Maths Solve the Following Question.(3 Marks)

event B: Bulb manufactured by machine B

event C: Bulb manufactured by machine C

event D: Bulb defective

$\begin{aligned} & \therefore P(A)=\frac{25}{100} \\ & P(B)=\frac{35}{100} \\ & P(C)=\frac{40}{100}\end{aligned}$

Machines A, B and C manufacture respectively 25%, 35% and 40% of the bulbs. Of their outputs, 5, 4, and 2 percent are respectively defective bulbs.

$P ( D / A )=\frac{5}{100}, P ( D / B )=\frac{4}{100}$

$P ( D / C )=\frac{2}{100}$

$\begin{aligned} & P(D)=P(A) P(D / A)+P(B) P(D / B) \\ &+P(C) P(D / C)\end{aligned}$

$=\frac{25}{100} \times \frac{5}{100}+\frac{35}{100} \times \frac{4}{100}+\frac{40}{100} \times \frac{2}{100}$

$=\frac{1}{100 \times 100}(125+140+80)=\frac{345}{100 \times 100}$

Required probability = P(B/D)

By Bayes’ theorem,

$P ( B / D )=\frac{ P ( B ) P ( D / B )}{ P ( D )}=\frac{\frac{35}{100} \times \frac{4}{100}}{\frac{345}{100 \times 100}}=\frac{28}{69}$

event $B_2$ : Selecting box II having two silver coins

event $B_3$ : Selecting box III having one silver and one gold coin,

event G: Coin is gold.

$P \left( B _1\right)= P \left( B _2\right)= P \left( B _3\right)=\frac{1}{3}$

$\begin{aligned} & P \left( G / B _1\right)=1, P \left( G / B _2\right)=0, P \left( G / B _3\right)=\frac{1}{2} \\ & P ( G )= P _{(}\left( B _1\right) P \left( G / B _1\right)+ P \left( B _2\right) P \left( G / B _2\right)\end{aligned}$

$+ P \left( B _3\right) P \left( G / B _3\right)$

$=\frac{1}{3}\left[1+0+\frac{1}{2}\right]=\frac{1}{3}\left(\frac{3}{2}\right)=\frac{1}{2}$

To find the probability that the other can in the box is also gold. Which is possible only when it is drawn from the box I.

∴ Required probability = P(B1/G)

By Bayes’ theorem,

$P\left(B_1 / G\right)=\frac{P\left(B_1\right) P\left(G / B_1\right)}{P(G)}$

$=\frac{\frac{1}{3}(1)}{\frac{1}{2}}=\frac{2}{3}$

event S: The event that machine produces a slightly defective part,

event D: The event that machine produces an obviously defective part.

$P ( G )=\frac{90}{100}=0.90, P ( S )=\frac{2}{100}=0.02$,

$P(D)=\frac{8}{100}=0.08$

Let $D^{ c }= G \cup S$. Then

$P \left( G / D ^{ c }\right)=\frac{ P \left( G \cap D ^{ c }\right)}{ P \left( D ^{ c }\right)}$

$=\frac{P(G)}{P(G \cup S)} \ldots[\because G \cap(G \cup S)=G]$

$=\frac{ P ( G )}{ P ( G )+ P ( S )}$

$\ldots[\because G$ and $S$ are disjoint sets $]$

$\begin{aligned} & =\frac{0.90}{0.90+0.02} \\ & =\frac{0.90}{0.92}=\frac{90}{92}=\frac{45}{46}\end{aligned}$

$\therefore$ The probability of a year being a leap year $=\frac{1}{4}$

$\therefore$ Probability of a year being a non-leap year $=1-\frac{1}{4}=\frac{3}{4}$

In a non-leap year, there are 52 weeks and one extra day, whereas a leap year has 52 weeks and 2 extra days.

$\therefore 53$ rd Wednesday's chance in a non-leap year $=\frac{1}{7}$

Two extra days of a leap year can be (Mon, Tue), (Tue, Wed), (Wed, Thu), (Thu, Fri), (Fri, Sat), (Sat, Sun), (Sun, Mon)

∴ There are 2 possibilities of 53rd Wednesday in a leap year.

