Maths 4 Marks

Probability distribution is given by $\bigg(\frac{1}{6} + \frac{5}{6}\bigg)^{4}$

Let X be the number of successes and P (X), the corresponding probability, where X takes values from 0 to 4

$\therefore$ The distribution is:

Let A be the event of a vehicle meeting an accident.

$\therefore \text{P} \text({E}_{1}) = \frac{1}{6}, \text{P} \text({E}_{2}) = \frac{1}{3}, \text{P} \text({E}_{3}) = \frac{1}{2}$

$\text{P} \bigg(\frac{A}{E_{1}}\bigg) = \frac{1}{100}, \text{P} \bigg(\frac{A}{E_{2}}\bigg) = \frac{3}{100}, \text{P} \bigg(\frac{A}{E_{3}}\bigg) = \frac{15}{100}$

$\text{P} \bigg(\frac{A}{E_{1}}\bigg) = \frac{P(E_{1}\big) \times P\bigg(\frac{A}{E_1}\bigg)}{\sum^{3}_ {i = 1} P (E_{i} \times \bigg(\frac{A}{E_{1}}\bigg)}, i = 1, 2, 3$

$= \frac{\frac{1}{6}\times \frac{1}{100}}{\frac{1}{6}\times\frac{1}{100} + \frac{1}{3} \times\frac{3}{100} + \frac{1}{2} \times\frac{15}{100}} \frac{\frac{1}{6}}{\frac{1}{6}+1+\frac{15}{2}} = \frac{1}{6}\times \frac{6}{52} = \frac{1}{52}$

Now,

Mean $=\sum\text{p}_\text{i}\text{x}_\text{i}=\text{ap}+\text{bq}$

Variance $=\sum\text{p}_\text{i}\text{x}_\text{i}^2-(\text{Mean})^2$

= a²p + b²q - (ap + bq)²

= a²p + b²q - a²p² - b²q² - 2abpq

= a²p - a²p² + b²q - b²q² - 2abpq

= a²p(1 - p) + b²q(1 - q) - 2abpq

= a²pq + b²qp - 2abpq $(\because\ \text{p}+\text{q}=1)$

= pq(a² + b² - 2ab)

= pq(a - b)²

Step Deviation $=\sqrt{\text{Variance}}$

$=\sqrt{\text{pq}(\text{a-b})^2}$

$=|\text{a}-\text{b}|\sqrt{\text{pq}}$

The man may get number greater than 4 in the first throw and then he quits the game. He may get a number less than equat to 4. In the first throw and in the second throw he may get the number greater than 4 and quits the game.

In the first two throws he gets a number less than equal to 4 and in the third throw he may get a number greater than 6. He may not get number greater than 4 in any one of three throws.

Let X be the amount he wins/looses.

Then, X can take values -3, 3, 4, 5 such that

P(X= 5) = P(Getting number greater than 4 in first throw) $=\frac{1}{3}$

(X= 4) = P(Getting number less than equal to 4 in the first throw and number greater than 4 in second throw) $=\frac{4}{6}\times\frac{2}{6}=\frac{2}{9}$

P (X = 3) = P(Getting number less than equal to 4 in the first two throws and number greater than 4 in third throw) $=\frac{4}{6}\times\frac{4}{6}\times\frac{2}{6}=\frac{4}{27}$

P(X = -3) = P(Getting number less than equal to 4 in all three throws) $=\frac{4}{6}\times\frac{4}{6}\times\frac{4}{6}=\frac{8}{27}$

$\text{E}(\text{X})=\Big(5\times\frac{1}{3}\Big)+4\Big(\frac{2}{9}\Big)+3\Big(\frac{4}{27}\Big)-3\Big(\frac{8}{27}\Big)$

$=\frac{1}{27}(45+24+12-24)$

$=\frac{57}{27}$

$=\frac{19}{9}$

Expected value of the amount he wins or loses is $=\frac{19}{9}$

1. $\because\text{P}(\text{A})=\text{P}(\text{A}\cap\text{B})+\text{P}(\text{A}\cap\bar{\text{B}})$

$\therefore\text{R.H.S.}=\text{P}(\text{A}\cap\text{B})+\text{P}(\text{A}\cap\bar{\text{B}})$

$=\text{P}(\text{A})\cdot\text{P}(\text{B})+\text{P}(\text{A})\cdot\text{P}\bar{(\text{B})}$

$=\text{P}(\text{A})\big[\text{P}(\text{B})+\text{P}\bar{(\text{B})}\big]$

$=\text{P}(\text{A})\big[\text{P}(\text{B})+1-\text{P}(\text{B})\big]$ $\big[\because\text{P}\bar{(\text{B})}=1-\text{P}(\text{B})\big]$

$= \text{P(A) = L. H. S} $ Hence proved.

