Probability

Given,

An experiment has 10 equally likely outcomes,

Let B have n no of outcomes,

So, $P(B) = {n \over {10}}$

$P(A) = {4 \over {10}}$

$P(A \cap B) = {4 \over {10}} \times {n \over {10}} = {{{{2n} \over 5}} \over {10}}$

Since, A and B are independent

n is an integer

n = 5 or 10

We have,

$ax + by = 0$

$cx + dy = 0$

since, the system of homogeneous equation is always consistent and has a solution.

Therefore, statement 2 is true.

Now, $\Delta = \left| {\matrix{ a & b \cr c & d \cr } } \right|$ and $a,b,c,d \in \{ 0,1\} $

$ = ad - bc$

No. of ways of selecting $a,b,c,d$ from the set {0, 1} is 2 $\times$ 2 $\times$ 2 $\times$ 2 = 16

If the system has unique solution,

Then $\Delta\ne0$

$\Rightarrow$ either $ad = 1,bc = 0$ or $ad = 0,bc = 1$ favourable case = 6

Therefore, probability that system of equation has unique solution is ${6 \over {16}} = {3 \over 8}$. Statement 1 is True.

Hence, the Statement 2 is True but is not a correct explanation of statement 1.

Probability of getting score 9 in a single throw

$ = {4 \over {36}} = {1 \over 9}.$

Probability of getting score 9 exactly twice

$ = {}^3{C_2} \times {\left( {{1 \over 9}} \right)^2} \times {8 \over 9} = {8 \over {243}}.$

The desired probability

= (0.7) (0.2) + (0.7) (0.8) (0.7) (0.2) + (0.7) (0.8) (0.7) (0.8) (0.7) (0.2) + .......

= 0.14 [1 + (0.56) + (0.56)² + .......]

= 0.14 $\left( {{1 \over {1 - 0.56}}} \right) = {{0.14} \over {0.44}} = {7 \over {22}}$ = 0.32

$\begin{aligned} & \mathrm{E}_{1} f_{1} G \text { are pairwise independent events. } \\\\ & \therefore \quad \mathrm{P}(\mathrm{E} \cap \mathrm{F})=\mathrm{P}(\mathrm{E}) \cdot \mathrm{P}(\mathrm{F}) \\\\ & \mathrm{P}(\mathrm{F} \cap \mathrm{G})=\mathrm{P}(\mathrm{F}) \cdot \mathrm{P}(\mathrm{G}) \\\\ & \mathrm{P}(\mathrm{G} \cap \mathrm{E})=\mathrm{P}(\mathrm{G}) \cdot \mathrm{P}(\mathrm{E}) \\\\ & \mathrm{P}\left(\frac{\mathrm{E}^{\mathrm{c}} \cap \mathrm{F}^{\mathrm{c}}}{\mathrm{G}}\right)=\mathrm{P}\left(\frac{\left(E^{\mathrm{c}} \cap \mathrm{F}^{\mathrm{c}}\right) \cap G}{\mathrm{P}(\mathrm{G})}\right) \\\\ &=\frac{\mathrm{P}(\mathrm{G})-\mathrm{P}(\mathrm{G} \cap \mathrm{E})-\mathrm{P}(\mathrm{G} \cap \mathrm{F})}{\mathrm{P}(\mathrm{G})} \\\\ &=1-\mathrm{P}(\mathrm{E})-\mathrm{P}(\mathrm{F}) \\\\ &=\mathrm{P}\left(\mathrm{E}^{\mathrm{c}}\right)-\mathrm{P}(\mathrm{F}) . \end{aligned}$

Let E = event when each American man is seated adjacent to his wife and

A = event when Indian man is seated adjacent to his wife.

Now,

$n(A\cap E)=(4!)\times(2!)^5$

Event when each American man is seated adjacent to his wife.

Again $n(E)=(5!)\times(2!)^4$

$p\left( {{A \over E}} \right) = {{n(A \cap E)} \over {n(E)}}$

$ = {{(4!) \times {{(2!)}^5}} \over {(5!) \times {{(2!)}^4}}} = {2 \over 5}$

Statement 1 : If P(H$_i$ $\cap$ E) = O for some, then

$P\left( {{{{H_i}} \over E}} \right) = P\left( {{E \over {{H_i}}}} \right) = 0$

If P(H$_i$ $\cap$ E) $\ne o~\forall i=1,2,3.......,n$ then

$P\left( {{{{H_i}} \over E}} \right) = {{P({H_i} \cap E)} \over {P({H_i})}} \times {{P({H_i})} \over {P(E)}}$

$ = {{P\left( {{E \over {{H_i}}}} \right) \times P({H_i})} \over {P(E)}}$

$ > P\left( {{E \over {{H_i}}}} \right) \times P({H_i})$

Hence, statement 1 may not always be true.

Statement 2 : Clearly, we can write as

H$_1$ $\cup$ H$_2$ $\cup$ H$_1$ $\cup$ ------------------ $\cup$ H$_n$ = S

$\Rightarrow$ P(H$_1$) + P(H$_2$) + ---------------- + P(H$_n$) = 1

Hence, statement 2 is true.