$\therefore 53$ rd Wednesday's chance in a leap year $=\frac{2}{7}$

Required probability = P(a non-leap year and Wednesday) + P(a leap year and Wednesday)

$\begin{aligned} & =\frac{3}{4} \times \frac{1}{7}+\frac{1}{4} \times \frac{2}{7} \\ & =\frac{5}{28}\end{aligned}$

∴ P(A ∩ B) = P(A) × P(B)

(i) P(A ∪ B) = P(A) + P(B) – P(A ∩ B)

∴ P(A ∪ B) = P(A) + P(B) – P(A) × P(B)

∴ 2P(B) – P(A) = P(A) + P(B) – P(A) × P(B) ……[∵ P(A ∪ B) = 2P(B) – P(A)]

$\begin{aligned} & \therefore 2 P(B)-\frac{1}{4}=\frac{1}{4}+P(B)-\frac{1}{4} \times P(B) \\ & \therefore 2 P(B)-P(B)+\frac{1}{4} P(B)=\frac{1}{4}+\frac{1}{4}\end{aligned}$

ii. $\quad \begin{aligned} P(A / B) & =\frac{P(A \cap B)}{P(B)} \\ \therefore \quad P(A / B) & =\frac{P(A) \times P(B)}{P(B)} \\ & =P(A)=\frac{1}{4}\end{aligned}$

iii. $\begin{aligned} P\left(B^{\prime} / A\right) & =\frac{P\left(B^{\prime} \cap A\right)}{P(A)} \\ & =\frac{P\left(B^{\prime}\right) \times P(A)}{P(A)} \\ & =P\left(B^{\prime}\right) \\ & =1-P(B) \\ & =1-\frac{2}{5}=\frac{3}{5}\end{aligned}$

Let $P\left(A^{\prime}\right)=P(A$ does not solve the problem $)=\frac{4}{4+3}=\frac{4}{7}$

So, the probability that A solves the problem = P(A) = 1 – P(A’)

$\begin{aligned} & =1-\frac{4}{7} \\ & =\frac{3}{7}\end{aligned}$

Similarly, let P(B) = P(B solves the problem) Since odds in favour of B solving the problem are 7 : 5.

$\therefore P ( B )=\frac{7}{7+5}=\frac{7}{12}$

Required probability P(A ∪ B) = P(A) + P(B) – P(A ∩ B)

Since A and B are independent events.

∴ P(A ∩ B) = P(A) . P(B)

∴ Required probability = P(A) + P(B) – P(A) . P(B)

$\begin{aligned} & =\frac{3}{7}+\frac{7}{12}-\frac{3}{7} \times \frac{7}{12} \\ & =\frac{16}{21}\end{aligned}$

In the words ASSISTANT, there are 9 letters out of which 2 letters are ‘A’, and in the word STATISTICS, there are 10 letters, out of which 1 letter is A.

$\therefore$ Probability of choosing A from both the letters $=\frac{{ }^2 C_1}{{ }^9 C_1} \times \frac{{ }^1 C_1}{{ }^{10} C_1}=\frac{2}{9} \times \frac{1}{10}=\frac{1}{45}$

Similarly,

Probability of choosing I from both the letters $=\frac{{ }^1 C _1}{{ }^9 C _1} \times \frac{{ }^2 C _1}{{ }^{10} C _1}=\frac{1}{9} \times \frac{2}{10}=\frac{1}{45}$

Probability of choosing $S$ from both the letters $=\frac{{ }^3 C _1}{{ }^9 C _1} \times \frac{{ }^3 C _1}{{ }^{10} C _1}=\frac{3}{9} \times \frac{3}{10}=\frac{1}{10}$

Probability of choosing $T$ from both the letters $=\frac{{ }^2 C_1}{{ }^9 C_1} \times \frac{{ }^3 C_1}{{ }^{10} C_1}=\frac{2}{9} \times \frac{3}{10}=\frac{1}{15}$

Required probability $=\frac{1}{45}+\frac{1}{45}+\frac{1}{10}+\frac{1}{15}=\frac{19}{90}$

$n ( S )=$ Total number of arrangements $=\frac{12 !}{2 ! 2 ! 2 ! 2 !}=\frac{12 !}{(2 !)^4}$

(i) A: Arrangement chosen at random begins with the letters EE. If the first and second places are filled with EE, there are 10 letters left in which 2A, 2N, 2R, G, M, T, S are there.