2. $\because\text{P}(\text{A}\cup\text{B})=\text{P}(\text{A}\cap\text{B})+\text{P}(\text{A}\cap\bar{\text{B}})+\text{P}(\bar{\text{A}}\cap\text{B})$

$\therefore\text{R.H.S.}=\text{P}(\text{A})\cdot\text{P}(\text{B})+\text{P}(\text{A})\cdot\text{P}\bar{(\text{B})}+\text{P}\bar{(\text{A})}\cdot\text{P}(\text{B})$

$=\text{P}(\text{A})\cdot\text{P}(\text{B})+\text{P}(\text{A})\cdot\big[1-\text{P}(\text{B})\big]+\big[1-\text{P}(\text{A})\big]\text{P}(\text{B})$

$=\text{P}(\text{A})\cdot\text{P}(\text{B})+\text{P}(\text{A})-\text{P}(\text{A})\cdot\text{P}(\text{B})+\text{P}(\text{B})-\text{P}(\text{A})\cdot\text{P}(\text{B})$

$=\text{P}(\text{A})+\text{P}(\text{B})-\text{P}(\text{A})\cdot\text{P}(\text{B})$

$=\text{P}(\text{A})+\text{P}(\text{B})-\text{P}(\text{A}\cap\text{B})$

$=\text{P}(\text{A}\cup\text{B})=\text{L.H.S.}$ Hence proved.

1. Let A be the set of favourable events. ⇒ n(A) = 2

$\therefore\ \text{P}(\text{A})=\frac{\text{n}(\text{S})}{\text{n}(\text{A})}=\frac{2}{6}=\frac{1}{3}\ \text{and}\ \text{P}(\overline{\text{A}})=1-\frac{1}{3}=\frac{2}{3}$

n = 2, r = 0, 1, 2

$\text{P}(\text{X}=0)=\text{P}(\overline{\text{A}}).\text{P}(\overline{\text{A}})=\frac{2}{3}\times\frac{2}{3}=\frac{4}{9}$

$\text{P}(\text{X}=1)=2\text{P}(\text{A}).\text{P}(\overline{\text{A}})=2\times\frac{1}{3}\times\frac{2}{3}=\frac{4}{9}$

$\text{P}(\text{X}=2)=\text{P}(\text{A}).\text{P}(\text{A})=\frac{1}{3}\times\frac{1}{3}=\frac{1}{9}$

Probability distribution

2. Let A represents that 6 appears on one die ⇒ A = {6} ⇒ n(A) = 1

$\therefore\ \text{P}(\text{A})=\frac{\text{n}(\text{S})}{\text{n}(\text{A})}=\frac{1}{6}\ \text{and}\ \text{P}(\overline{\text{A}})=1-\frac{1}{6}=\frac{5}{6}$

n = 2, r = 0, 1, 2

$\text{P}(\text{X}=0)=\text{P}(\overline{\text{A}}).\text{P}(\overline{\text{A}})=\frac{5}{6}\times\frac{5}{6}=\frac{25}{36}$

$\text{P}(\text{X}=1)=2\text{P}(\text{A}).\text{P}(\overline{\text{A}})=2\times\frac{1}{6}\times\frac{5}{6}=\frac{10}{36}$

$\text{P}(\text{X}=2)=\text{P}(\text{A}).\text{P}(\text{A})=\frac{1}{6}\times\frac{1}{6}=\frac{1}{36}$

P(at least on six) $=\frac{10}{36}\times\frac{1}{36}=\frac{11}{36}$

Probability distribution

Probability of solving the problem by B, P(B) $=\frac{1}{3}$

Since the problem is solved independently by A and B,

$\therefore\text{P}(\text{AB})=\text{P}(\text{A})\cdot\text{P}(\text{B})=\frac{1}{2}\times\frac{1}{3}=\frac{1}{6}$

