Maths M.C.Q (1 Marks)

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)

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

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

$=\frac{15}{29}$

4. $\frac{6}{25}$

Solution:

$\text{P(A)}=\frac{3}{8},\text{P(B)}=\frac{5}{8},\text{P}\Big({\text{A}}\cup{\text{B}}\Big)=\frac{3}{4}$

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

$\Rightarrow\ \frac{3}{4}=\frac{3}{8}+\frac{5}{8}-\text{P}(\text{A}\cap\text{B})$

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

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

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

$\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)\text{ P}\Big(\frac{\overline{\text{A}}}{\text{B}}\Big)=\frac{\frac{1}{4}}{\frac{5}{8}}\times\frac{\big(\frac{5}{8}-\frac{1}{4}\big)}{\frac{5}{8}}$

$\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)\text{ P}\Big(\frac{\overline{\text{A}}}{\text{B}}\Big)=\frac{6}{25}$

3. $\frac{48}{108}$

Solution:

Total number of balls = 5brown + 4white = 9

Required probability $=\frac{5}{9}\times\frac{4}{8}+\frac{4}{9}\times\frac{3}{8}=\frac{4}{9}$

$\Rightarrow\ \frac{4\times12}{9\times12}=\frac{48}{108}$

1. $\frac{5}{84}$

Solution:

Given:

Red balls = 2

Blue balls = 3

Black balls = 4

P(All three balls are of same colour) = P(all three are blue) + P(all three are black)

$=\frac{3}{9}\times\frac{2}{8}\times\frac{1}{7}+\frac{4}{9}\times\frac{3}{8}\times\frac{2}{7}$

$=\frac{1}{84}+\frac{4}{84}$

$=\frac{5}{84}$

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

Solution:

A Sample space when a die is thrown,

S₁ = {1, 2, 3, 4, 5, 6} ⇒ n(S₁) = 6

Let A be the event that getting even number.

A = {2, 4, 6} ⇒ n(A) = 3

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

A card is selected from a deck of 52 cards.

$\text{n}(\text{S}_2)= {^{52}}\text{C}_2=52$

Let B be the event that getting spade card.

$\text{n(B)}= {^{13}}\text{C}_2=13\Rightarrow\ \text{P(B)}=\frac{13}{52}=\frac{1}{4}$

Required probability = P(A) × P(B)

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

1. $\frac{^{7}\text{P}_5}{7_5}$

Solution:

Five persons can leave different floors

By ⁷P₅ ways.

Possible ways of leavinf the lift = 7⁵

Required probability $=\frac{^{7}\text{P}_5}{7^5}$

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. $\frac{2}{5}$

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. $\frac{8}{17}$

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. 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. 0.188

Solution:

Let:

A be the event of hitting the target by the person A,

B be the event of hitting the target by the person B and

C be the event of hitting the target by the person C

We have,

P(A) = 0.4, P(B) = 0.3 and P(C) = 0.2

Also,

$\text{P}(\overline{\text{A}})=1-\text{P(A)}=1-0.4=-0.6,$

$\text{P}(\overline{\text{B}})=1-0.3=0.7$ and

$\text{P}(\overline{\text{C}})=1-0.2=0.8$

Now,

$\text{P(Two hits)}=\text{P}(\text{AB}\overline{\text{C}})+\text{P}(\text{A}\overline{\text{B}}\text{C})+\text{P}(\overline{\text{A}}\text{BC})$

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

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

$=0.096+0.056+0.036$

$=0.188$

Hence, the correct alternative is option (d).

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. $\frac{5}{28}$

Solution:

We know a leap year is fallen within 4 years, So its probability is $\frac{25}{100}=\frac{1}{4}$

53rd Sunday leap year $=\frac{1}{4}\times\frac{2}{7}=\frac{2}{28}$

Similarly probability of 53rd Sunday in a non leap year $=\frac{75}{100}\times\frac{1}{7}=\frac{3}{4}\times\frac{1}{7}=\frac{3}{28}$

Required probability $=\frac{2}{28}+\frac{3}{28}=\frac{5}{28}$.

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

Solution:

Total balls in a box - 3orange + 3green + 2blue = 8

Three balls are drawn at random from the box then samplw space $\text{n(S)}= {^{8}}\text{C}_3=\frac{8\times7\times6}{3\times2\times1}=56$

Let A be the event that drawing 2 green and one blue ball.