$\therefore n ( A )=\frac{10 !}{2 ! 2 ! 2 !}=\frac{10 !}{(2 !)^3}$

B: Consonants (G, M, T, S, 2N, 2R) are together.

2A, 2E, and the group containing consonants form total 5 units. Which can be arranged in

$\frac{5 !}{2 ! 2 !}$ ways.

Also, 8 consonants can be arranged among themselves in $\frac{8 !}{2 ! 2 !}$ ways.

$\begin{aligned} n ( B ) & =\frac{5 !}{2 ! 2 !} \times \frac{8 !}{2 ! 2 !} \\ P ( B ) & =\frac{ n ( B )}{ n ( S )} \\ & =\frac{5 !}{2 ! \times 2 !} \times \frac{8 !}{2 ! \times 2 !} \times \frac{(2 !)^4}{(12) !} \\ & =\frac{5 !}{12 \times 11 \times 10 \times 9}\end{aligned}$

$=\frac{1}{99}$

∴ n(S) = 11!

Let event A: There will be exactly 4 letters between R and E. R, E can occur at (1, 6), (2, 7), ….,(6, 11) positions.

So, there are 6 possibilities.

Also, R and E can interchange their positions.

So, R, E can be arranged in 2 × 6 = 12 ways.

Remaining 9 letters can be arranged in 9! ways.

∴ n(A) = 12 × 9!

$\therefore P ( A )=\frac{ n ( A )}{ n ( S )}=\frac{12 \times 9 !}{11 !}=\frac{12 \times 9 !}{11 \times 10 \times 9 !}=\frac{6}{55}$

$\therefore P(A) \cdot P(B)=\frac{1}{2} \ldots \ldots( i )$

B and C are independent events.

∴ P(B ∩ C) = P(B) . P(C)

$\therefore P(B) \cdot P(C)=\frac{1}{3} \ldots \ldots($ ii $)$

A and C are independent events.

∴ P(A ∩ C) = P(A) . P(C)

$\therefore P(A) \cdot P(C)=\frac{1}{6} \ldots \ldots$ (iii)

Dividing (i) by (ii), we get

$\frac{ P ( A ) P ( B )}{ P ( B ) P ( C )}=\frac{\frac{1}{2}}{\frac{1}{3}}$

$P(A)=\frac{3}{2} P(C) \ldots .$. (iv)

Substituting equation (iv) in (iii), we get

$\begin{aligned} & \frac{3}{2} P(C) \cdot P(C)=\frac{1}{6} \\ & {[P(C)]^2=\frac{1}{9}} \\ & \therefore P(C)=\frac{1}{3}\end{aligned}$

Substituting $P(C)=\frac{1}{3}$ in equation (ii), we get $P(B)=1$

Substituting $P(B)=1$ in equation (i), we get $P(A)=\frac{1}{2}$

∴ n(S) = 8!

There are 5 vowels and 3 consonants.

(i) A: all vowels are together we need to arrange (E, U, A, I, O), Q, T, N

Let us consider all vowels as one unit.

So, there are 4 units, which can be arranged in 4! ways.

Also, 5 vowels can be arranged among themselves in 5! ways.

∴ n(A) = 4! × 5!

Required probability = P(A)

$\begin{aligned} & =\frac{n(A)}{n(S)} \\ & =\frac{4 ! \times 5 !}{8 !} \\ & =\frac{1}{14}\end{aligned}$

B: arrangement start with a vowel and ends with a consonant.

First and last places can be filled in 5 and 3 ways respectively. Remaining 6 letters are arranged in 6! Ways.

∴ n(B) = 5 × 3 × 6!