$\text{P}(\text{A}')=1-\text{P}(\text{A})=1-\frac{1}{2}=\frac{1}{2}$

$\text{P}(\text{B}')=1-\text{P}(\text{B})=1-\frac{1}{3}=\frac{2}{3}$

1. Probability that the problem is solved $=\text{P}(\text{A}\cup\text{B})$

$=\text{P}(\text{A})+\text{P}(\text{B})-\text{P}(\text{AB})$

$=\frac{1}{2}+\frac{1}{3}-\frac{1}{6}$

$=\frac{4}{6}$

$=\frac{2}{3}$

2. Probability that exactly one of them solves the problem is given by,

$\text{P}(\text{A}).\text{P}(\text{B}')+\text{P}(\text{B}).\text{P}(\text{A}')$

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

$=\frac{1}{3}+\frac{1}{6}$

$=\frac{1}{2}$

Two numbers are selected at random from integers 1 through 9.

A = Both numbers are odd

A = {(3, 1), (5, 1), (7, 1), (9, 1), (3, 5), (3, 7), (9, 3), (5, 3), (5, 7), (5, 9), (7, 3), (7, 5), (7, 9), (9, 3), (9, 5), (9, 7)}

B = Sum of both numbers is even

A = Sum of both numbers is 2, 4, 6, 8, 10, 12, 14, 16 or 18 = {(1, 3), (1, 5), (2, 4), (1, 7), (2, 6), (3, 5), (1, 9), (2, 8), (3, 7), (4, 6), (7, 5), (8, 4), (9, 3), (8, 6), (9, 5), (9, 7)}

$(\text{A}\cap\text{B})=$ {(1, 3), (1, 5), (1, 7), (3, 5), (1, 9), (3, 7), (7, 5), (9, 3), (9, 5), (9, 7)} 

$\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{\text{n}(\text{A}\cap\text{B})}{\text{n}(\text{B})}$

$=\frac{10}{16}$

Required probability $=\frac{10}{16}$

$\text{S} = \Bigg\{\begin{array}{c}(\text{H, 1}),\ (\text{H, 2}),\ (\text{H, 3}),\ (\text{H, 4}),\ (\text{H, 5}),\ (\text{H, 6}),\\ \text{(T, 1}),\ \text{(T, 2}),\ \text{(T, 3}),\ \text{(T, 4}),\ (\text{T, 5}),\ \text{(T, 6})\end{array}\Bigg\}$

Let A: Head appears on the coin

$\text{A}=\left\{(\text{H, 1}),\ (\text{H, 2}),\ (\text{H, 3}),\ (\text{H, 4}),\ (\text{H, 5}), (\text{H, 6})\right\}$

$\Rightarrow\text{P}(\text{A})=\frac{6}{12}=\frac{1}{2}$

B: 3 on die $=\left\{(\text{H, 3}),\ (\text{T, 3})\right\}$

$ \text{P}(\text{B})=\frac{2}{12}=\frac{1}{6}$

$\therefore\ \text{A}\cap\text{B}=\left\{(\text{H, 3})\right\}$

$\text{P}(\text{A}\cap\text{B})=\frac{1}{12}$

$\text{P}(\text{A}).\text{P}(\text{B})=\frac{1}{2}\times\frac{1}{6}=\text{P}(\text{A}\cap\text{B})$

Therefore, A and B are independent events.

Since there are 12000 persons, therefore,

$\text{Now}\ \ \ \text{P}(\text{E}_1)=\frac{2000}{12000}=\frac{1}{6},\ \text{P}(\text{E}_2)=\frac{4000}{12000}=\frac{1}{3},\ \text{P}(\text{E}_3)=\frac{6000}{12000}=\frac{1}{2}$

It is given that $\text{P}(\text{A}|\text{E}_1)$ = P(a person meets with an accident, he is a scooter driver) = 0.01

Similarly, $\text{P}(\text{A}|\text{E}_2)=0.03\ \text{and}\ \text{P}(\text{A}|\text{E}_3)=0.15$

To find: P(person meets with an accident that he was a scooter driver)

Therefore, by Bayes’ theorem,

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

$=\frac{\frac{1}{6}\times0.01}{\frac{1}{6}\times0.01+\frac{1}{3}\times0.03+\frac{1}{2}\times0.15}=\frac{1}{1+6+45}=\frac{1}{52}$

