Maths M.C.Q (1 Marks)

4. $\text{None of these}$

Solution:

The sample space = 15⁷ for selecting seven coupons from 15 coupons.

Maximum number on selected coupon is 9 can be made by 9⁷ ways.

A number selected on second card is less than 9 can be made by 8⁷ ways.

Required probability $=\frac{9^7-8^7}{15^7}$

2. $\frac{1}{81}$

Solution:

Given $\text{n}=4,\text{P(X}=0)=\frac{16}{81}$

$\text{P(X}=0)=\frac{16}{81}$

$\text{ }^5\text{C}_0\text{p}^0\text{q}^4=\frac{16}{81}$

$\text{q}^4=\frac{16}{81}$

$\text{q}=\frac{2}{3}\Rightarrow\text{p}=\frac{1}{3}$

$\Rightarrow\text{P(X}=4)=\text{ }^5\text{C}_4\big(\frac{1}{4}\big)^4=\frac{1}{81}$

2. 0.188

Solution:

We have

$\text{P}(\text{A})=0.4,\text{P}(\bar{\text{A}})=0.6,\text{P}(\text{B})=0.3,\text{P}(\bar{\text{B}})=0.7$

$\text{P}(\text{C})=0.2$ and $\text{P}(\bar{\text{C}})=0.8$

$\therefore$ Probability of two hits $=\text{P}_{\text{A}}\cdot\text{P}_{\text{B}}\cdot\text{P}_{\bar{\text{C}}}+\text{P}_{\text{A}}\cdot\text{P}{_\bar{\text{B}}}\cdot\text{P}_{\text{C}}+\text{P}{_\bar{\text{A}}}\cdot\text{P}_{\text{B}}\cdot\text{P}_{\text{C}}$

$=0.4\times0.3\times0.8+0.4\times0.7\times0.2+0.6\times0.3\times0.2$

$=0.096+0.056+0.036=0.188$

3. 0.67

Solution:

P(M and S) = 0.40

P(M) = 0.60

$\text{P(S|M})=\frac{\text{P (M and S)}}{\text{P(S)}}=\frac{0.40}{0.60}=\frac23=0.67$

1. 0.39

Solution:

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

$=0.25+0.5-0.14$

$0.61$

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

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

$=1-0.61$

$=0.39$

1. $\text{ }^5\text{C}_4(0.7)^4(0.3)$

Solution:

Given that a person is not a swimmer $\Rightarrow\text{q}=0.3$

$\Rightarrow\text{p}=0.7$

$\text{n = 5, X = 4}$

$\text{P(X}=4)=\text{ }^5\text{C}_4\times0.7^{4}\times0.3$

2. $\frac{4}{7}$

Solution:

Total number of balls = 5 red + 3 Blue = 8

Probability of getting exacctly two red balls given that first ball should be red

Required probability $=\text{P}\Big(\frac{\text{R}_2\text{B}_2}{\text{R}_1}\Big)+\text{P}\Big(\frac{\text{R}_1\text{B}_2}{\text{R}_1}\Big)$

Required probability $=\frac{4}{7}\times\frac{3}{6}+\frac{3}{7}\times\frac{4}{6}=\frac{4}{7}$

4. 0.96

Solution:

We have,

P(A) = 0.4, P(B) = 0.8 and P(B|A) = 0.6

As, P(B|A) = 0.6

$\Rightarrow\ \frac{\text{P}(\text{A}\cap\text{B})}{\text{P(A)}}=0.6$

$\Rightarrow\ \text{P}(\text{A}\cap\text{B})=0.6\times\text{P(A)}$

$\Rightarrow\ \text{P}(\text{A}\cap\text{B})=0.6\times0.4$

$\Rightarrow\ \text{P}(\text{A}\cap\text{B})=0.24$

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

$=0.4+0.8-0.24$

$=1.2-0.24$

$=0.96$

Hence, the correct alternative is option (d).