Required probability = P(B)

$\begin{aligned} & =\frac{n(B)}{n(S)} \\ & =\frac{5 \times 3 \times 6 !}{8 !} \\ & =\frac{15}{56}\end{aligned}$

Since the odds in favour of A winning a game against B are 3 : 2,

the probability of occurrence of event A and B is given by

$P(A)=\frac{3}{3+2}=\frac{3}{5}$ and $P(B)=\frac{2}{3+2}=\frac{2}{5}$

Let event E: A wins at least two games out of three games.

∴ P(E) = P(A) . P(A) . P(B) + P(A) . P(B) . P(A) + P(B) . P(A) . P(A) + P(A) . P(A) . P(A)

$\begin{aligned} & =\frac{3}{5} \times \frac{3}{5} \times \frac{2}{5}+\frac{3}{5} \times \frac{2}{5} \times \frac{3}{5}+\frac{2}{5} \times \frac{3}{5} \times \frac{3}{5}+\frac{3}{5} \times \frac{3}{5} \times \frac{3}{5} \\ & =\frac{18}{125}+\frac{18}{125}+\frac{18}{125}+\frac{27}{125}\end{aligned}$

$=\frac{81}{125}$

$\begin{aligned} P \left( E ^{\prime}\right) & =1- P ( E ) \\ & =1-\frac{81}{125} \\ & =\frac{44}{125}\end{aligned}$

∴ Odds in favour of A’s winning at least two games out of three are P(E) : P(E’)

$\begin{aligned} & =\frac{81}{125}: \frac{44}{125} \\ & =81: 44\end{aligned}$

$P(A)=\frac{4}{7+4}=\frac{4}{11}$

Since odds against B are 5 : 3,

$P(B)=\frac{3}{5+3}=\frac{3}{8}$

Since only one of the events A, B and C can happen,

P(A) + P(B) + P(C) = 1

$\begin{aligned} & \frac{4}{11}+\frac{3}{8}+P(C)=1 \\ & \therefore P(C)=1-\left(\frac{4}{11}+\frac{3}{8}\right) \\ & =1-\left(\frac{32+33}{88}\right) \\ & =\frac{23}{88}\end{aligned}$

$\begin{aligned} & \therefore P\left(C^{\prime}\right)=1-P(C) \\ & =1-\frac{23}{88} \\ & =\frac{65}{88} \\ & \therefore \text { Odds against the event } C \text { are } P\left(C^{\prime}\right): P(C) \\ & =\frac{65}{88}: \frac{23}{88} \\ & =65: 23\end{aligned}$

$\begin{aligned} & =\frac{4}{4+3} \\ & =\frac{4}{7}\end{aligned}$

So, the probability that John solves the problem

$P(A)=1-P\left(A^{\prime}\right)=1-\frac{4}{7}=\frac{3}{7}$

Similarly, Let P(B) = P(Rafi solves the problem) Since the odds in favour of Rafi solving the problem are 7 to 5,

$P(B)=\frac{7}{7+5}=\frac{7}{12}$

Required probability P(A ∪ B) = P(A) + P(B) – P(A ∩ B)

Since A, B are independent events,

P(A ∩ B) = P(A) . P(B)

∴ Required probability = P(A) + P(B) – P(A) . P(B)

$\begin{aligned} & =\frac{3}{7}+\frac{7}{12}-\frac{3}{7} \times \frac{7}{12} \\ & =\frac{85}{84}-\frac{1}{4} \\ & =\frac{85-21}{84}\end{aligned}$

$=\frac{64}{84}=\frac{16}{21}$

Given that $P ( A )=\frac{2}{7}, P ( C )=\frac{3}{7}, P ( T )=\frac{2}{7}$

$P ( L / A )=\frac{1}{2}, P ( L / C )=\frac{1}{4}, P ( L / T )=\frac{1}{4}$

$\begin{aligned} & P ( L )= P ( A \cap L )+ P ( C \cap L )+ P ( T \cap L ) \\ & = P ( A ) \cdot P ( L / A )+ P ( C ) \cdot P ( L / C )+ P ( T ) \cdot P ( L / T )\end{aligned}$

$\begin{aligned} & =\frac{2}{7} \times \frac{1}{2}+\frac{3}{7} \times \frac{1}{4}+\frac{2}{7} \times \frac{1}{4} \\ & =\frac{1}{7}+\frac{3}{28}+\frac{2}{28}=\frac{9}{28}\end{aligned}$

$P ( C / L )=\frac{ P ( L \cap C )}{ P ( L ) M }$

$\begin{aligned} & =\frac{P(C) P(L / C)}{P(L)} \\ & =\frac{\frac{3}{7} \times \frac{1}{4}}{\frac{9}{28}}=\frac{1}{3}\end{aligned}$

∴ n(S) = 100.