It is given that,

$\text{P}(\text{H})=60\%=\frac{60}{100}=\frac{6}{10}=\frac{3}{5}$

$\text{P}(\text{E})=40\%=\frac{40}{100}=\frac{2}{5}$

$\text{P}(\text{H}\cap\text{E})=20\%=\frac{20}{100}=\frac{1}{5}$

1. Probability that a student reads Hindi or English newspaper is,

$\text{P}(\text{H}\cup\text{E})'=1-\text{P}(\text{H}\cup\text{E})$

$=1-\big\{\text{P}(\text{H})+\text{P}(\text{E})-\text{P}(\text{H}\cap\text{E})\big\}$

$=1-\Big(\frac{3}{5}+\frac{2}{5}-\frac{1}{5}\Big)$

$=1-\frac{4}{5}$

$=\frac{1}{5}$

2. Probability that a randomly chosen student reads English newspaper, if she reads Hindi news paper, is given by P(E|H).

$\text{P}(\text{E}|\text{H})=\frac{\text{P}(\text{E}\cap\text{H})}{\text{P}(\text{H})}$

$=\frac{\frac{1}{5}}{\frac{3}{5}}$

$=\frac{1}{3}$

3. Probability that a randomly chosen student reads Hindi newspaper, if she reads English newspaper, is given by P(H|E).

$\text{P}(\text{H}|\text{E})=\frac{\text{P}(\text{H}\cap\text{E})}{\text{P}(\text{E})}$

$=\frac{\frac{1}{5}}{\frac{2}{5}}$

$=\frac{1}{2}$

1. Since, the sum of all the probabilities of a distribution is 1.

∴ P(X = 0) + P(X = 1) + …. + P(X = 7) = 1

⇒ 0 + k + 2k + 2k + 3k + k² + 2k² + 7k² + k = 1

⇒ 10k² + 9k - 1 = 0

⇒ (10k - 1) (k + 1) = 0

⇒ 10k - 1 = 0 or k + 1 = 0

 $\Rightarrow\ \text{k}=\frac{1}{10}$ or k = - 1

Since, k ≥ 0, therefore k = − 1 is not possible.

$\therefore\ \text{k}=\frac{1}{10}$

2. P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)

= 0 + k + 2k

$=3\text{k}=3\times\frac{1}{10}=\frac{3}{10}$

3. P(X > 6) = P(X = 7)

$=7\text{k}^2+\text{k}=7\Big(\frac{1}{10}\Big)^2+\frac{1}{10}=\frac{17}{100}$

4. P(0 < X < 3) = P(X = 1) + P(X = 2)

= k + 2k = 3k = $\frac{3}{10}$

P(X = 3) = P(larger number is 3) = {(2, 3), (3, 2)} $=\frac{2}{30}$

P(X = 4) = P(larger number is 4) = {(2, 4), (4, 2), (3, 4), (4, 3)} $=\frac{4}{30}$

P(X = 5) = P(larger number is 5) = {(2, 5), (5, 2), (3, 5), (5, 3), (4, 5), (5, 4)} $=\frac{6}{30}$

P(X = 6) = P(larger number is 6) = {(2, 6), (6, 2), (3, 6), (6, 3), (4, 6), (6, 4), (5, 6), (6, 5)} $=\frac{8}{30}$

P(X = 7) = P(larger number is 7) = {(2, 7), (7, 2), (3, 7), (7, 3), (4, 7), (7, 4), (5, 7), (7, 5), (6, 7), (7, 6)} $=\frac{10}{30}$

Thus, the probability distribution of random variable X is,

Mean $=\sum\text{x}_\text{i}\text{p}_\text{i}=\frac{17}{3}$

Variance $=\sum\text{x}_\text{i}\text{p}_\text{i}-\big(\sum\text{x}_\text{i}\text{p}_\text{i}\big)^2=\frac{101}{3}-\Big(\frac{17}{3}\Big)=\frac{14}{9}$

1. Let 'A' be the event that both the children born are girls. Let 'B' be the event that the youngest is a girls. We have to find conditional probability $\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)$.