1. $\frac{1}{36}$

Solution:

P(yellow face) $=\frac{3}{6}=\frac{1}{2}$

P(red face) $=\frac{2}{6}=\frac{1}{3}$

P(one face) $=\frac{1}{6}$

P(yellow face, red face and blue face appear in the required order) $=\frac{1}{2}\times\frac{1}{3}\times\frac{1}{6}=\frac{1}{36}$

3. $\frac{2}{3}$

Solution:

We know that the bag contains 5B (black), 4W(white) and 3R(red) balls.

Now,

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

$\text{P(R)}=\frac{3}{12}$

$\text{P}(\text{B or R})=\text{P(B)}+\text{P(R)}$

$=\frac{5}{12}+\frac{3}{12}$

$=\frac{8}{12}=\frac{2}{3}$

2. $\frac27$

Solution:

A leap year has 52 weeks and 2 days.

The 53^rd Sunday will be from these extra two days

These 2 days can be (Sunday, Monday) or (Mon, Tue) or (Tue, Wed).....(Sat, Sun)

There are 7 possibilities for these 2 days

Out of which Sunday is coming in 2 possibilities.

$\therefore$ P(2 sundays in leap year) $=\frac27$

3. $\frac{17}{19}$

Soluction:

Required probability that product of two integers should be even.

10 integers are odd out of first 20 integers.

Required probability = 1 - Probability of product is odd

Product of three integers is odd if two numbers are odd

Required probability $=1-\frac{10}{20}\times\frac{9}{19}\times\frac{8}{18}=\frac{17}{19}$

3. $\frac{7}{8}$

Solution:

$\text{P}(\text{A})=\frac{4}{5},\ \text{P}(\text{A}\cap\text{B})=\frac{7}{10}$

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

$=\frac{\frac{7}{10}}{\frac{4}{5}}=\frac{7}{8}$

2. $\frac25$

Solution:

Total number of outcomes = 5 × 4 × 3 × 2 = 120

The number of favourable cases = 2(4 × 3 × 2) = 48 (i.e., odd numbers)

Therefore,

Required probability $\frac{48}{120}=\frac25$

1. $\frac{7}{20}$

Soluction:

P(A speaks truth) = 0.75

P(A lies) = 1 - 0.75 = 0.25

P(B speaks truth) = 0.8

P(B lies) = 1 - 0.8 = 0.2

P(contradicting each other in a statement) = P(A speaks truth and lies) + P(B speaks truth and A lies)

= 0.75 × 0.2 + 0.8 × 0.25

= 0.15 + 0.2

= 0.35

$=\frac{35}{100}=\frac{7}{20}$

1. $\frac{14}{17}$

Solution:

Here, $\text{P}(\text{A}\cap\text{B})=\frac{7}{10}$ and $\text{P}(\text{B})=\frac{17}{20}$

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

$=\frac{\frac{7}{10}}{\frac{17}{20}}=\frac{14}{17}$

2. 0.4

Solution:

Let P(S) be the probability of automobile stolen.

And P(F) be the probability of automobile found.

According to the question,

$\text{P(S∩F) = 0.0006, P(S) = 0.0015}$

We know,

$\text{P}\Big(\frac{\text{F}}{\text{S}}\Big)=\frac{\text{P(F}\cap\text{S)}}{\text{P(S)}}=\frac{0.0006}{0. 0015}=0.4$

2. $\frac{27}{100}$

Solution:

$\text{P(A)}=\frac{1}{10},\text{P(B)}=\frac{3}{5},\text{P(C)}=\frac{1}{4}$

Required probability $=\text{P}(\overline{\text{A}}\cap\overline{\text{B}}\cap\overline{\text{C}})$

Required probability $=\text{P}(\overline{\text{A}})\text{ P}(\overline{\text{B}})\text{ P}(\overline{\text{C}})$

