Probability

Given in bag containing 5 black balls and 3 white balls.

Random variable $X$ denotes the number of white balls drawn $X(0)=$ No white ball is drawn

$ \begin{aligned} & P(0)=\frac{{ }^3 C_0 \times{ }^5 C_2}{{ }^8 C_2}=\frac{10}{28} \\ & X(1)=1 \text { white ball is drawn } \\ & P(X)=\frac{{ }^5 C_1 \times{ }^3 C_1}{{ }^8 C_2}=\frac{15}{28} \\ & X(2)=2 \text { white balls are drawn } \\ & P(X)=\frac{{ }^3 C_2 \times{ }^5 C_0}{{ }^8 C_2}=\frac{3}{28} \end{aligned} $

Mean of

$ \begin{aligned} X=\Sigma X_i P\left(X_i\right) & =0 \times \frac{10}{28}+1 \times \frac{15}{28}+2 \times \frac{3}{28} \\ & =\frac{21}{28}=\frac{3}{4} \end{aligned} $

We know that in binomial distribution

Mean $=n p$

Variance $=n p q$

$ \begin{aligned} \therefore \quad n p & =4 \text { and } n p q=\frac{4}{3} \\ \Rightarrow \quad q & =\frac{1}{3}, p=\frac{2}{3} \text { and } n=6 \\ P(X & =2)={ }^6 C_2 p^2 q^4 \\ & =\frac{6 \times 5}{2} \times\left(\frac{2}{3}\right)^2\left(\frac{1}{3}\right)^4=\frac{20}{243} \end{aligned} $

If we form a 4-digit number by using the digits 4, 5, 6, 7, 8, 9 allowing repetition, then we can categories the remainders on dividing by 3 as 0,1 or 2, so the probability to form a number which is exactly divisible by 3 is $1 / 3$.

$ \begin{aligned} &\text { We have, } P\left(E_r\right)=\frac{1}{1+r}\\ &\begin{aligned} & \therefore P\left(\bar{E}_r\right)=1-P\left(E_r\right)=1-\frac{1}{1+r}=\frac{r}{1+r} \\ & \therefore \text { Required probability } \\ & =P\left(E_1 \cup E_2 \cup E_3 \ldots \cup E_n\right) \\ & =1-P\left(\bar{E}_1 \cap E_2 \cap \bar{E}_3 \ldots \cap \bar{E}_n\right) \\ & =1-P\left(\bar{E}_1\right) P\left(\bar{E}_2\right) P\left(\bar{E}_3\right) \ldots P\left(\bar{E}_n\right) \\ & =1-\left(\frac{1}{1+1}\right)\left(\frac{2}{1+2}\right)\left(\frac{3}{1+3}\right) \ldots\left(\frac{n}{1+n}\right) \\ & =1-\frac{1}{2} \times \frac{2}{3} \times \frac{3}{4} \times \ldots \times \frac{n}{1+n} \\ & =1-\frac{1}{1+n}=\frac{n}{n+1} \end{aligned} \end{aligned} $

Let $A_i(i=1,2,3,4)$ be the event that the urn contains 2, 3, 4 or 5 white ball and the event that two white ball are drawn.

We have to find $P\left(A_4 / B\right)$

Since the four events $A_1, A_2, A_3, A_4$ are equally likely, we have

$ P\left(A_1\right)=P\left(A_2\right)=P\left(A_3\right)=P\left(A_4\right)=\frac{1}{4} $

$P\left(B / A_1\right)$ is the probability of event that the urn contains 2 white balls and both have been drawn.

Hence, $P\left(B / A_1\right)=\frac{2 C_2}{5 C_2}=\frac{1}{10}$

$ \begin{aligned} & \text { Similarly, } P\left(B / A_2\right)=\frac{3 C_2}{5 C_2}=\frac{3}{10} \\ & P\left(B / A_3\right)=\frac{4 C_2}{5 C_2}=\frac{6}{10}=\frac{3}{5} \\ & \Rightarrow \quad P\left(B / A_4\right)=\frac{5 C_2}{5 C_2}=1 \end{aligned} $

∴ From the Baye's theorem

$ \begin{aligned} P\left(A_4 / B\right) & =\frac{P\left(A_4\right) P\left(B / A_4\right)}{\sum_{i=1}^4 P\left(A_1\right) P\left(B / A_1\right)} \\ & =\frac{(1 / 4) \cdot 1}{\frac{1}{4}\left(\frac{1}{10}+\frac{3}{10}+\frac{3}{5}+1\right)} \\ P\left(A_4 / B\right) & =1 / 2 \end{aligned} $

