Maths M.C.Q (1 Marks)

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

Solution:

If A and B are two events such that $\text{A}\neq\phi, \text{B}=\phi$ then,

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

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

Solution:

Here, $\bar{\text{p}}=0.3\Rightarrow\text{p}=0.7$

and $\text{q}=0.3,\text{n}=5$ and $\text{r}=4$

$\therefore$ Required probability $={^5}\text{C}_4(0.7)^4(0.3)$

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

Solution:

Let X denote the number of defective bulbs.

Hence, the binomial distribution is given by

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

$\& \text{ q}=\frac{90}{100}=\frac{9}{10}$

Hence, the distribution is given by

$\text{P(X = r})=\text{ }^5\text{C}_{\text{r}}\big(\frac{1}{10}\big)^{\text{r}}\big(\frac{9}{10}\big)^{5-\text{r}}$

$\therefore\text{P(X}=0)=\big(\frac{9}{10}\big)^5$

2. $\text{ }^{16}\text{C}_6\big(\frac{1}{4}\big)^6\big(\frac{3}{4}\big)^{10}$

Solution:

$\text{np}=4,\text{npq}=3$

$\Rightarrow\text{q}=\frac{3}4{},\text{p}=\frac{1}{4},\text{n}=16$

$\text{P(X}=6)=\text{ }^{16}\text{C}_6\big(\frac{1}{4}\big)^6\big(\frac{3}{4}\big)^{10}$

4. $\frac{15}{16}$

Solution:

$\text{n}=4,\text{p = q}=\frac{1}{2}$

$\text{P(X}\geq1)=1-\text{P(X}=0)$

$\text{P(X}\geq1)=1-\big(\frac{1}{2}\big)^4$

$\text{P(X}\geq1)=\frac{15}{16}$

4. 10

Solution:

3. 32

Solution:

$\sum\limits_2^5\text{P}(\text{x})=1$

$\frac{5}{\text{k}}+\frac{7}{\text{k}}+\frac{9}{\text{k}}+\frac{11}{\text{k}}=1$

$\text{k}=32$

NOTE: Question is modified.

2. 3

Solution:

A fair coin is tossed $\Rightarrow\text{p = q}=\frac{1}{2}$

$\text{P(X}\geq1)\geq0.8$

$\Rightarrow1-\text{P}(0)\geq0.8$

$\Rightarrow\text{P(0)}=0.2$

$\Rightarrow\big(\frac{1}{2}\big)^{\text{n}}=0.2$

$\Rightarrow2^{-\text{n}}=0.2$

$\Rightarrow2^{\text{n}}\geq5$

$\Rightarrow\text{n}\geq3$

4. $\frac{1}{8},\frac{3}{4}$

Solution:

$\sum\limits_0^3\text{P}(\text{x})=1$

$\text{k}+3\text{k}+3\text{k}+\text{k}=1$

$\text{k}=\frac{1}{8}$

$\text{V(x)}=\text{E}(\text{x}^2)-[\text{E}(\text{x})^2]$

$\text{V(x)}=3-(1.5)^2$

$\text{V(x)}=0.75=\frac{3}{4}$

2. $\frac{7}{128}$

Solution:

$\text{n}=10,\text{p = q}=\frac{1}{2}$

$\text{P(X}\geq8)=\text{P(8) + P(9) + P(10)}$

$\text{P(X}\geq8)=\text{ }^{10}\text{C}_8\big(\frac{1}{2}\big)^{10}+\text{ }^{10}\text{C}_{9}\big(\frac{1}{2}\big)^{10}+\text{ }^{10}\text{C}_{10}\big(\frac{1}{2}\big)^{10}$

$\text{P(X}\geq8)=\frac{45+10+1}{2^8}$

$\text{P(X}\geq8)=\frac{56}{256}=\frac{7}{128}$

Let E be the event of getting an even prime number on each die.

$\therefore\text{E}=\left\{\left(2,\ 2\right)\right\}$

$\Rightarrow\text{P}(\text{E})=\frac{1}{36}$

4. 0.4

Solution:

Let:

P(X = 0) = m

P(X = 1) = k

Now,

P(X = 3) = 2k

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

⇒ 0 + k + 0.6 + 6k = 1.3

⇒ 7k = 1.6 - 0.6

$\Rightarrow\text{k}=\frac{0.7}{7}$

⇒ 0.1

We know that the sum of probabilities in a probability distribution is always 1.