Required probability $=(1-\text{P(A)})(1-\text{P(B)})(1-\text{P(C)})$

Required probability $=\Big(1-\frac{1}{10}\Big)\Big(1-\frac{3}{5}\Big)\Big(1-\frac{1}{4}\Big)$

Required probability $=\frac{27}{100}$

2. $\frac{7}{32}$

Solution:

We know that, probaility distribution $\text{P}(\text{X}=\text{r})={^\text{n}\text{C}}_\text{r}(\text{P})^{\text{r}}\text{q}^{\text{n}-\text{r}}$

Here, $\text{n}=8,\text{r}=3,\text{p}=\frac{1}{2}$ and $\text{q}=\frac{1}{2}$

$\therefore$ Reuaired probability $={^8}\text{C}_3\Big(\frac{1}{2}\Big)^3\Big(\frac{1}{2}\Big)^{8-3}=\frac{8!}{5!3!}\Big(\frac{1}{2}\Big)^8 $

$=\frac{8\cdot7\cdot6}{3\cdot2}\cdot\frac{1}{2^8}=\frac{7}{32}$

4. $\frac{1}{5}$

Solution:

P(A) = 0.4, P(B) = 0.3 and $\text{P}(\text{A}\cup\text{B})=0.5$

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

$\text{P}(\overline{\text{B}}\cap\text{A})=\text{P(A)}-\big[\text{P(A)}+\text{P(B)}-\text{P}(\text{A}\cup\text{B})\big]$

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

$\text{P}(\overline{\text{B}}\cap\text{A})=0.5-0.3$

$\text{P}(\overline{\text{B}}\cap\text{A})=0.2$

$\text{P}(\overline{\text{B}}\cap\text{A})=\frac{1}{5}$

4. $\frac{23}{24}$

Soluction:

4 letter can be placed in 4 envelopes in 4! ways = 24ways

Now, there is only one method, by which all the letters are placed in the right envelope.

P(all letters are placed in the envelopes) $=\frac{1}{24}$

P(all letters are not placed in the right envelopes) = 1 - P(all letters are placed in the right envelopes)

$=1-\frac{1}{24}=\frac{23}{24}$

2. $\frac{7}{12}$

Solution:

$\text{P(B)}=\frac{3}{10},\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{2}{5},\text{P}(\text{A}\cup\text{B})=\frac{3}{5}$

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

$\text{P}(\text{A}\cap\text{B})=\frac{3}{10}+\frac{2}{5}-\frac{3}{5}$

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

$\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)+\text{P}\Big(\frac{\text{B}}{\text{A}}\Big)=\frac{\frac{1}{10}}{\frac{2}{5}}+\frac{\frac{1}{10}}{\frac{3}{10}}$

$\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)+\text{P}\Big(\frac{\text{B}}{\text{A}}\Big)=\frac{7}{12}$

Note: Option is modified.

4. $\frac{1}{3}$

Solution:

S = {GBB, GGB, GBG, GGG, BGG, BGB, BBG, BBB}

Let E₁ be the event that choosing a family with a girl as eldest child. E₂ be the event that choosing a family with at least one girl.

E₁ = {GBB, GGB, GBG, GGG}

E₂ = {GBB, GGB, GBG, GGG, BGG, BGB, BBG}

$\text{n}(\text{E}_1)=4,\text{n}(\text{E}_2)=7,\text{n}(\text{A}\cap\text{B})=4$

$\Rightarrow\ \text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{\text{n}(\text{A}\cap\text{B})}{\text{n(B)}}=\frac{4}{7}$

1. $\frac{1}{13}\times\frac{13}{13}$

Solution:

Required probability $=\frac{4}{52}\cdot\frac{4}{52}$

$=\frac{1}{13}\times\frac{1}{13}$ [with replacement]

4. 0.96

Solution:

Here, P(A) = 0.4, P(B) = 0.8

and $\text{P}\Big(\frac{\text{B}}{\text{A}}\Big)=0.6,$ 

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

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

$=0.6\times0.4=0.24$

$=1.2-0.24=0.96$

2. $\frac{\text{x}}{12}$

Solution:

Total number of balls = 12

Number of white balls = x

P (white ball) $=\frac{\text{x}}{12}$

4. $\frac{1}{5}$

Solution:

Here, $\text{P}(\text{A}) = 0.4,\text{P}(\text{B}) = 0.3$ and $\text{P}(\text{A}\cup\text{B})=0.5$

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

$\Rightarrow\text{P}(\text{A}\cap\text{B})=0.4+0.3-0.5=0.2$

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

$=0.4-0.2=0.2=\frac{1}{5}$

2. $\frac{5}{9}$

Solution:

We have,

$\text{P(A)}=\frac{7}{13},\text{P(B)}=\frac{9}{13}$ and $\text{P}(\text{A}\cap\text{B})=\frac{4}{13}$

As, $\text{P}(\overline{\text{A}}\cap\text{B})=\text{P(B)}-\text{P}(\text{A}\cap\text{B})$

$=\frac{9}{13}-\frac{4}{13}$

$=\frac{5}{13}$

Now,

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

$=\frac{\Big(\frac{5}{13}\Big)}{\Big(\frac{9}{13}\Big)}$

$=\frac{5}{9}$

Hence, the correct alternative is option (a).

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

Solution:

$\text{S}=\begin{Bmatrix} (1, 1),(1, 2),(1, 3),(1, 4),(1, 5),(1, 6),\$2, 1),(2, 2),(2, 3),(2, 4),(2, 5),(2, 6),\$3, 1),(3, 2),(3, 3),(3, 4),(3, 5),(3, 6),\$4, 1),(4, 2),(4, 3),(4, 4),(4, 5),(4, 6),\$5, 1),(5, 2),(5, 3),(5, 4),(5, 5),(5, 6),\$6, 1),(6, 2),(6, 3),(6, 4),(6, 5),(6, 6) \end{Bmatrix}$

$\text{n(S)}=36$

Let A be the event that sum of the numbers on dice was less than 6.

$\text{A} =\{(1, 1), (1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (4, 1)\}$

$\text{n(A)} = 10$

Let B be the event that getting sum 3.

$\text{B}=\{(1, 2), (2, 1)\}\Rightarrow\text{n(B)}=2$

$\text{A}\cap\text{B}=\{(1,2),(2,1)\}\Rightarrow\text{n}(\text{A}\cap\text{B})=2$

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

$\Rightarrow\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{2}{10}=\frac{1}{5}$

3. $\frac{25}{42}$

Solution:

$\text{P(A)}=\frac{2}{5},\text{P(B)}=\frac{3}{10}$ and $\text{P}(\text{A}\cap\text{B})=\frac{1}{5}$

$\text{P}(\text{A}\cap\text{B})=\frac{1}{5}$

$\text{P(A)}+\text{P(B)}-\text{P}(\text{A}\cup\text{B})=\frac{1}{5}$

$\frac{2}{5}+\frac{3}{10}-\text{P}(\text{A}\cup\text{B})=\frac{1}{5}$

$\text{P}(\text{A}\cup\text{B})=\frac{1}{2}$

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

$\text{P}\Big(\frac{\overline{\text{A}}}{\overline{\text{B}}}\Big)\text{ P}\Big(\frac{\overline{\text{B}}}{\overline{\text{A}}}\Big)=\frac{\big[\text{P}(\overline{\text{A}\cup\text{B}})\big]^2}{\text{P}(\overline{\text{B}})\text{ P}(\overline{\text{A}})}$

$\text{P}\Big(\frac{\overline{\text{A}}}{\overline{\text{B}}}\Big)\text{ P}\Big(\frac{\overline{\text{B}}}{\overline{\text{A}}}\Big)=\frac{\big[1-\text{P}(\text{A}\cup\text{B})\big]^2}{\text{P}(\overline{\text{B}})\text{ P}(\overline{\text{A}})}$