$\text{n(A)}={^{3}}\text{C}_2\times{^{2}}\text{C}_2=6$

$\text{P(A)}=\frac{6}{56}=\frac{3}{28}$

3. $\frac{11}{16}$

Solution:

Rusted items = 3 + 5 = 8

Rusted nails = 3

Total nails = 6

P(getting a rusted item or a nail) = P(getting a rusted item) + P(getting a nail) - P(getting a rusted item and a nail)

$=\frac{8}{16}+\frac{6}{16}-\frac{3}{16}$

$=\frac{8+6-3}{16}$

$=\frac{11}{16}$

3. P(A) P(B)

Solution:

$\text{P}(\text{A}\cap\text{B})=\text{P(A)} \text{ P(B)}$ for independent events.

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

Solution:

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

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

$\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{\frac{7}{10}}{\frac{17}{20}}=\frac{14}{17}$

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).

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

Solution:

$\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\text{p},\text{P(A)}=\text{p},\text{P(B)}=\frac{1}{3},\text{P}(\text{A}\cup\text{B})=\frac{5}{9}$

Consider,

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

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

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

$\Rightarrow\frac{\text{p}+\frac{1}{3}-\frac{5}{9}}{\frac{1}{3}}=\text{p}$

$\Rightarrow\text{p}+\frac{1}{3}-\frac{5}{9}=\frac{\text{p}}{3}$

$\Rightarrow\frac{-2}{9}=\frac{\text{p}}{3}-\text{p}$

$=\frac{-2}{3}\text{p}=\frac{-2}{9}$

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

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

Solution:

We have,

The total number of batteries = 8

The number of dead batteries = 3

Let A be the event of selecting the first dead battery and B be the event of selecting the second dead battery.

Now,

P(both dead batteries are selected) $=\text{P}(\text{A}\cap\text{B})$

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

$=\frac{3}{8}\times\frac{2}{7}$

$=\frac{3}{28}$

Hence, the correct alternative is option (a).

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

Solution:

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

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

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

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

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

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

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

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

1. $\frac{64}{64}$

Solution:

P(good item) $=\frac{10}{16}$

P(defected item) $=\frac{6}{16}$

P(eitherr good or defected item) = P(good item) + P(defected item)

$=\frac{10}{16}+\frac{6}{16}$

$=\frac{16}{16}$

$=1$

$=\frac{64}{64}$

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

Solution:

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

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

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

$\frac{\text{P}(\text{A}\cap\text{B})}{\frac{1}{3}}=\frac{1}{4}$

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

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

$\Rightarrow\ \text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{\frac{1}{12}}{\frac{1}{3}}=\frac{1}{4}$

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

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

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

$\text{P}(\overline{\text{A}}\cap\overline{\text{B}})=1-\Big[\frac{1}{2}+\frac{1}{3}-\frac{1}{12}\Big]$

$\text{P}(\overline{\text{A}}\cap\overline{\text{B}})=\frac{1}{4}$

1. $\frac{44}{85\times49}$

Solution:

Total cards = 52 There are four suits of cards in a pack, i.e. diamond, heart, spade and club.

Pall 4 cards are of same suit = Pall 4 cards are of diamond + Pall 4 cards are of heart + Pall 4 cards are of spade + Pall 4 cards are of club.

$=4\times\frac{13}{52}\times\frac{12}{51}\times\frac{11}{50}\times\frac{10}{59}$

$=4\times\frac{11}{85\times49}$

$=\frac{44}{85\times49}$

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

Solution:

We have,

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

Now,

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

$=\frac{\Big(\frac{7}{10}\Big)}{\Big(\frac{4}{5}\Big)}$

$=\frac{7\times5}{10\times4}$

$=\frac{7}{8}$

Hence, the correct alternative is option (c).

3. $\frac{1}{70}$

Solution:

We have,

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

As, A and B are independent events

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

$=0.3\times\text{P(B)}$

$=0.3\text{ P(B)}\ .....\text{(i)}$

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

$\Rightarrow 0.5 = 0.3+\text{P(B)}-0.3\text{ P(B)}$ [Using (i)]

$\Rightarrow 0.5-0.3 = 0.7\text{ P(B)}$

$\Rightarrow0.7\text{ P(B)}=0.2$

$\Rightarrow\text{ P(B)}=\frac{0.2}{0.7}$

$\Rightarrow\text{ P(B)}=\frac{2}{7}$

Using (i), we get

$\text{P}(\text{A}\cap\text{B})=0.3\times\frac{2}{7}=\frac{6}{70}$

Now,

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

$=\frac{\Big(\frac{6}{70}\Big)}{\Big(\frac{2}{7}\Big)}-\frac{\Big(\frac{6}{70}\Big)}{0.3}$

$=\frac{6\times7}{70\times2}-\frac{6}{70\times0.3}$

$=\frac{3}{10}-\frac{2}{7}$

$=\frac{21-20}{70}$

$=\frac{1}{70}$

Hence, the correct alternative is option (c).