Required probability

$ = P(X = 0) + P(X = 1)$

$ = {{{e^{ - 5}}} \over {0!}}{5^0} + {{{e^{ - 5}}} \over {1!}}{5^1}$

$ = {e^{ - 5}} + 5{e^{ - 5}}$

$ = {6 \over {{e^5}}}$

$\because \mathrm{P}\left(u_{i}\right) \propto i$

$\mathrm{P}\left(u_{i}\right)=\mathrm{K} i$

We know that $\sum \mathrm{P}\left(\mathrm{u}_{i}\right)=1$

$\begin{aligned} \mathrm{K} \frac{n(n+1)}{2} & =1 \\ \mathrm{~K} & =\frac{2}{n(n+1)} \\ \mathrm{P}\left(u_{i}\right) & =\frac{2 i}{n(n+1)} \\ \mathrm{P}\left(\frac{w}{u_{2 i}}\right) & =\frac{2 i}{n+1} \end{aligned}$

Now, $\quad \mathrm{P}(w)=\sum_\limits{1^{i}=1}^{n} \mathrm{P}\left(u_{i}\right) \cdot \mathrm{P}\left(\frac{w}{u_{i}}\right)$

$ = \sum\limits_{{1^i} = 1}^n {{{2i} \over {n(n + 1)}}{i \over {(n + 1)}}} $

$ = \sum\limits_{{1^i} = 1}^n {{{2{i^2}} \over {n{{(n + 1)}^2}}}} $

$ = {2 \over {n{{(n + 1)}^2}}}\sum\limits_{{1^i} = 1}^n {{i^2}} $

$ = {2 \over {n{{(n + 1)}^2}}}{{n(n + 1)(2n + 1)} \over 6}$

$ = {{2n + 1} \over {3(n + 1)}}$

Now, $\mathop {\lim }\limits_{n \to \infty } {{2n + 1} \over {3(n + 1)}} = \mathop {\lim }\limits_{n \to \infty } {{\left( {2 + {1 \over n}} \right)} \over {3\left( {{1 \over n} + 1} \right)}} = {2 \over 3}$

Since, $\mathrm{P}\left(u_{i}\right)=\mathrm{C}$ means

All events are equally likely

$\mathrm{P}\left(u_{i}\right)=\mathrm{C}=\frac{1}{n} \quad$ for $i=1,2,3, \ldots . n$

$\mathrm{P}(u_n)=\frac{1}{n}$

$\mathrm{P}\left(\frac{w}{u_{n}}\right)=\frac{n}{n+1}$

$\begin{aligned} \mathrm{P}(w) & =\sum \mathrm{P}\left(u_{i}\right) \cdot \mathrm{P}\left(\frac{w}{u_{n}}\right) \\ & =\sum\left(\frac{1}{n} \frac{2^{i}}{(n+1)}\right)=\frac{1}{n(n+1)} \cdot \frac{n(n+1)}{2}=\frac{1}{2} \\ \mathrm{P}\left(\frac{u_{\mathrm{n}}}{w}\right) & =\frac{\mathrm{P}\left(u_{\mathrm{n}}\right) \times \mathrm{P}\left(\frac{w}{u_{\mathrm{n}}}\right)}{\mathrm{P}(w)} \\ & =\frac{\frac{1}{n} \times \frac{n}{(n+1)}}{\frac{1}{2}}=\frac{2}{n+1} \end{aligned}$

Total number of ball

Even numbered urns $=\frac{n}{2}(n+1)$

Total no. of white balls = $2+4+6+....+n$.

$=2(1+2+3+...+m)$

where $m=\frac{n}{2}$

$=2 \times \frac{m(m+1)}{2}$

$=\frac{n}{2} \cdot\left(\frac{n}{2}+1\right)=\frac{n(n+2)}{4}$

$\therefore \quad \mathrm{P}\left(\frac{w}{\mathrm{E}}\right)=\frac{\frac{n}{4}(n+2)}{\frac{n}{2}(n+1)}$

$=\frac{(n+2)}{2(n+1)}$

Person 1st has three options to apply.

Similarly, person 2nd has three options t apply

and person 3rd has three options to apply.

Total cases = 3³

Now, favourable cases = 3 (An either all has applied for house 1 or 2 or 3)

So, probability $ = {3 \over {{3^3}}} = {1 \over 9}$.

Given that,

$P(\overline {A \cup B} ) = {1 \over 6}$, $P(A \cap B) = {1 \over 4}$, $P(\overline A ) = {1 \over 4}$

$\because$ $P(\overline {A \cup B} ) = {1 \over 6}$

$ \Rightarrow 1 - P(A \cup B) = {1 \over 6}$

$ \Rightarrow 1 - P(A) - P(B) + P(A \cap B) = {1 \over 6}$

$ \Rightarrow P(\overline A ) - P(B) + {1 \over 4} = {1 \over 6}$

$ \Rightarrow P(B) = {1 \over 4} + {1 \over 4} - {1 \over 6}$

$ \Rightarrow P(B) = {1 \over 3}$ and $P(A) = {3 \over 4}$

Clearly, $P(A \cap B) = P(A)P(B)$,

so, the events A and B are independent events but not equally likely.