$\therefore$ P(X = 0) + P(X = 1) + P(X = 3) = 1

⇒ m + 0.1 + 0.3 + 0.2 = 1

⇒ m + 0.6 = 1

⇒ m = 0.4

$\therefore\ \text{P}(\text{A}|\text{B})=\frac{\text{P}(\text{A}\cap\text{B})}{\text{P}(\text{B})}=\frac{0}{0}=\text{not defined}$

1. $\frac{1}{10}$

Solution:

We know that the sum of probabilities in a probability distribution is always 1.

$\therefore$ P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) +P(X = 6) + P(X = 7) +P(X = 8) = 1

⇒ 0 + 2p + 2p + 3p + p² + 2p² + 7p² + 2p = 1

⇒ 10p² + 9p - 1 = 0

⇒ (10p - 1)(p + 1) = 0

$\Rightarrow\text{p}=\frac{1}{10}\text{ or }-1$ (Negleting -1 as the value of the probability cannot be negative)

3. 12

Solution:

$\text{p}=\frac{1}{4},\sqrt{\text{npq}}=3$

$\Rightarrow\text{q}=\frac{3}{4},\text{npq}=9$

$\Rightarrow\text{Mean = np}=\frac{9}{\text{q}}$

$\Rightarrow\text{Mean}=9\times\frac{4}{3}=12$

1. 0.77

Solution:

P(E) = P(2) + P(3) + P(5) + P(7)

P(E) = 0.23 + 0.12 + 0.20 + 0.07

P(E) = 0.62

And

P(F) = P(1) + P(2) + P(3)

P(F) = 0.15 + 0.23 + 0.12

P(F) = 0.5

Also,

$\text{P}(\text{E}\cap\text{F})=\text{P}(2)+\text{P}(3)$

$\text{P}(\text{E}\cap\text{F})=0.23+0.12$

$\text{P}(\text{E}\cap\text{F})=0.35$

$\text{P}(\text{E}\cup\text{F})=\text{P}(\text{E})+\text{P(F)}-\text{P}(\text{E}\cap\text{F})$

$\text{P}(\text{E}\cup\text{F})=0.62+0.5-0.35$

$\text{P}(\text{E}\cup\text{F})=0.77$

3. $\frac{15}{29}$

Solution:

For sum of two integers to be odd, one integer should be even and the other should be odd. In 30 consecutive integers, 15 are even and 15 are odd.

P(Sum is odd) = P(first integer is odd and second is even) + P(first integer is even and second integer is odd)

$=\frac{15}{30}\times\frac{15}{29}+\frac{15}{30}\times\frac{15}{29}$

$=\frac{450}{30\times29}$

$=\frac{15}{29}$

2. $\frac{105}{512}$

Solution:

$\text{n}=10,\text{x}=6,\text{p = q}=\frac{1}{2}$

$\text{P(X}=6)=\text{ }^{10}\text{C}_6\big(\frac{1}{2}\big)^{10}​=\frac{105}{512}$

Distracter Rationale.

1. Let P(A) = m, P(B) = n, 0 < m, n < 1

A and B are mutually exclusive.

$\therefore\text{A}\cap\text{B}=\phi$

$\Rightarrow\text{P}(\text{AB})=0$

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

$\therefore\text{P}(\text{A}).\text{P}(\text{B})\neq\text{P}(\text{AB})$

2. Consider the result given in alternative.

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

$\Rightarrow\text{P}(\text{A}'\cap\text{B}')=1-\text{P}(\text{A})-\text{P}(\text{B})+\text{P}(\text{A}).\text{P}(\text{B})$

$\Rightarrow1-\text{P}(\text{A}\cup\text{B})=1-\text{P}(\text{A})-\text{P}(\text{B})+\text{P}(\text{A}).\text{P}(\text{B})$

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

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

$\Rightarrow\text{P}(\text{AB})=\text{P}(\text{A}).\text{P}(\text{B})$

This implies that A and B are independent, if $\text{P}(\text{A}'\text{B}')=\big[1-\text{P}(\text{A})\big]\big[1-\text{P}(\text{B})\big]$