$\text{P}\Big(\frac{\overline{\text{A}}}{\overline{\text{B}}}\Big)\text{ P}\Big(\frac{\overline{\text{B}}}{\overline{\text{A}}}\Big)=\frac{\Big[1-\frac{1}{2}\Big]^2}{\frac{7}{10}\times\frac{3}{5}}$

$\text{P}\Big(\frac{\overline{\text{A}}}{\overline{\text{B}}}\Big)\text{ P}\Big(\frac{\overline{\text{B}}}{\overline{\text{A}}}\Big)=\frac{25}{42}$

4. 12

Solution:

X represents number of males.

$\text{p = q}=\frac{1}{2}$

$\text{p(n}-1)=\frac{3}{2^{10}}$

$\text{ }^{\text{n}}\text{C}_{\text{n}-1}\text{p}^{\text{n}-1}\text{q}=\frac{3}{2^{10}}$

$\text{n}\big(\frac{1}{2}\big)^{\text{n}-1}\big(\frac{1}{2}\big)^{\text{n}}=\frac{3}{2^{10}}$

$\text{n}\big(\frac{1}{2}\big)^{\text{n}}=\frac{1}{4}\times\frac{3\times4}{2^{10}}$

$\text{n}\big(\frac{1}{2}\big)^{\text{n}}=12\big(\frac{1}{2}\big)^{12}$

$\Rightarrow\text{n}=12$

1. 49, 50

Solution:

When a coin is tossed $\text{p = q}=\frac{1}{2}$

$\Rightarrow\text{P(X = r})^{\text{ }^{\text{n}}}\text{C}_{\text{r}}\times0.5^{\text{n}}$

Coin is tossed 99 times.

For odd number of n maximum terms at

$\text{r}=\frac{\text{n}-1}{2}$ and $\text{r}=\frac{\text{n}+1}{2}$

$\text{n}=99\Rightarrow\text{r}=49 \text{ or }50$

2. $\frac{\text{ }^7\text{C}_2\times5^5}{6^8}$

Solution:

probability of getting $6=\text{p}=\frac{1}{6},\text{q}=\frac{5}{6}$

probability of getting third six in eight throw.

= probability of getting 2 sixes in first seven throw + probability of getting six in eight throw

$=\Big(\text{ }^7\text{C}_2\big(\frac{1}{6}\big)^2\big(\frac{5}{6}\big)^5\Big)\big(\frac{1}{6}\big)$

$=\frac{\text{ }^7\text{C}_2\times5^5}{6^8}$

3. $\frac{84\times5^6}{6^{10}}$

Solution:

A fair die is thrown then probebility of getting 6 isp $=\frac{1}{6}.$

$\Rightarrow\text{q}=\frac{5}{6}$

To find probability that on tenth throw 4^th six appears, in the first nine throw 3 six should appear.

Required probability = P(3 six in first 9 throw) × P(a six in tenth throw)

Required probability $=\text{ }^9\text{C}_3\big(\frac{1}{6}\big)^3\big(\frac{5}{6}\big)^6\times\frac{1}{6}$

Required probability $=\frac{84\times5^6}{6^{10}}$

2. Dependent.

Solution:

S = [(HHH), (HHT), (HTH), (HTT), (THH), (THT), (TTH), (TTT)]

$\text{P(A)}=\text{P}(2\text{heads})=\frac{3}{8}$

$\text{P(B)}=\text{P}(\text{last one is heads})=\frac{4}{8}$

$\text{P}(\text{A}\cap\text{B})=\frac{2}{8}=\frac{1}{4}\neq\text{P(A) P(B)}$

Thus, A and B are dependent.