In a position distribution,

$P(X = r) = {{{e^{ - \lambda }}{\lambda ^r}} \over {r!}}$ ($\lambda$ = mean).

Now, $P(X = r > 1.5) = P(2) + P(3) + $ ..... $\infty$

$ = 1 - \{ P(0) + P(1)\} $

$ = 1 - \left( {{e^{ - 2}} + {{{e^{ - 2}} \times 2} \over 1}} \right) = 1 - {3 \over {{e^2}}}$

$\begin{aligned} P\left(E_{\text {car }}\right)=\frac{1}{7} \quad & P\left(E_{S_{\text {cooter }}}\right)=\frac{3}{2} \\ P\left(E_{\text {bus}}\right)=\frac{2}{7} \quad & P\left(E_{\text {train }}\right)=\frac{1}{7} \\ P\left(\frac{E_{\text {late }}}{E_{\text {car }}}\right) & =\frac{2}{9} \\ P\left(\frac{E_{\text {late }}}{E_{\text {scooter }}}\right) & =\frac{1}{9} \\ E\left(\frac{E_{\text {late }}}{E_{\text {bus }}}\right) & =\frac{4}{9} \\ E\left(\frac{E_{\text {late }}}{E_{\text {train }}}\right) & =\frac{1}{9}\end{aligned}$

Using Total probability theorem.

$\begin{aligned} & P\left(E_{\text {late }}\right)=P\left(E_{\text {car }}\right) \times P\left(\frac{E_{\text {late }}}{E_{\text {car }}}\right)+P\left(E_{\text {scooter }}\right) \\ & \times P\left(\frac{E_{\text {late }}}{E_{\text {scooter }}}\right)+P\left(E_{\text {Bus }}\right) \times P\left(\frac{E_{\text {late }}}{E_{\text {Bus }}}\right) \\ & +P\left(E_{\text {train }}\right) \times P\left(\frac{E_{\text {late }}}{E_{\text {train }}}\right) \\ & =\frac{1}{7} \times \frac{2}{9}+\frac{3}{2} \times \frac{1}{9}+\frac{2}{7} \times \frac{4}{9}+\frac{1}{7} \times \frac{1}{9} \\ & =\frac{2}{63}+\frac{3}{18}+\frac{8}{63}+\frac{1}{63}=\frac{11}{63}+\frac{3}{18}=\frac{43}{126} \\ & \mathrm{P}\left(\mathrm{E}_{\text {ontime }}\right)=1-\frac{43}{126}=\frac{83}{126} \\ & P\left(\frac{E_{\text {car }}}{E_{\text {ontime }}}\right)=\frac{P\left(E_{\text {car }} \cap E_{\text {ontime }}\right)}{P\left(E_{\text {ontime }}\right)} \\ & =\frac{P\left(E_{\text {car }}\right) \times P\left(E_{\text {ontime }}\right)}{P\left(E_{\text {ontime }}\right)}=P\left(E_{\text {car }}\right)=\frac{1}{7} \end{aligned}$

The probability of speaking truth by A, P(A) = ${4 \over 5}$. The probability of not speaking truth by A, P($\overline A $) = 1 $-$ ${4 \over 5} = {1 \over 5}$.

The probability of speaking truth by B, P(B) = ${3 \over 4}$. The probability of not speaking truth of B, P($\overline B $) $ = {1 \over 4}$.

The probability that they contradict each other

$ = P(A) \times P(\overline B ) + P(\overline A ) \times P(B)$

$ = {4 \over 5} \times {1 \over 4} + {1 \over 5} \times {3 \over 4}$

$ = {1 \over 5} + {3 \over {20}} = {7 \over {20}}$

Given that mean = 4 = np = 4 and variance = 2

npq = 2 $\Rightarrow$ 4q = 2

$ \Rightarrow q = {1 \over 2}$

$\therefore$ $p = 1 - q = 1 - {1 \over 2} = {1 \over 2}$

Also, n = 8

Probability of 2 successes $ = P(X = 2) = {}^8{C_2}{p^2}{q^6}$

$ = {{8!} \over {2! \times 6!}} \times {\left( {{1 \over 2}} \right)^2} \times {\left( {{1 \over 2}} \right)^6} = 28 \times {1 \over {{2^8}}}$

$ = {{28} \over {256}}$

X : Mr. A selected wining horse

$\overline X $ : Mr. A did not select wining horse

Mr. A selected two horses, now probability of not wining the first horse which Mr. A choses = ${4 \over 5}$

And probability of not wining the second horse also which Mr. A choses = ${3 \over 4}$ (Here Mr. A out of remaining 4 horses choses one horse among 3 horses which did not win)

$\therefore$ $P(\overline X ) = {4 \over 5} \times {3 \over 4}$

$\therefore$ $P(X) = 1 - P(\overline X )$

$ = 1 - {4 \over 5} \times {3 \over 4}$

$ = 1 - {3 \over 5}$

$ = {2 \over 5}$