $\frac{5^3}{6^4}$

Solution:

$\text{P}\text{(X}=4)=\frac{5}{6}\times\frac{5}{6}\times\frac{5}{6}\times\frac{1}{6}=\frac{5^3}{6^4}$

3. (c) $\frac{5}{6}$

Solution:

$\text{P}(\text{X}\ge2)=1-\text{P}(\text{X}<2)$

$=1-\text{P}(\text{X}=1)=1-\frac{1}{6}=\frac{5}{6}$

4. (a) $\frac{5^5}{6^5}$

Solution:

$\text{P}(\text{X}\ge6)=(\frac{5}{6})^5\times\frac{1}{6}+(\frac{5}{6})^6\times\frac{1}{6}+...\infty$

$=\frac{5^5}{6^6}[1+\frac{5}{6}+(\frac{5}{6})^2+...\infty]=\frac{5^5}{6^6}[\frac{1}{1\frac{5}{6}}]=(\frac{5}{6})^5$

5. (d) $\frac{5^3}{6^3}$

Solution:

$\text{P}\text{(X}\ge4)=(\frac{5}{6})^3.\frac{1}{6}+(\frac{5}{6})^4.\frac{1}{6}+.....$

$=\frac{5^3}{6^4}[1+(\frac{5}{6})+(\frac{5}{6})^2+....]$

$=\frac{5^3}{6^4}[\frac{1}{1-(\frac{5}{6})}]=(\frac{5^3}{6^4}).6=\frac{5^3}{6^3}$

$\therefore\text{P}(\text{B})=\frac{12}{35},\text{P}(\text{R})=\frac{8}{35},\text{P}(\text{Y})=\frac{10}{35}$ and $\text{P}(\text{G})=\frac{5}{35}$

1. (c) $\frac{6}{119}$ 

Solution:

$\text{P}(\text{G}\cap\text{B})=\text{P}(\text{B}).\text{P}(\text{G}|\text{B})=\frac{12}{35}.\frac{5}{34}=\frac{6}{119}$

2. (b) $\frac{8}{119}$ 

Solution:

$\text{P}(\text{R}\cap\text{Y})=\text{P}(\text{Y}).\text{P}(\text{R}|\text{Y})=\frac{10}{35}.\frac{8}{34}=\frac{8}{119}$

3. (a) $\frac{4}{85}$ 

Solution:

Let E = event of drawing a first red ball and F = event of drawing a second red ball

Here, $\text{P}(\text{E})=\frac{8}{35}$ and $\text{P}(\text{E})=\frac{7}{34}$

$\therefore\text{P}(\text{F}\cap\text{E})=\text{P(E)}.\text{P(F|E)}=\frac{8}{35}.\frac{7}{34}=\frac{4}{85}$

4. (c) $\frac{12}{119}$ 

Solution:

$\text{P}(\text{Y}'\cap\text{G})=\text{P}\text{(G)}.\text{(Y}'|\text{G)}=\frac{5}{35}.\frac{24}{34}=\frac{12}{119}$

5. (d) $\frac{253}{595}$ 

Solution:

Let E = event of drawing a first non-blue ball and F = event of drawing a second non-blue ball,

Here, $\text{P}(\text{F}\cap\text{E})=\text{P}(\text{E}).\text{P}(\text{F}|\text{E})=\frac{23}{35}.\frac{22}{34}=\frac{253}{595}$

$\text{P}(\text{A}\cup\text{B}|\text{C})=\frac{\text{P(A}\cup\text{B)}\cap\text{C}}{\text{P(C)}}=\frac{\frac{3}{6}}{\frac{3}{6}}=1$ 

Sample space is given by,

S = {BBBB, BBBG, BBGB, BGBB, BBGG, BGBG, BGGB, BGGG,GBBB,GBBG,GBGB,GBGG,GGBB,GGBG, GGGB,GGGG}

1. (d) None of these.

Solution:

Let E = All children are girls.

$\therefore$ E = ( GGGG} i.e., n(E) = 1

F = Elder child is a boy

$\therefore$ F = {BBBB, BBBG, BBGB, BGBB, BBGG, BGBG, BGGB, BGGG) i.e., n(F) = 8

Now, $\text{n(E}\cap\text{F)}=0$

$\therefore\text{P(E}|\text{F)}=\frac{\text{E}\cap\text{F}}{\text{n(F})}=0$

2. (a) $\frac{1}{4}$

Solution:

Let E = All are boys.

