Concept:

P(A ∪ B) = P(A) + P(B) - P(A ∩ B)

For mutually exclusive events A and B, P(A ∩ B) = 0

Calculation:

We know, for mutually exclusive events A and B,  P(A ∩ B) = 0

∴ P(A ∪ B) = P(A) + P(B) - 0

=  P(A) + P(B)

Hence, option (2) is correct.

Explanation:

Leap Year

A year that has 29 days in February is said to be a leap year.

Such a year has 366 days.

It has 52 weeks and 2 days.

1 week has 7 days, ∴ 52 weeks has 52 × 7 = 364 days.

Days left = 366 – 364 ⇒ 2 days.

Sample space = {(Mon,Tue) – (Tue,Wed) – (Wed,Thu) – (Thu,Fri) – (Fri, Sat) – (Sat,Sun) – (Sun,Mon)}

⇒ Sample space = 7 days

Favourable condition = {(Tue,Wed) – (Wed,Thu)} ⇒ 2 days

$\therefore \mathrm{P}\mathrm{r}\mathrm{o}\mathrm{b}\mathrm{a}\mathrm{b}\mathrm{i}\mathrm{l}\mathrm{i}\mathrm{t}\mathrm{y}\mathrm{o}\mathrm{f}\mathrm{h}\mathrm{a}\mathrm{v}\mathrm{i}\mathrm{n}\mathrm{g}53\mathrm{W}\mathrm{e}\mathrm{d}\mathrm{n}\mathrm{e}\mathrm{s}\mathrm{d}\mathrm{a}\mathrm{y}\mathrm{i}\mathrm{s}=\frac{\mathrm{F}\mathrm{a}\mathrm{v}\mathrm{o}\mathrm{u}\mathrm{r}\mathrm{a}\mathrm{b}\mathrm{l}\mathrm{e}\mathrm{c}\mathrm{o}\mathrm{n}\mathrm{d}\mathrm{i}\mathrm{t}\mathrm{i}\mathrm{o}\mathrm{n}}{\mathrm{S}\mathrm{a}\mathrm{m}\mathrm{p}\mathrm{l}\mathrm{e}\mathrm{s}\mathrm{p}\mathrm{a}\mathrm{c}\mathrm{e}}$
$\Rightarrow \frac{2}{7}$

Non-Leap Year

A year that has 28 days in February is said to be a non-leap year.

Such a year has 365 days.

It has 52 weeks and 1 day.

1 week has 7 days, ∴ 52 weeks has 52 × 7 = 364 days.

Days left = 365 – 364 ⇒ 1 day.

Sample space = {(Mon, Tue, Wed, Thus, Fri, Sat, Sun)}

⇒ Sample space = 7 days

Favourable condition = {Wednesday} ⇒ 1 day.

$\therefore \mathrm{P}\mathrm{r}\mathrm{o}\mathrm{b}\mathrm{a}\mathrm{b}\mathrm{i}\mathrm{l}\mathrm{i}\mathrm{t}\mathrm{y}\mathrm{o}\mathrm{f}\mathrm{h}\mathrm{a}\mathrm{v}\mathrm{i}\mathrm{n}\mathrm{g}53\mathrm{W}\mathrm{e}\mathrm{d}\mathrm{n}\mathrm{e}\mathrm{s}\mathrm{d}\mathrm{a}\mathrm{y}\mathrm{i}\mathrm{s}=\frac{\mathrm{F}\mathrm{a}\mathrm{v}\mathrm{o}\mathrm{u}\mathrm{r}\mathrm{a}\mathrm{b}\mathrm{l}\mathrm{e}\mathrm{c}\mathrm{o}\mathrm{n}\mathrm{d}\mathrm{i}\mathrm{t}\mathrm{i}\mathrm{o}\mathrm{n}}{\mathrm{S}\mathrm{a}\mathrm{m}\mathrm{p}\mathrm{l}\mathrm{e}\mathrm{s}\mathrm{p}\mathrm{a}\mathrm{c}\mathrm{e}}$

$\Rightarrow \frac{1}{7}$

Concept:

The sample space when a coin is tossed three times is S = 2³ ⇒ {TTT, TTH, THT, HTT, THH, HTH, HHT, HHH}

Calculation:

Given:

Case-1: Probability of getting atleast two tails.

Favourable cases = {TTH, THT, HTT, TTT}

Probability of getting at least two cases = $\frac{Favourablecases}{Samplespace}\Rightarrow \frac{4}{8}=\frac{1}{2}$

Case-2: Probability of getting almost two tails.

Favourable cases = {TTH, THT, HTT, THH, HTH, HHT, HHH}

Probability of getting almost two tails = $\frac{Favourablecases}{Samplespace}\Rightarrow \frac{7}{8}$

Concept:

Probability Mass Function: 

Probability Mass Function gives the probability of a discrete random variable being exactly equal to some value.

Total probability = Σ p(x) = 1.