3. Let A: Event of getting an odd number on throw of a die = {1, 3, 5}

$\Rightarrow\text{P}(\text{A})=\frac{3}{6}=\frac{1}{2}$

B: Event of getting an even number on throw of a die = {2, 4, 6}

$\text{P}(\text{B})=\frac{3}{6}=\frac{1}{2}$

Here, $\text{A}\cap\text{B}=\phi$

$\therefore\text{P}(\text{AB})=0 $

$\text{P}(\text{A}).\text{P}(\text{B})=\frac{1}{4}\neq0$

$ \Rightarrow\text{P}(\text{A}).\text{P}(\text{B})\neq\text{P}(\text{AB})$

4. From the above example, it can be seen that,

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

However, it cannot be inferred that A and B are independent.

Thus, the correct answer is B.

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

Solution:

Consider,

$\text{P(X = r})=\text{kP(X = n}-\text{r})$

Using $\text{ }^{\text{n}}\text{C}_{\text{r}}=\text{ }^{\text{n}}\text{C}_{\text{n}-\text{r}},\text{q}=1-\text{p}$

$\text{p}^{\text{r}}\text{q}^{\text{n}-\text{r}}=\text{kp}^{\text{n}-\text{r}}\text{q}^{\text{r}}$

$\text{p}^{\text{r}}(1-\text{p})^{\text{n}-\text{r}}=\text{kp}^{\text{n}-\text{r}}(1-\text{p})^{\text{r}}$

$\text{P}^{2\text{r}-\text{n}}=\text{k}(1-\text{p})^{2\text{r}-\text{n}}$

$\big(\frac{\text{p}}{\text{q}}\big)^{2\text{r}-\text{n}}=\text{k}$

when $\text{p = q}$ then $\text{k}=1$

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

3. 32.

Solution:

We know that, $\sum\text{P}\text{X}=1$

$\Rightarrow\frac{5}{\text{k}}+\frac{7}{\text{k}}+\frac{9}{\text{k}}+\frac{11}{\text{k}}=1$

$\Rightarrow\frac{32}{\text{k}}=1$

$\therefore\text{k}=32$

4. $\text{None of these}$

Solution:

If last digit is either O or 5 then the number is divisible by 5.

Case : 1

Last digit is 0.

First three places can be selected by 9 × 9 × 9 = 729 ways.

If c = 0 then three places can be selected by 9 × 8 × 1 = 72

If C ≠ 0 then 729 - 72 = 657

Fourth place has 8 choices = 657 × 8 = 5256

Total = 72 + 5256 = 5904

Case : 2

If C = 5

First place other than 5

then first three places can be filled in 8 × 8 × 1 = 64

If first place is 5 then first three places can be filled in 1× 9 × 1 = 9 ways.

If third place is other than 5 then 729 - 64 - 9 = 656 ways.

For fourth place has 8 choices.

As per required condition = (64 + 9) × 9 + 656 × 8 = 5905

required probability $=\frac{5904+5905}{9\times10\times10\times10\times10}=\frac{11809}{90000}$

NOTE: Answer not matching with back answer.

4. 3

Solution:

A fair coin is tossed $\Rightarrow\text{p = q}=\frac{1}{2}$

$\text{P(X}\geq1)\geq0.8$

$\Rightarrow1-\text{P}(0)\geq0.8$

$\Rightarrow\text{P}(0)=0.2$

$\Rightarrow\big(\frac{1}{2}\big)^{\text{n}}=0.2$

$\Rightarrow2^{-\text{n}}=0.2$

$\Rightarrow2^{\text{n}}\geq5$

$\Rightarrow\text{n}\geq3$

4. $\frac{\pi}{8}$

Solution:

$\text{I}=\int\limits^1_0\sqrt{\text{x}(1-\text{x})}\text{ dx} $

$\text{I}= \int\limits^1_0\sqrt{\text{x}-\text{x}^2}\text{dx}$

$\text{I}=\int\limits^1_0\sqrt{\frac{1}{4}+\text{x}-\text{x}^2+\frac{1}{4}}\text{dx}$

$\text{I}=\int\limits^1_0\sqrt{\frac{1}{4}-\Big(\text{x}^2-\text{x}+\frac{1}{4}\Big)}\text{dx}$

$\text{I}=\int\limits^1_0\sqrt{\Big(\frac{1}{2}\Big)^2-\Big(\text{x}-\frac{1}{2}\Big)^2}\text{dx}$