$\therefore$ E = (BBBB) i.e., n(E) = 1

F = Two elder children are boys

$\therefore$ F = {BBBB, BBBG, BBGB, BBGG} i.e., n(F) = 4

Now, $\text{n}(\text{E}\cap\text{F})=1$

$\therefore\text{P}(\text{E}|\text{F})=\frac{\text{n(E}\cap\text{F)}}{\text{n(F)}}=\frac{1}{4}$

3. (c) $\frac{1}{4}$

Solution:

Let E = Two middle children are boys.

$\therefore$ E = (BBBB, BBBG, GBBB, GBBG) i.e., n(E) = 4

F = Eldest child is a girl

$\therefore$ F = ( GBBB, GBBG, GBGB, GBGG, GGBB, GGBG, GGGB, GGGG) i.e., n(F) = 8

Now, $\text{n}(\text{E}\cap\text{F})=2$

$\therefore\text{n(E}|\text{F})=\frac{2}{8}=\frac{1}{4}$

4. (b) $\frac{1}{5}$

Solution:

Let E = All are boys.

E = {BBBB} i.e., n{E} = 1

F = At most one child is girl.

$\therefore$ F = (BBBB, BBBG, BBGB, BGBB, GBBB)

i.e., n(F) = 5

New, $\text{n(F}\cap\text{F})=1$

$\therefore\text{n(E}|\text{F})=\frac{1}{5}$

5. (a) $\frac{1}{5}$

Solution:

Let E = All are boys.

E = (BBBB) i.e., n(E) = 1

F = At least three of the children are boys.

F = {BBBB, BBBG, BBGB, BGBB, G BBB) i.e., n (F) = 5

Now, $\text{n(E}\cap\text{F})=1$

$\therefore\text{P(E}|\text{F})=\frac{1}{5}$ 

1. (c) 0.3

Solution:

It is given that if India loose any match, then the probability that it wins the next match is 0.3.

$\therefore$ Required probability = 0.3

2. (d) 0.7

Solution:

It is given that, if India loose any match, then the probability that it wins the next match is 0.3.

$\therefore$ Required probability = 1 - 0.3 = 0.7

3. (b) 0.28

Solution:

Required probability = P(India losing first match).

P(India losing second match when India has already lost first match)

= 0.4 × 0.7 = 0.28

4. (d) 0.096

Solution:

Required probability = P(India winning first match) · P(India winning second match if India has already won first match).

 P(India winning third match if india has already won first two matches)

= 0.6 × 0.4 × 0.4 = 0.096

5. (c) 0.408

Solution:

Required probability = P(Win 1^st match) P(Lose 2^nd match) P(Lose 3^rd match) + P (Lose 1^st match) P(Win 2^nd match) P(Lose 3^rd match) + P(Lose 1^st match) P(Lose 2^nd match) P(Win 3^rd match),

= 0.6 × ( 1 - 0.4).(1 - 0.3) + (1 - 0.6)(0.3) (1 - 0.4) + (1 - 0.6) (1 - 0.3) (0.3)

= 0.6 × 0.6 × 0. 7 + 0.4 × 0.3 × 0.6 + 0.4 × 0.7 × 0.3

= 0.252 + 0.072 + 0.084 = 0.408

1. (a) 0.1

Solution:

We know that $\sum\text{p}_\text{i}=1$

Then 0.2 + k + 2k + 3k + 2k + 0 = 1

⇒ 8k = 1 - 0.2 = 0.8 ⇒ k = 0.1

2. (b) 0.3

Solution:

P(Average study time is not more than 1 hour)

$=\text{P}\text{(X}\leq1)$

= P(X = 0) + P(X = 1) = 0.2 + 0.1 = 0.3

3. (a) 0.5

Solution:

P(Average study time is at least 3 hours)

$=\text{P}\text{(X}\leq1)$

= P(X = 3) + P(X = 4) = 0.3 + 0.2 = 0.5

4. (d) 0.2

Solution:

P(Average study time is exactly 2 hours)

= P(X = 2) = 0.2

5. (c) 0.8

Solution:

P(Average study time is at least 1 hour)

= 1 - P(X = 0) = 1 - 0.2 = 0.8

1. (b) 0.3

Solution:

Since, it is given that during prime time husband is watching T.V. 70% of the time,

Required probability,

= 1 - P(husband is watching television during prime time)

= I - 0.7 = 0.3

2. (b) $\frac{7}{11}$

Solution:

Let H be the event that husband is watching T.V., W be the event that wife is watching T. V.