Let event A: The child is sick with flu,

event B: The child is sick with measles,

event C: The child is sick with rash.

∴ n(A) = 80 and n(B) = 20

$\begin{aligned} & \therefore P ( A )=\frac{n(A)}{n(S)}=\frac{80}{100}=\frac{4}{5} \\ & P ( B )=\frac{n(B)}{n(S)}=\frac{20}{100}=\frac{1}{5}\end{aligned}$

Since the chances of having rashes, if the child is suffering from measles is 0.95 and the chances of having rashes if the child has flu is 0.08,

$\begin{aligned} & P(C / B)=0.95=\frac{95}{100} \text { and } \\ & P(C / A)=0.08=\frac{8}{100}\end{aligned}$

Required probability = P(B/C)

By Bayes’ theorem,

$\begin{aligned} P(B / C) & =\frac{P(B) \cdot P(C / B)}{P(A) \cdot P(C / A)+P(B) \cdot P(C / B)} \\ & =\frac{\left(\frac{1}{5}\right)\left(\frac{95}{100}\right)}{\left(\frac{4}{5}\right)\left(\frac{8}{100}\right)+\left(\frac{1}{5}\right)\left(\frac{95}{100}\right)}\end{aligned}$

$=\frac{95}{32+95}$

$=\frac{95}{127}=0.748$

event $J_2$ : Ball drawn from jar II.

$\begin{aligned} & P \left(J_1\right)= P (\text { head })=\frac{1}{2} \\ & P \left(J_2\right)= P (\text { tail })=\frac{1}{2}\end{aligned}$

Let event W: Ball drawn is white. In Jar I, there are total 12 balls, out of which 5 balls are white.

∴ Probability that the ball drawn is white under the condtion that it is drawn from Jar I.

$P \left( W / J_1\right)=\frac{{ }^5 C_1}{{ }^{12} C_1}=\frac{5}{12}$

Similarly, $P \left( W / J _2\right)=\frac{{ }^3 C_1}{{ }^{15} C_1}=\frac{3}{15}=\frac{1}{5}$

Required probability $= P \left( J _2 / W \right)$

By Bayes’ theorem,

$\begin{aligned} P \left( J _2 / W \right) & =\frac{ P \left( J _2\right) P \left( W / J _2\right)}{ P \left( J _1\right) P \left( W / J _1\right)+ P \left( J _2\right) P \left( W / J _2\right)} \\ & =\frac{\frac{1}{2} \times \frac{1}{5}}{\frac{1}{2} \times \frac{5}{12}+\frac{1}{2} \times \frac{1}{5}}\end{aligned}$

$=\frac{\frac{1}{5}}{\frac{25+12}{60}}=\frac{12}{37}$

$\therefore P\left(E_1\right)=P\left(E_2\right)=\frac{1}{2}$

$ \begin{aligned} & P\left(A / E_1\right)=\frac{P\left(A \cap E_1\right)}{P\left(E_1\right)}=0.2=\frac{2}{10} \\ & P\left(A / E_2\right)=\frac{P\left(A \cap E_2\right)}{P\left(E_2\right)}=0.3=\frac{3}{10} \end{aligned} $ By Bayes' theorem,

$\begin{aligned} P\left(E_1 / A\right) & =\frac{P\left(E_1\right) \cdot P\left(A / E_1\right)}{P\left(E_1\right) \cdot P\left(A / E_1\right)+P\left(E_2\right) \cdot P\left(A / E_2\right)} \\ & =\frac{\frac{1}{2} \cdot\left(\frac{2}{10}\right)}{\frac{1}{2}\left(\frac{2}{10}\right)+\frac{1}{2}\left(\frac{3}{10}\right)} \cdots\left[\because P\left(E_1\right)=P\left(E_2\right)\right]\end{aligned}$

$\begin{aligned} & =\frac{2}{2+3} M \\ & =\frac{2}{5}=0.4\end{aligned}$

∴ n(S) = 8 Let event A: Getting three heads.