2. $\frac{119}{128}$

Solution:

$\text{n = 8,}\text{p}=\frac{1}{2}=\text{q}$

$\text{P(|X}-4|)\leq2$

$\Rightarrow-2\leq\text{x}-4\leq2$

$\Rightarrow4-2\leq\text{x}\leq2+4$

$\Rightarrow2\leq\text{x}\leq6$

$\text{P}(2\leq\text{x}\leq6)=\text{P(2)+P(3)+P(4)+P(5)+P(6)}$

$\text{P(2}\leq\text{x}\leq6)=\text{ }^8\text{C}_2\Big(\frac{1}{2^8}\Big)+\text{ }^8\text{C}_3\Big(\frac{1}{2^8}\Big)\\+\text{ }^8\text{C}_4\Big(\frac{1}{2^8}\Big)\text{ }^8\text{C}_5\Big(\frac{1}{2^8}\Big)+\text{ }^8\text{C}_6\Big(\frac{1}{2^8}\Big)$

$=\frac{119}{128}$

1. Event A and B are mutually exclusive, exhaustive and complementary events.

Solution:

Since P(A) + P(B) = 1

$\therefore\text{A}\cap\text{B}=0.$

Thus, event A and B are mutually exclusive, exhaustive and complementary events.

3. $\frac{15}{2^{13}}$

Solution:

Let X be the number of heads.

$\text{p}=\frac{1}{2}\Rightarrow\text{q}=\frac{1}{2}\dots(1)$

$\text{P(X}=7)=\text{P(X}=9)$

$\text{ }^{\text{n}}\text{C}_7\text{p}^7\text{q}^{\text{n}-7}=\text{ }^{\text{n}}\text{C}_9\text{p}^9\text{q}^{\text{n}-9}$

$\frac{\text{ }^{\text{n}}\text{C}_7}{\text{ }^{\text{n}}\text{C}_9}=\frac{\text{q}^{\text{n}-9}}{\text{q}^{\text{n}-7}}\times\frac{\text{p}^9}{\text{p}^7}$

$\frac{\frac{\text{n}!}{7!(\text{n}-7)!}}{\frac{\text{n}!}{9!(\text{n}-9)!}}=\text{q}^{-2}\text{p}^2$

$\frac{9!(\text{n}-9)!}{7!(\text{n}-7)!}=\frac{\text{p}^2}{\text{q}^2}$

$\frac{9\times8\times7!(\text{n}-9)!}{7!(\text{n}-7)(\text{n}-8)(\text{n}-9)!}=1\dots\big[\because\text{from (1)}\big]$

$9\times8=(\text{n}-7)(\text{n}-8)$

Comparing both sides,

$\text{n}-7=9\Rightarrow\text{n}=16$

$\Rightarrow\text{P(X}=2)=\text{ }^{16}\text{C}_2\times0.5^2\times0.5^{14}$

$\Rightarrow\text{P(X}=2)=\frac{15}{2^{13}}$

3. $\text{P}(\text{A})-\text{P}(\text{B})$

Solution:

If A and B are independent, then $\text{P}(\text{A}\cap\text{B})=\text{P}(\text{A})\cdot\text{P}(\text{B})$

1. $\frac{1}{2}$

Solution:

Here, $\text{n}=100$

Let X denote the number of times a tail is obtained.

Here, $\text{p = q}=\frac{1}{2}$

$\text{P(X = odd})=\text{P(X}=1,3,5,\dots99)$

$=\big(\text{ }^{100}\text{C}_1+\text{ }^{100}\text{C}_3+\dots+\text{ }^{100}{\text{C}}_{99}\big)\big(\frac{1}{2}\big)^{100}$

= Sum of odd coefficients in binomial expansion in $(1+\text{x})^{100}\big(\frac{1}{2}\big)^{100}$

$=\frac{2^{(100-1)}}{2^{100}}$

$=\frac{1}{2}$

3. $\frac{1-\text{P}(\text{A}\cup\text{B})}{\text{P}(\overline{\text{B}})}$

Solution:

We have,

$\text{P(A)}\neq0$ and $\text{P(B)}\neq1$

Now,

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

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

$=\frac{1-\text{P}(\text{A}\cup\text{B})}{\text{P}(\overline{\text{B}})}$

Hence, the correct alternative is option (C).