$\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{\text{P}(\text{A}\cap\text{B})}{\text{P}(\text{B})}$

$\text{A}\subset\text{B} \Rightarrow\text{A}\cap\text{B}=\text{A}$

$\Rightarrow\ \text{P}(\text{A}\cap\text{B})=\text{P(A)}=\text{P(GG)}=\frac{1}{2}\times\frac{1}{2}=\frac{1}{4}$

$\text{P(B)}=\text{P(BG)}+\text{P(GG)}=\frac{1}{2}\times\frac{1}{2}+\frac{1}{2}\times\frac{1}{2} \\=\frac{1}{4}+\frac{1}{4}=\frac{1}{2}$

Hence, $\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{\frac{1}{4}}{\frac{1}{2}}=\frac{1}{2}$

2. Let 'A' be the event that both the children born are girls. Let 'B' be the event that at least one is a girl. We have to find the conditional probability $\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)$.

$\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{\text{P}(\text{A}\cap\text{B})}{\text{P(B)}}$

$\text{A}\subset\text{B}\Rightarrow\text{A}\cap\text{B}=\text{A}$

$\Rightarrow\ \text{P}(\text{A}\cap\text{B})=\text{P(A)}=\text{P(GG)}=\frac{1}{2}\times\frac{1}{2}=\frac{1}{4}$

$\text{P(B)}=1-\text{P(BB)}=1-\frac{1}{2}\times\frac{1}{2}=1-\frac{1}{4}=\frac{3}{4}$

Hence, $\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{\frac{1}{4}}{\frac{3}{4}}=\frac{1}{3}$

A = getting sum 8 or more

= Getting sum 8, 9, 10, 11 or 12 on the pair of dice

A = {(2, 6), (3, 5), (4, 4), (5, 3), (6, 2), (3, 6)

(4, 5), (5, 4), (6, 3), (4, 6), (5, 5), (6, 4)

(5, 6), (6, 5), (6, 6)

B = 4 on first die

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

$(\text{A}\cap\text{B})=\{(4, 4), (4, 5), (4, 6)\}$

$\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{\text{n}(\text{A}\cap\text{B})}{\text{n}(\text{B})}$

$=\frac{3}{6}$

$=\frac{1}{2}$

Required probability $=\frac{1}{2}$

1. $\therefore\text{P}(\text{E})=\text{P}(\text{A}_1)\cdot \text{P}\Big(\frac{\text{E}}{\text{A}_1}\Big)+\text{P}(\text{A}_2)\cdot\text{P}\Big(\frac{\text{E}}{\text{A}_2}\Big) +\text{P}(\text{A}_3)\cdot\text{P}\Big(\frac{\text{E}}{\text{A}_3}\Big)$

$=\frac{4}{10}\cdot\frac{45}{100}+\frac{4}{10}\cdot\frac{60}{100}+\frac{2}{10}\cdot\frac{35}{100}$

$=\frac{180}{1000}+\frac{240}{1000}+\frac{70}{1000}$

$=\frac{490}{1000}=0.49$

2. $\text{P}\Big(\frac{\text{E}'}{\text{A}_3}\Big)=1-\text{P}\Big(\frac{\text{E}}{\text{A}_3}\Big)$

$=1-\frac{35}{100}=\frac{65}{100}$

3. $\text{P}\Big(\frac{\text{A}_2}{\text{E}'}\Big)=\frac{\text{P}(\text{A}_2)\cdot\text{P}\Big(\frac{\text{E}'}{\text{A}_2}\Big)}{\text{P}(\text{A}_1)\cdot\text{P}\Big(\frac{\text{E}'}{\text{A}_1}\Big)+\text{P}(\text{A}_2)\cdot\text{P}\Big(\frac{\text{E}'}{\text{A}_2}\Big)+\text{P}(\text{A}_3)\cdot\text{P}\Big(\frac{\text{E}}{\text{A}_3}\Big)}$

$=\frac{\frac{4}{10}\cdot\frac{40}{100}}{\frac{4}{10}\cdot\frac{55}{100}+\frac{4}{10}\cdot\frac{40}{100}+\frac{2}{10}\cdot\frac{65}{100}}$

$=\frac{16}{51}$

1. P(none) = P(X = 0)

$=\ ^5\text{C}_{0}(0.95)^{5}.(0.05)^{0}$

$=1\times(0.95)^{5}$

$=(0.95)^{5}$

2. P(not more than one) = P(X ≤ 1)

$\text{P}(\text{X}=0)+\text{P}(\text{X}=1)$

$=\ ^5\text{C}_0(0.95)^{5}\times(0.05)^{0}+\ ^5\text{C}_1(0.95)^{4}\times(0.05)^{1}$