$\text{P}(\text{V})=0.7,\text{P}(\overline{\text{H}})=0.3$

$\therefore$ Required probability $=\text{P}(\text{H}|\text{W})$

$=\frac{\text{P}\text{(H)}.\text{P}(\text{W}|\text{H})}{\text{P}\text{(H}).\text{P}\text{(W}|\text{H)}+\text{P}(\text{H}})\text{P}(\text{W}|\overline{\text{H}})$

$=\frac{0.7\times0.3}{0.7\times0.3+0.4\times0.3}=\frac{0.21}{0.33}=\frac{7}{11}$ 

3. (a) 0.21

Solution:

Required probability $\text{P}(\text{H}\cap\text{W})$

= P(H) P(W | H) = 0.7 × 0.3 = 0.21

4. (b) 0.33

Solution:

Required probability $=\text{P(W)}=\frac{\text{P(H}\cap\text{W)}}{\text{P(H|}\text{W)}}$

$=\frac{0.21}{\frac{7}{11}}=\frac{21}{100}\times\frac{11}{7}=\frac{33}{100}=0.33$

5. (b) $\frac{4}{11}$

Solution:

Required probability $=\text{P}(\overline{\text{H}}|\text{W)}$

$=1-\text{P}(\text{H)}|\text{W})=1-\frac{7}{11}=\frac{4}{11}$

1. (a) $\frac{1}{5}$

Solution:

Clearly, P(T₂ winning a match against T₁)

= P(T₁ losing) $=\frac{1}{5}$

2. (d) $\frac{3}{10}$

Solution:

Clearly, P(T₂ drawing a match against T₁)

= P(T₁ drawing) $=\frac{3}{10}$

3. (d) $\frac{11}{20}$

Solution:

According to given information, we have the following possibilities for the values of X and Y.

Now, P(X > Y) = P(X = 4, Y = 0) + P(X = 3, Y = 1)

= P(T₁ win) P(T₁ win) + P(T₁ win) P(match draw) + P(match draw) P(T₁ win)

$=\frac{1}{2}\times\frac{1}{2}+\frac{1}{2}\times\frac{3}{10}+\frac{3}{10}\times\frac{1}{2}$

$=\frac{5+3+3}{20}=\frac{11}{20}$

4. (c) $\frac{29}{100}$

Solution:

P(X = Y) = P(X = 2, Y = 2)

= P(T₁ win) P(T₂ win) + P(T₂ win) P(T₁ win) + P(match draw) P(match draw)

$=\frac{1}{2}\times\frac{1}{5}+\frac{1}{5}\times\frac{1}{5}+\frac{3}{10}\times\frac{3}{10}$

5. (a) $0$

Solution:

From the given information, it is clear that maximum sum of X and Y can be 4, therefore P(X + Y = 8) = 0.

$\therefore\text{P}(\text{E})=\frac{50}{100}=\frac{1}{2},\text{P}(\text{M})=\frac{35}{100}=\frac{7}{20}$

1. (c) 0.48

Solution:

$\text{P}(\text{A}\cap\text{B})=\text{P}(\text{A}).\text{P}(\text{B})=(0.6)(0.8)=0.48$

2. (a) 0.92

Solution:

$\text{P}(\text{A}\cup\text{B})=\text{P}(\text{A}).\text{P}(\text{B})-\text{P(A}\cap\text{B})\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =0.6+0.8-0.48=0.92$

3. (d) 0.8

Solution:

P(B | A) = P(B) ($\because$ A and B are independent) = 0.8

4. (a) 0.6

Solution:

P(A | B) = P(A) ($\because$ A and B are independent) = 0.6

5. (c) 0.08

Solution:

P(not A and B) $=\text{P(A}'\cap\text{B}')=\text{P}(\text{A}\cup\text{B})'$

$=1-\text{P(A}\cup\text{B})=1-0.92=0.08$

1. (b) $\frac{2}{7}$

Solution:

P(A₂ | E) =Probability that the doctor arriving late and he comes by metro,

$=\frac{\text{P(A}_2)\text{P(E}|\text{A}_2)}{\sum\text{P}(\text{A}_i)\text{P}(\text{E}|\text{A}_\text{i})}$

$=\frac{(0.2)(0.3)}{(0.3)(0.25)+(0.2)(0.3)+(0.1)(0.35)+(0.4)(0.1)}$

$=\frac{0.06}{0.21}=\frac{2}{7}$

2. (c) $\frac{5}{14}$

Solution:

P(A₁ | E) = Probability that the doctor arriving late and he comes by cab,

$=\frac{\text{P(A}_1)\text{P(E}|\text{A}_1)}{\sum\text{P}(\text{A}_i)\text{P}(\text{E}|\text{A}_\text{i})}$

$=\frac{(0.3)(0.25)}{(0.3)(0.25)+(0.2)(0.3)+(0.1)(0.35)+(0.4)(0.1)}$

$=\frac{0.075}{0.21}=\frac{5}{14}$

3. (d) $\frac{1}{6}$

Solution:

P(A₃ | E) = Probability that the doctor arriving late and he comes by bike,

$=\frac{\text{P(A}_3)\text{P(E}|\text{A}_3)}{\sum\text{P}(\text{A}_i)\text{P}(\text{E}|\text{A}_\text{i})}$

$=\frac{(0.1)(0.35)}{(0.3)(0.25)+(0.2)(0.3)+(0.1)(0.35)+(0.4)(0.1)}$

$=\frac{0.035}{0.21}=\frac{1}{6}$

4. (c) $\frac{4}{21}$

Solution:

P(A₄ | E) = Probability that the doctor arriving late and he comes by other means of transport,

$=\frac{\text{P(A}_4)\text{P(E}|\text{A}_4)}{\sum\text{P}(\text{A}_i)\text{P}(\text{E}|\text{A}_\text{i})}$

$=\frac{(0.4)(0.1)}{(0.3)(0.25)+(0.2)(0.3)+(0.1)(0.35)+(0.4)(0.1)}$

$=\frac{0.04}{0.21}=\frac{4}{21}$

5. (a) 1

Solution:

Probability that the doctor is late by any means,

$=\frac{2}{7}+\frac{5}{14}+\frac{1}{6}+\frac{3}{21}=1$

1. (b) 0.04

Solution:

Required probability = P(A|E₂)

$=\frac{\text{P}(\text{A}\ \cap\ \text{E}_2)}{\text{P}(\text{E}_2)}$

$=\frac{\Big(0.04\times\frac{20}{100}\Big)}{\Big(\frac{20}{100}\Big)}=0.04$

2. (c) 0.008

Solution:

Required probability $=\text{P}(\text{A}\ \cap\ \text{E}_2)$

$=0.04\times\frac{20}{100}=0.008$

3. (b) 0.047

Solution:

Total probability is given by

P(A) = P(E₁) × P(A|E₁) + P(E₂) × P(A|E₂) + P(E₃) × P(A|E₃).

$=\frac{50}{100}\times0.06+\frac{20}{100}\times0.04+\frac{30}{100}\times0.03$

$=\ 0.047$

4. (d) $\frac{17}{47}$

Solution:

Using Bayes' theorem, we have

$\text{P}(\text{E}_1\ |\ \text{A})=\frac{\text{P}(\text{E}_1)\times\text{P}(\text{A}\ |\ \text{E}_1)}{\text{P}(\text{E}_1)\ \times\ \text{P}(\text{A}\ |\ \text{E}_1)\ +\ \text{P}(\text{E}_2)\ \times\ \text{P}(\text{A}\ |\ \text{E}_2)\ +\ \ \text{P}(\text{E}_3)\ \times\ \text{P}(\text{A}\ |\ \text{E}_3)}$