Calculation:

Given that $\mathrm{p}(\mathrm{x})={\displaystyle \frac{\mathrm{k}\mathrm{x}}{3}},\mathrm{x}=1,2,3$.

Therefore, $\mathrm{p}(1)={\displaystyle \frac{\mathrm{k}}{3}}$, $\mathrm{p}(2)={\displaystyle \frac{2\mathrm{k}}{3}}$ and $\mathrm{p}(3)={\displaystyle \frac{3\mathrm{k}}{3}}$.

We know that Σ p(x) = 1.

⇒ p(1) + p(2) + p(3) = 1

⇒ ${\displaystyle \frac{\mathrm{k}}{3}}+{\displaystyle \frac{2\mathrm{k}}{3}}+{\displaystyle \frac{3\mathrm{k}}{3}}=1$

⇒ ${\displaystyle \frac{6\mathrm{k}}{3}}=1$

⇒ $\mathrm{k}={\displaystyle \frac{1}{2}}$.

Given:

P(A) = 0.3 and P(A ∪ B) = 0.8

Formula used:

P(A ∪ B)  = P(A) + P(B) - P(A) × P(B)

Two events are independent if the incidence of one event does not affect the probability of the other event.

Calculation:

We know that,

P(A ∪ B) = P(A) + P(B) - P(A) × P(B)

⇒ 0.8 = 0.3 + P(B)  - 0.3 × P(B)

⇒ 0.5 = P(B)[1 - 0.3] = 0.7P(B)

⇒ P(B) = 0.5/0.7

∴ P(B) = 5/7

Concept:

E = the event a man reports that it is even

E1 = the event of an even number is picked

E2 = the event of an odd number is picked

To find : The probability that it is actually even i.e. P(E1|E)

P(E1|E) = ${\displaystyle \frac{\mathrm{P}(\mathrm{E}|{\mathrm{E}}_1).\mathrm{P}({\mathrm{E}}_1)}{\mathrm{P}(\mathrm{E}|{\mathrm{E}}_1).\mathrm{P}({\mathrm{E}}_1)+\mathrm{P}(\mathrm{E}|{\mathrm{E}}_2).\mathrm{P}({\mathrm{E}}_2)}}$

Calculations:

Given, a man speaks truth 2 out of 3 times. He picks one of the natural numbers in the set S = {1, 2, 3, 4, 5, 6, 7} and reports that it is even.

E₁ = the event of an even number is picked

E₂ = the event of an odd number is picked

⇒P(E1) = ${\displaystyle \frac{3}{7}}$

⇒P(E2) = ${\displaystyle \frac{4}{7}}$

E = the event a man reports that it is even

 A man speaks truth 2 out of 3 times.

⇒P(E|E₁) = ${\displaystyle \frac{2}{3}}$

⇒P(E|E2) = ${\displaystyle \frac{1}{3}}$

To find : The probability that it is actually even i.e. P(E₁|E)

P(E1|E) = ${\displaystyle \frac{\mathrm{P}(\mathrm{E}|{\mathrm{E}}_1).\mathrm{P}({\mathrm{E}}_1)}{\mathrm{P}(\mathrm{E}|{\mathrm{E}}_1).\mathrm{P}({\mathrm{E}}_1)+\mathrm{P}(\mathrm{E}|{\mathrm{E}}_2).\mathrm{P}({\mathrm{E}}_2)}}$

⇒P(E1|E) = ${\displaystyle \frac{({\displaystyle \frac{2}{3}}).({\displaystyle \frac{3}{7}})}{({\displaystyle \frac{2}{3}}).({\displaystyle \frac{3}{7}})+({\displaystyle \frac{1}{3}}).({\displaystyle \frac{4}{7}})}}$

⇒P(E1|E) = ${\displaystyle \frac{6}{10}}$

⇒⇒P(E1|E) = ${\displaystyle \frac{3}{5}}$

Hence, a man speaks truth 2 out of 3 times. He picks one of the natural numbers in the set S = {1, 2, 3, 4, 5, 6, 7} and reports that it is even. The probability that it is actually even is ${\displaystyle \frac{3}{5}}$

Explanation

A and B are independent events 

⇒ P(A ∩ B) = P(A) × P(B)

According to question

⇒ P(A) × P(B) = 0.16   -----(A)

⇒ P(A') × P(B') = 0.36      ----(I)

⇒ [(1 - P(A)} × {1 - P(B)} = 0.36

⇒ [1 - P(B) - P(A) + P(A)P(B)] = 0.36

⇒ [1 - 0.36 + 0.16] = P(B) + P(A)

⇒ [1 - 0.36 + 0.16] = P(B) + P(A)

P(A) + P(B) = 0.8

By option elimination, we can conclude

P(A) = 0.4 and P(B) = 0.4

Concept:

The probability density function is always positive f(x) ≥ 0, and it follows the below condition.