$=1\times(0.95)^{5}+5\times(0.95)^{4}\times (0.05)$

$=(0.95)^{5}+(0.25)(0.95)^4$

$=(0.95)^{4}[0.95+0.25]$

$=(0.95)^{4}\times1.2$

3. P(more than 1) = P(X > 1)

$=1-\text{P}(\text{X}\leq1)$

= 1 - P(not more than 1)

$=1-(0.95)^4\times1.2$

4. P(at least one) = P(X ≥ 1)

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

$=1-\text{P}(\text{X}=0)$

$=1-\ ^5\text{C}_0(0.95)^5\times(0.05)^0$

$=1-1\times(0.95)^5$

$=1-(0.95)^5$

1. Second ball is black and third is red (E₁).
2. Second ball is black and third is also black (E₂). 
3. Second ball is red and third is black (E₃). 

$\therefore\text{P}(\text{E}_1)=\text{P}(\text{R}_1)\cdot\text{P}\Big(\frac{\text{B}_1}{\text{R}_1}\Big)\cdot\text{P}\Big(\frac{\text{R}_2}{\text{R}_1\text{B}_1}\Big)$

$=\frac{5}{8}\cdot\frac{3}{7}\cdot\frac{4}{6}=\frac{60}{336}=\frac{5}{28}$

$\text{P}(\text{E}_2)=\text{P}(\text{R}_1)\cdot\text{P}\Big(\frac{\text{B}_1}{\text{R}_1}\Big)\cdot\text{P}\Big(\frac{\text{B}_2}{\text{R}_1\text{B}_1}\Big)$

$\frac{5}{8}\cdot\frac{3}{7}\cdot\frac{2}{6}=\frac{30}{336}=\frac{5}{56}$

And $\text{P}(\text{E}_3)=\text{P}(\text{R}_1)\cdot\text{P}\Big(\frac{\text{R}_2}{\text{R}_1}\Big)\cdot\text{P}\Big(\frac{\text{B}_1}{\text{R}_1\text{R}_2}\Big)$

$=\frac{5}{8}\cdot \frac{4}{7}\cdot\frac{3}{6}=\frac{60}{336}=\frac{5}{28}$

$\therefore\text{P(E)}=\text{P(E}_1)+\text{P(E}_2)+\text{P(E}_3)=-\frac{5}{28}+\frac{5}{56}+\frac{5}{28}$

$=\frac{10+5+10}{56}=\frac{25}{56}$

1. We have to find the smallest value of n for which P(X = 0) is less than $\frac{1}{4}$

$\text{P(X = 0)}<\frac{1}{4}$

$\text{ }^\text{n}\text{C}_0\big(\frac{1}{1}\big)^0\big(\frac{3}{4}\big)^{\text{n}-0}<\frac{1}{4}$

$\big(\frac{3}{4}\big)^{\text{n}}<\frac{1}{4}$

Put  $\text{n}=1, \big(\frac{3}{4}\big)\nless\frac{1}{4}$

$\text{n}=2, \big(\frac{3}{4}\big)^2\nless\frac{1}{4}$

$\text{n}=3, \big(\frac{3}{4}\big)^3\nless\frac{1}{4}$

So, smallest value of n = 3

$\therefore$ we must draw card at least 3 times

2. Given the probability of drawing a heart $>\frac{3}{4}$

$1-\text{P(X = 0)}>\frac{3}{4}$

$1-\text{ }^{\text{n}}\text{C}_0\big(\frac{1}{4}\big)^0\big(\frac{3}{4}\big)^{\text{n}-0}>\frac{3}{4}$

$1-\big(\frac{3}{4}\big)^{\text{n}}>\frac{3}{4}$

$1-\frac{3}{4}>\big(\frac{3}{4}\big)^{\text{n}}$

$\frac{1}{4}>\big(\frac{3}{4}\big)^{\text{n}}$

For $\text{n}=1,\big(\frac{3}{4}\big)^1$ not less than $\frac{1}{4}$

$\text{n}=2,\big(\frac{3}{4}\big)^2$ not less than $\frac{1}{4}$

$\text{n}=3,\big(\frac{3}{4}\big)^3$ not less than $\frac{1}{4}$

$\text{n}=4,\big(\frac{3}{4}\big)^4$ not less than $\frac{1}{4}$

$\text{n}=5,\big(\frac{3}{4}\big)^5$ not less than $\frac{1}{4}$

1. P(winning at least once) $=\text{P(X}\geq0)$

$=1-\text{P(X}-0)$

$=1-\big(\frac{99}{100}\big)^{50}$

2. P(winning exactly once) $=\text{P(X}=1)$

$=\text{ }^{50}\text{C}_1\big(\frac{1}{100}\big)^1\big(\frac{99}{100}\big)^{50-1}$