$=\frac{0.5\times0.06}{0.5\times0.06+0.2\times0.04+0.3\times0.03}=\frac{30}{47}$

$\therefore$ Required probability $=\ \text{P}(\bar{\text{E}\ |\ \text{A}})$

$=1-\text{P}(\text{E}_1\ |\ \text{A})=1-\frac{30}{47}=\frac{17}{47}$

5. (d) 1

Solution:

$\sum\limits^3_\text{i=1}\ \text{P}(\text{E}_\text{i}\ |\ \text{A})=\text{P}(\text{E}_1\ |\ \text{A})\ +\ \text{P}(\text{E}_2\ |\ \text{A})\ +\ \text{P}(\text{E}_3\ |\ \text{A})\ =1$

[ $\therefore$ Sum of posterior probabilities is 1)

1. (b) $\frac{12}{49}$

Solution:

Total number of tickets = 50

Let event A = First ticket shows even number and B = Second ticket shows even number,

Now, P(Both tickets show even number) $=\text{P(A)}.\text{P(B}|\text{A)}=\frac{25}{50}.\frac{24}{49}=\frac{12}{49}$

2. (c) $\frac{12}{49}$

Solution:

Let the event A = First ticket shows odd number and B = Second ticket shows odd number,

P(Both tickets show odd number) $=\frac{25}{50}\times\frac{24}{49}=\frac{12}{49}$

3. (c) $\frac{24}{245}$

Solution:

Required probability = P(one number is a multiple of 4 and other is a multiple of 5).

= P(multiple of 5 on first ticket and multiple of 4 on second ticket) + P(multiple of 4 on first ticket and multiple of 5 on second ticket).

$=\frac{10}{50}.\frac{12}{49}+\frac{12}{50}\times\frac{10}{49}=\frac{12}{245}=\frac{24}{245}$

4. (d) $\frac{36}{245}$

Solution:

Required probability = P(one ticket with prime number and other ticket with a multiple of 4),

$=2(\frac{15}{50}\times\frac{12}{49})=\frac{36}{245}$

5. (b) $\frac{25}{98}$

Solution:

Let the event A = First ticket shows even number and B = Second ticket shows odd number.

Now, P(First ticket shows an even number and second ticket shows an odd number) $=\text{P(A)}.\text{P}(\text{B}|\text{A})=\frac{25}{50}\times\frac{25}{49}=\frac{25}{98}$

$\therefore\text{n(S)}=24$

1. (b) 0.5

Solution:

We have, $\text{P(A}\cup\text{B}\cup\text{C})=0.5$

$\Rightarrow1-\text{P}(\text{A}\cup\text{B}\cup\text{C})=0.5$

$\Rightarrow1-\text{P}(\bar{\text{A}}\cap\bar{\text{B}}\cap\bar{\text{C}})=0.5$

$\Rightarrow1-\text{P}(\bar{\text{A}}).\text{P}(\bar{\text{B}}).\text{P}(\bar{\text{C}})=0.5$

$\Rightarrow1-(1-\text{a})(1-\text{b})(1-\text{c})=0.5$

$\Rightarrow1-[(1-\text{a}-\text{b}+\text{ab})(1-\text{c})]=0.5$

$\Rightarrow1-[1-\text{c}-\text{a}+\text{ac}-\text{b}+\text{bc}+\text{ab}-\text{abc}]=0.5$

$\Rightarrow\text{a}+\text{b}+\text{c}-\text{ab}-\text{bc}-\text{ca}+\text{abc}=0.5\ ...(\text{i})$

2. (c) 0.4

Solution:

We have, P(selection in at least two competitive exams) = 0.4

$\Rightarrow\text{P}(\text{A}\cap\text{B}\cap\bar{\text{C}})+\text{P}(\bar{\text{A}}\cap\text{B}\cap\text{C})+\text{P}(\text{A}\cap\bar{\text{B}}\cap\text{C})+\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{P}(\text{A}\cap\text{B}\cap\text{C})=0.4$

$\Rightarrow\text{ab}(1-\text{c})+(1-\text{a})\text{bc}+\text{a}(1-\text{b})\text{c}+\text{abc}=0.4$

$\Rightarrow\text{ab}-\text{abc}+\text{bc}-\text{abc}+\text{ac}-\text{abc}+\text{abc}=0.4$