$\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}f\left(x\right)dx=1$

Calculation:

Given:

$f\left(x\right)=\{\begin{array}{c}2k{x}^4,0\le x\le 1,\\ 0,otherwise;\end{array}$

We know that $\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}f\left(x\right)dx=1$

$\therefore \underset{-\mathrm{\infty }}{\overset{0}{\int }}f\left(x\right)dx+\underset{0}{\overset{1}{\int }}f\left(x\right)dx+\underset{1}{\overset{\mathrm{\infty }}{\int }}f\left(x\right)dx=1$

$\therefore \underset{0}{\overset{1}{\int }}2k{x}^{4}dx=1$

$\therefore 2k{\left[\frac{x^5}{5}\right]}_0^1=1$

∴ 2k = 5

$\therefore k=\frac{5}{2}$

Concept:

Conditional probability:

It gives the probability of happening of any event if the other has already occurred.

$\mathrm{P}\left(\frac{{\mathrm{E}}_1}{{\mathrm{E}}_2}\right)=\mathrm{P}\mathrm{r}\mathrm{o}\mathrm{b}\mathrm{a}\mathrm{b}\mathrm{i}\mathrm{l}\mathrm{i}\mathrm{t}\mathrm{y}\mathrm{o}\mathrm{f}\mathrm{g}\mathrm{e}\mathrm{t}\mathrm{t}\mathrm{i}\mathrm{n}\mathrm{g}\mathrm{t}\mathrm{h}\mathrm{e}\mathrm{e}\mathrm{v}\mathrm{e}\mathrm{n}\mathrm{t}{\mathrm{E}}_1\mathrm{w}\mathrm{h}\mathrm{e}\mathrm{n}{\mathrm{E}}_2\mathrm{i}\mathrm{s}\mathrm{a}\mathrm{l}\mathrm{r}\mathrm{e}\mathrm{a}\mathrm{d}\mathrm{y}\mathrm{o}\mathrm{c}\mathrm{c}\mathrm{u}\mathrm{r}\mathrm{e}\mathrm{d}.$

$P\left(\frac{E_1}{E_2}\right)=\frac{P\left(E_1\cap E_2\right)}{P\left(E_2\right)}$

Calculation:

Given:

P (E₁) = Probability of stopping at the gas station and ask for tyre checked = 0.12

P (E₂) = Probability of stopping at the gas station and ask for oil checked = 0.29

P (E₁∩ E₂) = Probability of both checked = 0.07

$\mathrm{P}\left(\frac{{\mathrm{E}}_2}{{\mathrm{E}}_1}\right)=\mathrm{P}\mathrm{r}\mathrm{o}\mathrm{b}\mathrm{a}\mathrm{b}\mathrm{i}\mathrm{l}\mathrm{i}\mathrm{t}\mathrm{y}\mathrm{o}\mathrm{f}\mathrm{p}\mathrm{e}\mathrm{r}\mathrm{s}\mathrm{o}\mathrm{n}\mathrm{w}\mathrm{h}\mathrm{o}\mathrm{h}\mathrm{a}\mathrm{s}\mathrm{h}\mathrm{i}\mathrm{s}\mathrm{t}\mathrm{y}\mathrm{r}\mathrm{e}\mathrm{c}\mathrm{h}\mathrm{e}\mathrm{c}\mathrm{k}\mathrm{e}\mathrm{d}\mathrm{w}\mathrm{i}\mathrm{l}\mathrm{l}\mathrm{a}\mathrm{l}\mathrm{s}\mathrm{o}\mathrm{h}\mathrm{a}\mathrm{v}\mathrm{e}\mathrm{o}\mathrm{i}\mathrm{l}\mathrm{c}\mathrm{h}\mathrm{e}\mathrm{c}\mathrm{k}\mathrm{e}\mathrm{d}$   

∵ $P\left(\frac{E_2}{E_1}\right)=\frac{P\left(E_1\cap E_2\right)}{P\left(E_1\right)}$

∴ $P\left(\frac{E_2}{E_1}\right)=\frac{0.07}{0.12}=0.58$

Concept:

For a random variable X = xi with probabilities P(X = x) = pi:

• Mean/Expected Value: μ = ∑pixi.
• Variance: Var(X) = σ² = ∑pi(xi)2 - (∑pixi)2 = ∑pi(xi)2 - μ2.
• Standard Deviation: σ = $\sqrt{\mathrm{V}\mathrm{a}\mathrm{r}(\mathrm{X})}$.

Calculation:

We have x1 = 1, x2 = 2, x3 = 3 and p1 = 0.3, p2 = 0.4, p3 = 0.3.

Now, ∑pixi = (0.3 × 1) + (0.4 × 2) +(0.3 × 3)

= 0.3 + 0.8 + 0.9

= 2

And, ∑pi(xi)2 = (0.3 × 1²) + (0.4 × 2²) +(0.3 × 3²)

= 0.3 + 1.6 + 2.7

= 4.6

∴ Var(X) = σ2 = ∑pi(xi)2 - (∑pixi)2 = 4.6 - 22 = 4.6 - 4 = 0.6.

The variance of the distribution is Var(X) = 0.6.