$\text{I}=\Bigg[\frac{\text{x}-\frac{1}{2}}{2}\sqrt{\text{x}(1-\text{x})}+\frac{1}{2}\times\frac{1}{4}\sin^{-1}(2\text{x}-1)\Bigg]^1_0$

$\text{I}=0+\frac{1}{8}\big(\sin^{-1}(1)-\sin^{-1}(-1)\big)$

$\text{I}= \frac{1}{8}\Big(\frac{\pi}{2}-\Big(\frac{\pi}{2}\Big)\Big)$

$\text{I}= \frac{\pi}{8}$

4. $\frac{3}{28}$

Solution:

Required probability $=\text{P}_{\text{D}}\cdot\text{P}_{\text{D}}$

$=\frac{3}{8}\cdot\frac{2}{7}=\frac{3}{28}$

3. $\frac{1}{21}$

Solution:

10 persons can sit around a table in 9! ways.

Consider the particular four persons as one unit.

Now, the entities are 6 + 1 = 7

These 7 entities can be arranged in 6! ways.

In the entities itself they can be arranged in 4! ways.

The required number of arrangements = 6!4!

Probability $= \text{nm}​ = \frac{!6!4}{9!} ​= \frac1{21}​$

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

Solution:

Consider,

$\text{P(X = r) = kP(X = n}-\text{r})$

Using $\text{ }^{\text{n}}\text{C}_{\text{r}}=\text{ }^{\text{n}}\text{C}_{\text{n}-\text{r}},\text{q}=1-\text{p}$

$\text{p}^{\text{r}}\text{q}^{\text{n}-\text{r}}=\text{kp}^{\text{n}-\text{r}}\text{q}^{\text{r}}$

$\text{p}^{\text{r}}(1-\text{p})^{\text{n}-\text{r}}=\text{kp}^{\text{n}-\text{r}}(1-\text{p})^{\text{r}}$

$\text{p}^{2\text{r}-\text{n}}=\text{k}(1-\text{p})^{\text{2r}-\text{n}}$

$\big(\frac{\text{p}}{\text{q}}\big)^{\text{2r}-\text{n}}=\text{k}$

when p = q then k = 1

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

2. $\frac{12}{17}$

Solution:

We define the following events:

A₁​: Selecting a pair of consecutive letters from the word LONDON

A₂​: Selecting a pair of consecutive letters from the word CLIFTON

E: Selecting a pair of letters ON

Then ${\text{P(A}_1​∩\text{E})=\frac52},$​ as there are 5 pairs of consecutive letters out of which 2 are ON.

$\text{P(A}_2​∩\text{E})=\frac61,$​ as there are 6 pairs of consecutive letters of which 1 is ON.

So, required probability $\text{P}=\Big(\frac{\text{A}_1}{\text{E}}\Big)$

$\Rightarrow\Big(\frac{\text{A}_1}{\text{E}}\Big)=\frac{\text{P}(\text{A}_1\cap\text{E})}{\text{P}(\text{A}_1\cap\text{E}) + \text{P}(\text{A}_1\cap\text{E})}=\frac{\frac25}{\frac25+\frac16}=\frac{12}{17}$

1. $\frac{5}{6}$

Solution:

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

$\text{P}\Big(\frac{\text{A}'}{\text{B}'}\Big)=\frac{\text{P}(\text{A}'\cap\text{B}')}{\text{P}(\text{B}')}+\frac{1-\text{P}(\text{A}\cap\text{B})}{1-\text{P}(\text{B})}$

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

$=\frac{1-\Big(\frac{2}{5}+\frac{3}{10}-\frac{1}{5}\Big)}{1-\frac{3}{10}}$

$=\frac{1-\Big(\frac{4+3-2}{10}\Big)}{\frac{7}{10}}-\frac{1-\frac{1}{2}}{\frac{7}{10}}$

$=\frac{5}{7}$

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

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

$=\frac{1-\frac{1}{2}}{1-\frac{2}{5}}$ $\Big[\because\text{P}(\text{A}\cup\text{B})=\frac{1}{2}\Big]$

$=\frac{\frac{1}{2}}{\frac{3}{5}}=\frac{5}{6}$

$\therefore\text{P}\Big(\frac{\text{A}'}{\text{B}'}\Big)\cdot\text{P}\Big(\frac{\text{B}'}{\text{A}'}\Big)=\frac{5}{7}\cdot\frac{5}{6}=\frac{25}{42}$

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

Solution:

Total number of vowels in English alphabet = 5 which are a, e, i, o, u

So, probability of u when a vowel is selected $=\frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}=\frac15$

4. None of the above is correctIf two. events are independent, then.

Solution:

Let A and B are two independent events, Then,

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

As, $\text{P}(\text{A}\cap\text{B})\neq0\text{ or }\text{P(A)}+\text{P(B)}\neq1$

So, both are neither mutually exclisive nor their sum of probability is 1.