$\Rightarrow\text{ab}+\text{bc}+\text{ac}-2\text{abc}=0.4\ ...(\text{ii})$

3. (a) 0.1

Solution:

We have, P(selection in exactly two examinations) = 0.3

$\Rightarrow\text{P}(\text{A}\cap\text{B}\cap\bar{\text{C}})+\text{P}(\bar{\text{A}}\cap\text{B}\cap\text{C})+\text{P}(\text{A}\cap\bar{\text{B}}\cap\text{C})=0.3$

$\Rightarrow\text{ab}(1-\text{c})+(1-\text{a})\text{bc}+\text{a}(1-\text{b})\text{c}=0.3$

$\Rightarrow\text{ab}+\text{bc}+\text{ac}-3\text{abc}=0.3\ ...(\text{iii})$

On subtracting (iii) from (ii), we get

abc = 0.1

4. (b) 0.6

Solution:

On substituting the value of abe in (ii), we get ab + be + ac = 0.6

5. (a) 1

Solution:

On substituting the value of ab+ bc + ac and abe in (i), we get

a + b + e - 0.6 + 0.1 = 0.5

⇒ a + b + c = 1

1. (b) $\frac{22}{425}$

Solution:

Required probability $=\frac{^{12}\text{C}_2}{^{51}\text{C}_2}$

$=\frac{12\times11}{51\times50}=\frac{22}{425}$

2. (a) $\frac{26}{425}$

Solution:

Required probability $=\frac{^{13}\text{C}_2}{^{51}\text{C}_2}$

$=\frac{12\times13}{51\times50}=\frac{26}{425}$

3. (b) 0.24

Solution:

Clearly, $=\frac{^{12}\text{C}_2}{^{51}\text{C}_2}=\frac{22}{425}$

$\text{P}(\text{A}|\text{E}_2)=\frac{^{13}\text{C}_2}{^{51}\text{C}_2}=\frac{26}{425}$

$\text{P}(\text{A}|\text{E}_3)=\text{P}(\text{A}|\text{E}_4)=\frac{26}{425}$

$\therefore\displaystyle\sum_{\text{i}=1}^{4}\text{P(A|E}_\text{i})=\frac{22}{425}+\frac{26}{425}+\frac{26}{425}+\frac{26}{425}=\frac{100}{425}=0.24$

4. (b) $\frac{1}{221}$

Solution:

P(getting both king) $=\frac{^{4}\text{C}_2}{^{52}\text{C}_2}=\frac{4\times3}{52\times51}=\frac{1}{221}$

5. (a) $\frac{144}{169}$

Solution:

P(drawing a king) $=\frac{4}{52}=\frac{1}{13}$

$\therefore$ P(not drawing a king) $=1-\frac{1}{13}=\frac{12}{13}$

$\therefore$ Required probability $=\frac{12}{13}\times\frac{12}{13}=\frac{144}{169}$

1. (d) $\frac{1}{61}$

Solution:

Since, it rained only 6 days each year, therefore, probability that it rains on chosen day is,

$\frac{6}{366}=\frac{1}{61}$

2. (c) $\frac{360}{366}$

Solution:

The probability that it does not rain on chosen day $=1-\frac{1}{61}=\frac{60}{61}=\frac{360}{366}$

3. (c) $\frac{4}{5}$

Solution:

It is given that, when it actually rains, the weather man correctly forecasts rain 80% of the time.

$\therefore\text{}$ Required probability $=\frac{80}{100}=\frac{8}{10}=\frac{4}{5}$

4. (a) 0.0625

Solution:

Let A₁ be the event that it rains on chosen day, A₂ be the event that it does not rain on chosen day and E be the event the weatherman predict rain.