∴ A = {HHH}

Let event B: Getting at least two heads.

∴ B = {HHT, HTH, THH, HHH}

∴ n(B) = 4

$\begin{aligned} & \therefore P ( B )=\frac{n(B)}{n(S)}=\frac{4}{8} \\ & \text { Now, } A \cap B =\{ HHH \} \\ & \therefore n ( A \cap B )=1 \\ & \therefore P ( A \cap B )=\frac{n(A \cap B)}{n(S)}=\frac{1}{8}\end{aligned}$

∴ Probability of getting three heads, given that at least two coins show heads, is given by

$\begin{aligned} & P(A / B)=\frac{P(A \cap B)}{P(B)} \\ & =\frac{\frac{1}{8}}{\frac{4}{8}} \\ & =\frac{1}{4}\end{aligned}$

Two balls are selected from $(9+m)$ balls in ${ }^{9+m} C_2$ ways.

$\therefore n (S)={ }^{9+m} C _2$

Let event A: The two balls selected are green.

$\therefore 2$ balls can be selected from $m$ balls in ${ }^{ m } C _2$ ways.

$\begin{aligned} & \therefore n(A)={ }^m C_2 \\ & \begin{aligned} P(A) & =\frac{n(A)}{n(S)} \\ \frac{1}{7} & =\frac{{ }^m C_2}{{ }^{9+m} C_2} \\ & =\frac{\frac{m !}{(m-2) ! 2 !}}{(9+m) !} \\ & =\frac{(9+m-2) ! 2 !}{(9+m)(8+m)(7+m) !}\end{aligned}\end{aligned}$

$\frac{1}{7}=\frac{m(m-1)}{(9+m)(8+m)}$

(9 + m)(8 + m) = 7m(m – 1)

72 + 9m + 8m + m2 = 7m2 – 7m

6m2 – 24m – 72 = 0

m2 – 4m – 12 = 0

(m – 6)(m + 2) = 0

m = 6 or m = -2

Since number of balls cannot be negative, m ≠ -2 ∴ m = 6

P(A ∪ B) = P(A) + P(B) – P(A ∩ B)

∴ P(A ∩ B) = P(A) + P(B) – P(A ∪ B)

$\begin{aligned} & =\frac{3}{8}+\frac{1}{2}-\frac{5}{8} \\ & =\frac{1}{4}\end{aligned}$

2. P(A’ ∩ B’) = P(A ∪ B)’ = 1 – P(A ∪ B)

$\begin{aligned} & =1-\frac{5}{8} \\ & =\frac{3}{8}\end{aligned}$

3. P(A’ ∪ B’) = P(A ∩ B)’ = 1 – P(A ∩ B)

$\begin{aligned} & =1-\frac{1}{4} \\ & =\frac{3}{4}\end{aligned}$

S = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}

∴ n(S) = 36

Let event A: The number on the upper face of the first die is 3.

∴ A = {(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6)}

∴ n(A) = 6

$\therefore P ( A )=\frac{n(A)}{n(S)}=\frac{6}{36}$

Let event B: Sum of the numbers on their upper faces is 6.

∴ B = {(1, 5), (2, 4), (3, 3), (4, 2), (5, 1)}

∴ n(B) = 5

$\therefore P(B)=\frac{n(B)}{n(S)}=\frac{5}{36}$

Now, A ∩ B = {(3, 3)}

∴ n(A ∩ B) = 1

$\therefore P ( A \cap B )=\frac{ n ( A \cap B )}{ n ( S )}=\frac{1}{36}$

∴ Required probability

P(A ∪ B) = P(A) + P(B) – P(A ∩ B)

$\begin{aligned} & =\frac{6}{36}+\frac{5}{36}-\frac{1}{36} \\ & =\frac{10}{36} \\ & =\frac{5}{18}\end{aligned}$

i.e., 7 + 4 = 11 balls.