4. None of the above is correct.

Solution:

If two events A and B are independent, then we know that

$\text{P}(\text{A}\cap\text{B})=\text{P}(\text{A})\cdot\text{P}(\text{B}),\text{P}(\text{A})\neq0,\text{P}(\text{B})\neq0$

Since, A and B have a common outcome.

Further, mutually exclusive events never have a common outcome.

In other words, two independents events having non-zero probabilities of occurrence cannot be mutually exclusive and conversely, i.e., two mutually exclusive events having non-zero probabilities of outcome cannot be independent.

2. $\text{P}(\text{A}\cap\text{B})=\text{P}(\text{A}).\text{P}(\text{B})$

Solution:

Since, A and B are independent events

$\therefore\text{P}(\text{A}\cap\text{B})=\text{P}(\text{A}).\text{P}(\text{B})$

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

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

2. $\frac{51}{101}$

Solution:

Let X denote the number of coins showing head.

Therefore, X follows a binomial distribution with p and n as parameters.

Given that $\text{P(X}=50)=\text{P(X}=51)$

$\Rightarrow\text{ }^{100}\text{C}_{50}\text{p}^{50}\text{q}^{50}=\text{ }^{100}\text{C}_{51}\text{p}^{51}\text{q}^{49}$

on simplifying we get,

$\frac{51}{50}=\frac{\text{p}}{\text{q}}$

$\Rightarrow\frac{51}{50}=\frac{\text{p}}{1-\text{p}}$ (Since p + q = 1)

$\Rightarrow\text{p}=\frac{51}{101}$

4. $\big(\frac{9}{10}\big)^5+\frac{1}{2}\big(\frac{9}{10}\big)^4$

Solution:

$\text{p}=\frac{10}{100}=\frac{1}{10},\text{q}=\frac{90}{100}=\frac{9}{10},\text{n}=5$

$\text{P(X}\leq1)=\text{P(0)}+\text{P(1)}$

$\text{P(X}\leq1)=\big(\frac{9}{10}\big)^{5}+\text{ }^5\text{C}_1\big(\frac{1}{10}\big)\big(\frac{9}{10}\big)^{4}$

$\text{P(X}\leq1)=\big(\frac{9}{10}\big)^{5}+\big(\frac{1}{2}\big)\big(\frac{9}{10}\big)^4$

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

Solution:

If $\text{A}\neq\phi,\text{B}\neq\phi,$ then $\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{\text{P}(\text{A}\cap\text{B})}{\text{P}(\text{B})}$

4. $\Big(\frac{9}{10}\Big)^5+\frac{1}{2}\Big(\frac{9}{10}\Big)^4$

Solution:

We have, $\text{n}=5,\text{P}=\frac{10}{100}=\frac{1}{10}$ and $\text{q}=\frac{9}{10}$

$\text{r}<1\Rightarrow\text{r}=0,1 $

Also, $\text{P}(\text{X}=\text{r})={^\text{n}}\text{C}_\text{r}\text{P}^\text{r}\text{q}^{\text{n}-\text{r}}$

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

$={^5}\text{C}_0\Big(\frac{1}{10}\Big)^0\Big(\frac{9}{10}\Big)^5+{^5}\text{C}_1\Big(\frac{1}{10}\Big)^1\Big(\frac{9}{10}\Big)^4$

$=\Big(\frac{9}{10}\Big)^5+5\cdot\frac{1}{10}\cdot\Big(\frac{9}{10}\Big)^4$

$=\Big(\frac{9}{10}\Big)^5+\frac{1}{2}\Big(\frac{9}{10}\Big)^4$

1. Event A and B are mutually exclusive, exhaustive and complementary events.

Solution:

Since P(A) + P(B) = 1

$∴ \text{A ∩ B} =0.$

Thus, event A and B are mutually exclusive, exhaustive and complementary events.