$=\frac{1}{2}\big(\frac{99}{100}\big)^{49}$

3. P(winning at least twice) $=\text{P(X}\geq2)$

$=1-\text{P(X}=0)-\text{P(X}=1)$

$=1-\big(\frac{99}{100}\big)^{50}-\text{ }^{50}\text{C}_1\times\frac{1}{100}\times\big(\frac{99}{100}\big)^{49}$

$=1-\frac{99^{49}\times149}{100^{50}}$

1. Required probability $=\text{P}\Big(\frac{\text{E}_1}{\text{A}}\Big)=\frac{\text{P}(\text{E}_1)\text{ P}\Big(\frac{\text{A}}{\text{E}_1}\Big)}{\text{P}(\text{E}_1)\text{ P}\Big(\frac{\text{A}}{\text{E}_1}\Big)+\text{P}(\text{E}_2)\text{ P}\Big(\frac{\text{A}}{\text{E}_2}\Big)+\text{P}(\text{E}_3)\text{ P}\Big(\frac{\text{A}}{\text{E}_3}\Big)}$

$=\frac{\frac{1}{4}\times\frac{2}{100}}{\frac{1}{4}\times\frac{2}{100}+\frac{1}{3}\times\frac{3}{100}+\frac{5}{12}\times\frac{4}{100}}$

$=\frac{\frac{1}{2}}{\frac{1}{2}+1+\frac{5}{3}}=\frac{\frac{1}{2}}{\frac{3+6+10}{6}}=\frac{3}{19}$

2. Required probability $=\text{P}\Big(\frac{\text{E}_2}{\text{A}}\Big)=\frac{\text{P}(\text{E}_2)\text{ P}\Big(\frac{\text{A}}{\text{E}_2}\Big)}{\text{P}(\text{E}_1)\text{ P}\Big(\frac{\text{A}}{\text{E}_1}\Big)+\text{P}(\text{E}_2)\text{ P}\Big(\frac{\text{A}}{\text{E}_2}\Big)+\text{P}(\text{E}_3)\text{ P}\Big(\frac{\text{A}}{\text{E}_3}\Big)}$

$=\frac{\frac{1}{3}\times\frac{3}{100}}{\frac{1}{4}\times\frac{2}{100}+\frac{1}{3}\times\frac{3}{100}+\frac{5}{12}\times\frac{4}{100}}$

$=\frac{\frac{1}{2}}{\frac{1}{2+1+\frac{5}{3}}}=\frac{\frac{1}{2}}{\frac{3+6+10}{6}}=\frac{6}{19}$

3. Required probability $\text{P}\Big(\frac{\text{E}_3}{\text{A}}\Big)=\frac{\text{P}(\text{E}_3)\text{ P}\Big(\frac{\text{A}}{\text{E}_3}\Big)}{\text{P}(\text{E}_1)\text{ P}\Big(\frac{\text{A}}{\text{E}_1}\Big)+\text{P}(\text{E}_2)\text{ P}\Big(\frac{\text{A}}{\text{E}_2}\Big)+\text{P}(\text{E}_3)\text{ P}\Big(\frac{\text{A}}{\text{E}_3}\Big)}$

$=\frac{\frac{5}{15}\times\frac{4}{100}}{\frac{1}{4}\times\frac{2}{100}+\frac{1}{3}\times\frac{3}{100}+\frac{5}{12}\times\frac{4}{100}}$

$=\frac{\frac{5}{3}}{\frac{1}{2}+1+1\frac{5}{3}}=\frac{\frac{5}{3}}{\frac{3+6+10}{6}}=\frac{10}{19}$

1. Selecting bag I and then drawing a white and a red ball from it.
2. Selecting bag II and then drawing a white and a red ball from it.
3. Selecting bag III and then drawing a white and a red ball from it.