$\therefore 3$ balls can be drawn out of 11 balls in ${ }^{11} C _3$ ways.

$\therefore n ( S )={ }^{11} C _3$

(i) Let event A: All 3 balls drawn are black.

There are 7 black balls.

$\therefore 3$ black balls can be drawn from 7 black balls in ${ }^7 C _3$ ways.

$\therefore n ( A )={ }^7 C _3$

$\therefore P ( A )=\frac{ n ( A )}{ n ( S )}=\frac{{ }^7 C _3}{{ }^{11} C _3}=\frac{7 \times 6 \times 5}{11 \times 10 \times 9}=\frac{7}{33}$

Let event B: Out of 3 balls drawn, one is black and two are red.

There are 7 black and 4 red balls.

$\therefore$ One black ball can be drawn from 7 black balls in ${ }^7 C _1$ ways and 2 red balls can be

drawn from 4 red balls in ${ }^4 C _2$ ways.

$\begin{aligned} & \therefore n ( A )={ }^7 C _1 \times{ }^4 C _2 \\ & \therefore P ( A )=\frac{n(A)}{n(S)}=\frac{{ }^7 C_1 \times{ }^4 C_2}{{ }^{11} C_3}=\frac{7 \times \frac{4 \times 3}{2}}{\frac{11 \times 1 \times 9}{3 \times 2 \times 1}}=\frac{14}{55}\end{aligned}$

Number of defective bulbs = 6

∴ Number of non-defective bulbs = 4

3 bulbs can be selected out of 10 bulbs in ${ }^{10} C _3$ ways.

$\therefore n(S)={ }^{10} C_3$

(i) Let event A: The room is dark.

For event A to happen the bulbs should be selected from the 6 defective bulbs. This can

be done in ${ }^6 C _3$ ways.

$\begin{aligned} & \therefore n ( A )={ }^6 C _3 \\ & \therefore P ( A )=\frac{n(A)}{n(S)}=\frac{{ }^6 C_3}{{ }^{10} C_3}=\frac{6 \times 5 \times 4}{10 \times 9 \times 8}=\frac{1}{6}\end{aligned}$

ii. Let event A’: The room is lit.

∴ P(Room is lit) = 1 – P(Room is not lit)

$\therefore P\left(A^{\prime}\right)=1-P(A)=1-\frac{1}{6}=\frac{5}{6}$

One ball can be drawn from 20 balls in ${ }^{20} C _1$ ways.

$\therefore n ( S )={ }^{20} C _1=20$

Let event A: Ball drawn is red. There are total 10 red balls.

$\therefore 1$ red ball can be drawn from 10 red balls in ${ }^{10} C_1$ ways.

$\begin{aligned} & \therefore n ( A )={ }^{10} C _1=10 \\ & \therefore P ( A )=\frac{n(A)}{n(S)}=\frac{10}{20}=\frac{1}{2}\end{aligned}$

Let event B: The ball drawn is blue or black. There are 4 blue and 6 black balls.

$\therefore 1$ blue ball can be drawn from 4 blue balls in ${ }^4 C _1$ ways

or 1 black ball can be drawn from 6 black balls in ${ }^6 C_1$ ways.

$\begin{aligned} & \therefore n ( B )={ }^4 C _1+{ }^6 C _1=4+6=10 \\ & \therefore P ( B )=\frac{n(B)}{n(S)}=\frac{10}{20}=\frac{1}{2}\end{aligned}$

Let event C: Ball drawn is not black,

i.e., ball drawn is red or blue.

There are total 14 red and blue balls.

$\therefore 1$ ball can be drawn from 14 balls in ${ }^{14} C _1$ ways.

$\begin{aligned} & \therefore n (C)={ }^{14} C _1=14 \\ & \therefore P ( C )=\frac{n(C)}{n(S)}=\frac{14}{20}=\frac{7}{10}\end{aligned}$