2. $\frac{10}{9}$

Solution:

$\text{P}(\text{A}\cup\text{B})=\frac{5}{9},\text{P}(\overline{\text{A}}\cup\overline{\text{B}})=\frac{2}{3}$

Consider,

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

$\Rightarrow\text{P}(\overline{\text{A}}\cup\overline{\text{B}})=\frac{2}{3}$

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

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

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

$\Rightarrow\ \text{P(A)}+\text{P(B)}-\frac{5}{9}=\frac{1}{3}$

$\Rightarrow\ \text{P(A)}+\text{P(B)}=\frac{8}{9}$

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

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

$\Rightarrow\text{P}(\overline{\text{A}})+\text{P}(\overline{\text{B}})=2-\frac{8}{9}$

$\Rightarrow\text{P}(\overline{\text{A}})+\text{P}(\overline{\text{B}})=\frac{10}{9}$

1. $\frac{13}{32}$

Solution:

Total balls in first box = 3 white + 1 black = 4

Total balls in second box = 2 white + 2black = 4

Total balls in third box = 1white + 3black = 4

Probability of 2 white and 1 black

= P(WWB) + P(WBW) + P(BWW)

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

$=\frac{18+6+2}{64}=\frac{13}{32}$

1. $\frac{6}{25}$

Solution:

Number of cards divisible by 6 = 16

$\Rightarrow\ \text{P(A)}=\frac{16}{100}$

Number of cards divisible by 8 = 12

$\Rightarrow\ \text{P(B)}=\frac{12}{100}$

Number of cards divisible by 24 = 4

$\Rightarrow\ \text{P}(\text{A}\cap\text{B})=\frac{4}{100}$

$\text{P}(\text{A}\cup\text{B})=\frac{16}{100}+\frac{12}{100}-\frac{4}{100}$

$\text{P}(\text{A}\cup\text{B})=\frac{6}{25}$

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. 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.

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) $=\text{P}(\text{A}\cup\text{B})'$

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

$=1-0.61$

$=0.39$

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

Solution:

$\text{P(A)}=\frac{3}{5},\text{P(B)}=\frac{4}{9}$

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

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

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

$\text{P}(\overline{\text{A}}\cap\overline{\text{B}})=1-\Big[\frac{3}{5}+\frac{4}{9}-\frac{3}{5}\times\frac{4}{9}\Big]$

($\because$ A and B are independent)

$\text{P}(\overline{\text{A}}\cap\overline{\text{B}})=1-\frac{7}{9}$

$\text{P}(\overline{\text{A}}\cap\overline{\text{B}})=\frac{2}{9}$

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

Soluction:

Non-leap year has 365 days = 52 weeks + 1

366 days in leap year.

We want to find probability of 53 Fridays or 53 Saturday.

Favourable cases = {(Thursday, Friday), (Friday, Saturday), (Saturday, Sunday)}

Required probability $=\frac{3}{7}$

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

Solution:

Two cards are drawn from 52 cards.

Let, E₁ be the event that getting queen in first draw and E₂ be the event that getting queen in second draw,

$\text{P}(\text{E}_1\cap\text{E}_2)=\frac{4}{52}\times\frac{4}{52}=\frac{1}{13}\times\frac{1}{13}$

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

Solution:

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

Consider,

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

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

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

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

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

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

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

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

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

$\text{P}\Big(\frac{\text{B}}{\overline{\text{A}}}\Big)=\frac{3}{5}$

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

Solution:

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

$\Rightarrow\ \text{P(B)}=3\text{P(A)}\ .....(\text{i})$

A and B are mutually exclusive events.

$\Rightarrow\ \text{P}(\text{A}\cap\text{B}) = 0$

Now,

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

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

$\Rightarrow\ \text{P(A)}+3\text{P(A)}=1$ [From (i)]

$\Rightarrow\ 4\text{P(A)}=1$

$\Rightarrow\ \text{P(A)}=\frac{1}{4}$

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}

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

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

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) $=\frac{1}{2}\times\frac{1}{3}\times\frac{1}{6}=\frac{1}{36}$

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. $\frac{1}{36}$

Solution:

Required probability = Probability of ace in first throw + Probability of ace in second throw

$=\frac{1}{6}\times\frac{1}{6}=\frac{1}{36}$

4. 0

Solution:

$\text{P(A)}=0.2,\text{P(B)}=0.4,\text{P}(\text{A}\cup\text{B})=0.6$

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

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

$\Rightarrow\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{0.2+0.4-0.6}{\text{P(B)}}$

$\Rightarrow\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=0$

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,

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

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 $=\frac{27}{100}$