Hence, the correct alternative is option (d).

3. $\frac{6}{35}$

Solution:

Bag A has 4 green balls and 6 red balls

⇒ probability of choosing green ball from A is p(greenA​) $= \frac4{10}$

​Bag A has 3 green balls and 4 red balls

⇒ probability of choosing green ball from B is p(green_B​) $=\frac37$

​On choosing one ball from each bag probability  that both are green = p(green_A​) × p(green_B​)

$=\frac4{10}\times\frac37=\frac{6}{35}$

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

Solution:

Let A be the event that students failed in Physics. B be the event that students failed in Mathematics.

Given that, $\text{P(A)}=30\%=\frac{30}{100}$

$\text{P(B)}=25\%=\frac{25}{100}$

$\text{P}(\text{A}\cap\text{B})=10\%=\frac{10}{100}$

Required probability is given by $\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)$

$\Rightarrow\ \text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{\text{P}(\text{A}\cap\text{B})}{\text{P(B)}}=\frac{\frac{10}{100}}{\frac{25}{100}}=\frac{2}{5}$

4. 1

Solution:

$\text{P(B)}=\frac{3}{5},\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{1}{2},\text{P}\Big({\text{A}}\cup{\text{B}}\Big)=\frac{4}{5}$

Consider,

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

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

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

$\Rightarrow\ \text{P}(\text{A}\cap\text{B})=\frac{3}{10}$

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

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

$=1-\frac{3}{10}+\frac{3}{5}-\frac{3}{10}$

$=1$

2. $\frac{11}{35}$

Solution:

Out of the 70 numbers, numbers that are a multiple of 5 or 7 are 5, 7, 10, 14, 15, 20, 21, 25, 28, 30, 35, 40, 42, 45, 49, 50, 55, 56, 60, 63, 65, 70

So, probability that the number is even $=\frac{22}{70}=\frac{22}{70}=\frac{11}{35}$

2. $\frac{8}{17}$

Solution:

9 can be obtained from throw of two dice in only 4 cases as given below:

{(3, 6), (4, 5), (5, 4), (6, 3)]

$\Rightarrow\ \text{P(getting }9)=\frac{4}{36}=\frac{1}{9}$

$\text{P(not getting }9)=\frac{32}{36}=\frac{8}{9}$

Now,

P(B is winning) = P(getting 9 in 2^nd throw) + P(getting p in 4^th throw) + P(getting 9 in 6^th throw) + .....

$=\frac{8}{9}\times\frac{1}{9}+\frac{8}{9}\times\frac{8}{9}\times\frac{8}{9}\times\frac{1}{9}+\ .....$

$=\frac{8}{81}\Big[1+\frac{64}{81}+\Big(\frac{64}{81}\Big)^2+\ ......\Big]$

$=\frac{8}{81}\times\frac{1}{1-\frac{64}{81}}$

$=\frac{8}{81}\times\frac{81}{17}$

$=\frac{8}{17}$

4. $\frac{10}{13}$

Solution:

Let E₁ = Event that both A and B solve the problem

$\therefore\text{P}(\text{E}_1)=\frac{1}{3}\times\frac{1}{4}=\frac{1}{12}$

Let E₂ = Event that both A and B got incorrect solution of the problem

$\therefore\text{P}(\text{E}_2)=\frac{2}{3}\times\frac{3}{4}=\frac{1}{2}$

Here, $\text{P}\Big(\frac{\text{E}_1}{\text{E}}\Big)=1,\text{P}\Big(\frac{\text{E}}{\text{E}_2}\Big)=\frac{1}{20}$

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

$=\frac{\frac{1}{12}\times1}{\frac{1}{12}\times1+\frac{1}{2}\times\frac{1}{20}}=\frac{10}{30}$