Then we have, $\text{P(A}_2)=\frac{6}{366},\text{P(A}_2)=\frac{360}{366},$

$\text{P(E}|\text{A}_1)=\frac{8}{10}$ and $\text{P(E}|\text{A}_2)=\frac{2}{10}$

Required probability = P(A₁ | E)

$=\frac{\text{P(A}_1).\text{P(E}|\text{A}_1)}{\text{P(A}_1).\text{P(E}|\text{A}_1)+\text{P(A}_2).\text{P(E}|\text{A}_2)}$

$=\frac{\frac{6}{366}\times\frac{8}{10}}{\frac{6}{366}\times\frac{8}{10}+\frac{360}{366}\times\frac{2}{10}}=\frac{48}{768}=0.0625$

5. (a) 0.94

Solution:

Required probability =1 - P(A₁ | E)

$= 1 - 0.0625 = 0.9375\approx0.94$

Then $\text{P}(\text{E}_1)=\frac{30}{100}=\frac{3}{10},\text{P}(\text{E}_2)=\frac{20}{100}=\frac{1}{5}$

and $\text{P}(\text{E}_3)=\frac{50}{100}=\frac{1}{2}$

Also, let E be the event of drawing a defective bolt.

Then $\text{P}(\text{E}|\text{E}_1)=\frac{5}{100}=\frac{1}{20},\text{P}(\text{E}|\text{E}_2)=\frac{2}{100}=\frac{1}{50},$

$\text{P}(\text{E}|\text{E}_1)=\frac{5}{100}=\frac{1}{20},\text{P}(\text{E}|\text{E}_2)=\frac{2}{100}=\frac{1}{50},$

$\text{P}(\text{E}|\text{E}_3)=\frac{4}{100}=\frac{1}{25}$

1. (b) $\frac{5}{13}$

Solution:

Probability that the defective bolt is manufactured by machine A = P(E₁ | E)

$=\frac{\text{P}(\text{E}|\text{E}_1)\times\text{P}(\text{E}_1)}{\text{P}(\text{E}|\text{E}_1).\text{P}(\text{E}_1)+\text{P}(\text{E}|\text{E}_2).\text{P}(\text{E}_2)+\text{P}(\text{E}|\text{E}_3).\text{P}(\text{E}_3)}$

[Using Bayes' Theorem]

$=\frac{\frac{1}{20}\times\frac{3}{10}}{\frac{1}{20}\times\frac{3}{10}+\frac{1}{50}\times\frac{1}{5}+\frac{1}{25}\times\frac{1}{2}}=\frac{3}{200}\times\frac{1000}{39}=\frac{5}{13}$

2. (b) 0.1

Solution:

Probability that the defective bolt is manufactured by machine B = P(E₂ | E)

$=\frac{\text{P}(\text{E}|\text{E}_2)\times\text{P}(\text{E}_2)}{\text{P}(\text{E}|\text{E}_1).\text{P}(\text{E}_1)+\text{P}(\text{E}|\text{E}_2).\text{P}(\text{E}_2)+\text{P}(\text{E}|\text{E}_3).\text{P}(\text{E}_3)}$

$=\frac{\frac{1}{50}\times\frac{1}{5}}{(\frac{1}{20}\times\frac{3}{10})+(\frac{1}{50}\times\frac{1}{5})+(\frac{1}{25}\times\frac{1}{2})}$

$=\frac{1}{250}\times\frac{1000}{39}=\frac{4}{39}\approx0.1$

3. (c) $\frac{20}{39}$

Solution:

Probability that the defective bolt is manufactured by machine C = P(E₃ | E)

$=\frac{\text{P}(\text{E}|\text{E}_3)\times\text{P}(\text{E}_3)}{\text{P}(\text{E}|\text{E}_1).\text{P}(\text{E}_1)+\text{P}(\text{E}|\text{E}_2).\text{P}(\text{E}_2)+\text{P}(\text{E}|\text{E}_3).\text{P}(\text{E}_3)}$

$=\frac{\frac{1}{25}\times\frac{1}{2}}{(\frac{1}{20}\times\frac{3}{10})+(\frac{1}{50}\times\frac{1}{5})+(\frac{1}{25}\times\frac{1}{2})}$

$=\frac{1}{50}\times\frac{1000}{39}=\frac{20}{39}$

4. (a) $\frac{35}{39}$

Solution:

Probability that the defective bolt is not manufactured by machine B i.e., it is manufactured by machine A or C $=\frac{15}{39}+\frac{20}{39}=\frac{35}{39}$

5. (c) 0.5

Solution:

Required probability $=\frac{15}{39}+\frac{4}{39}=\frac{19}{39}\approx0.5$