4. 0.9

Solution:

If $\text{P}(\text{A}\cup\text{B})=0.8\text{ P}(\text{A}\cap\text{B})=0.3,$

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

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

$\Rightarrow\ \text{P(A)}+\text{P(B)}=1.1$

$\Rightarrow\ \text{P}(\overline{\text{A}})+\text{P}(\overline{\text{B}})=1-\text{P(A)}+1-\text{P(B)}$

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

$\Rightarrow\ \text{P}(\overline{\text{A}})+\text{P}(\overline{\text{B}})=2-1.1$

$\Rightarrow\ \text{P}(\overline{\text{A}})+\text{P}(\overline{\text{B}})=0.9$

4. $\frac{8}{15}$

Solution:

A white ball can be drawn in two mutually exclusive ways:

1. Selecting bag X and then drawing a white ball from it.
2. Selecting bag Y ane then drawing a white ball from it.

Let E₁, E₂ and A be the three evenes as defined below:

E₁ = Selecting abg X

E₂ = Selecting bag Y

A = Drawing a white ball

We know that one bag is selected randomly.

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

Solution:

Given that,

$\text{P(A)}=0.6,\text{P(B)}=0.2,\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=0.5$

Consider,

$\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=0.5$

$\Rightarrow\frac{\text{P}(\text{A}\cap\text{B})}{\text{P(B)}}=0.5$

$\Rightarrow\frac{\text{P}(\text{A}\cap\text{B})}{0.2}=0.5$

$\Rightarrow \text{P}(\text{A}\cap\text{B})=0.1$

$\Rightarrow \text{P(A)}+\text{P(B)}-\text{P}(\text{A}\cup\text{B})=0.1$

$\Rightarrow \text{P}(\text{A}\cup\text{B})=\text{P(A)}+\text{P(B)}-0.1$

$\Rightarrow\text{P}(\text{A}\cup\text{B})=0.7$

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

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

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

To find

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

$\text{P}\Big(\frac{\overline{\text{A}}}{\overline{\text{B}}}\Big)=\frac{0.3}{0.8}$

$\text{P}\Big(\frac{\overline{\text{A}}}{\overline{\text{B}}}\Big)=\frac{3}{8}$

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

Solution:

From the diagram, we get $\text{A}\cap\text{B}$ and $\overline{\text{A}}\cap\text{B}$ are mutually exclusive events such that $(\text{A}\cap\text{B})\cup(\overline{\text{A}}\cap\text{B})=\text{B}.$ therefore by

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

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

3. $\frac{1}{8}$

Solution:

Let E₁ = Event for getting an even number on the die

And E₂ = Event that a spade card is selected.

$\therefore\text{P}(\text{E}_1)=\frac{3}{6}=\frac{1}{2}$ and $\text{P}(\text{E}_2)=\frac{13}{52}=\frac{1}{4}$

Then, $\text{P}(\text{E}_1\cap\text{E}_2)=\text{P}(\text{E}_1)\cdot\text{P}(\text{E}_2)$

$=\frac{1}{2}\cdot\frac{1}{4}=\frac{1}{8}$

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

Solution:

We have, $\text{P}(\text{A})=\frac{1}{2},\text{P}(\text{B})=\frac{1}{3}$ and $\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{1}{4}$

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

$=\frac{1}{4}\cdot\frac{1}{3}=\frac{1}{12}$

Now, $\text{P} ({\text{A}'}\cap{\text{B}'})=1-\text{P}(\text{A}\cup{\text{B}})$

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

$=1-\Big[\frac{1}{2}+\frac{1}{3}-\frac{1}{12}\Big]=1-\frac{9}{12}$

$=\frac{3}{12}=\frac{1}{4}$

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

Solution:

We have, $\text{P}(\text{B})=\frac{3}{5},\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{1}{2}$ and $\text{P}(\text{A}\cup\text{B})=\frac{4}{5}$

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

$=\frac{1}{2}\cdot\frac{3}{5}=\frac{3}{10}$

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

$\Rightarrow\frac{4}{5}=\text{P}(\text{A})+\frac{3}{5}-\frac{3}{10}$

$\therefore\text{P}(\text{A})=\frac{4}{5}-\frac{3}{5}+\frac{3}{10}=\frac{1}{2}$