Given, probability function of random variable $X$ is

$ P(X=n)=\frac{k(n+1)}{3^n} \text { for } n \in \mathbf{N} \cup\{0\} $

$ \begin{aligned} &\begin{array}{ll} \therefore & \sum_{n=0}^{\infty} P(X=n)=1 \\ \Rightarrow & k \sum_{n=0}^{\infty} \frac{(n+1)}{3^n}=1 \\ \Rightarrow & k\left[1+\frac{2}{3}+\frac{3}{3^2}+\frac{4}{3^3}+\ldots\right]=1 \end{array}\\ &\text { ∵ Sum of infinite AGP }\\ &\begin{array}{ll} & a+(a+d) r+(a+2 d) r^2 \\ & +\ldots .=\frac{a}{1-r}+\frac{d r}{(1-r)^2} \\ \therefore & k\left[\frac{1}{1-1 / 3}+\frac{1 / 3}{(1-1 / 3)^2}\right]=1 \\ \Rightarrow & k\left[\frac{3}{2}+\frac{3}{4}\right]=1 \Rightarrow k=\frac{4}{9} \\ \therefore & P(X<2)=P(X=0)+P(X=1) \\ & =k\left[1+\frac{2}{3}\right]=\frac{4}{9} \times \frac{5}{3}=\frac{20}{27} \end{array} \end{aligned} $

The average arrival rate, $\lambda$ is 240 veh/h or 1/15 vehicles per second. According to poisson distribution,

$ P(n)=\frac{(\lambda t)^n e^{-\lambda t}}{n!} $

Where, $P(n)=$ probability of having $n$ vehicles arrive in time $t$, $\lambda=$ Average vehicle flow or arrival rate in per unit time

$t$ = duration of the time interval over which vehicles are counted $e=$ base of natural logarithm

$ \begin{aligned} &\begin{aligned} & P(0)=\frac{\left(\frac{1}{15} \times 30\right)^0 \cdot e^{-\left(\frac{1}{15}\right) \times(30)}}{0!}=e^{-2} \\ & P(1)=\frac{\left(\frac{1}{15} \times 30\right)^1 \cdot e^{-\frac{1}{15} \times 30}}{1!}=2 e^{-2} \\ & P(2)=\frac{\left(\frac{1}{15} \times 30\right)^2 \cdot e^{-\frac{1}{15} \times 30}}{2!}=2 e^{-2} \end{aligned}\\ &\text { For more than two vehicles, }\\ &\begin{aligned} P(n>2) & =1-P(n \leq 2) \\ & =1-\left\{e^{-2}+2 e^{-2}+2 e^{-2}\right\} \\ P(n>2) & =\frac{e^2-5}{e^2} \end{aligned} \end{aligned} $

Let $A$ be events greater than 3 when a dice is thrown

$ \therefore \quad P(A)=\frac{3}{6}=\frac{1}{2} $

and $B$ is events gets 5

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

Required probability

$ \begin{aligned} & =P(B)+P(A) \cdot P(B)+P(A)^2 P(B)+\ldots \\ & =\frac{1}{6}+\frac{1}{2} \times \frac{1}{6}+\left(\frac{1}{2}\right)^2 \times \frac{1}{6}+\ldots \\ & =\frac{1}{6}\left(1+\frac{1}{2}+\frac{1}{2^2}+\ldots\right)=\frac{1 / 6}{1-1 / 2}=\frac{1}{3} \end{aligned} $

Consider the events,

$E_1=$ Person suffering from a certain disease

$E_2=$ Person are not suffering from a certain disease

$A=$ Diagnostic test is positive

$ \begin{aligned} P\left(E_1\right) & =0.5 \% \quad P\left(E_2\right)=99.5 \% \\ P\left(A / E_1\right) & =0.95 \quad P\left(A / E_2\right)=0.10 \end{aligned} $

Required probability

$ \begin{aligned} P\left(E_1 / A\right) & P\left(E_1\right) \times P\left(A / E_1\right) \\ & =\frac{}{P\left(E_1\right) \times P\left(A / E_1\right)+P\left(E_2\right) \times P\left(A / E_2\right)} \\ & =\frac{0.005 \times 0.95}{(0.005 \times 0.95)+0.995 \times 0.10} \\ & =\frac{0.00475}{0.00475+0.0995} \\ & =\frac{0.00475}{0.10425}=\frac{475}{10425} \\ & =0.0455 \end{aligned} $

$P_1, P_2, P_3$ are three independent events Probability of happening at least one event is

$ \begin{aligned} & 1-\left[\left(\bar{P}_1\right) \cdot\left(\bar{P}_2\right)\left(\bar{P}_3\right)\right] \\ & \quad=1-\left(1-\bar{P}_1\right)\left(1-\bar{P}_2\right)\left(1-\bar{P}_3\right) \end{aligned} $

$\therefore A$ is true.

If there independent events are $A, B$ and $C$

$ \begin{aligned} \therefore P & (A \cup B \cup C) \\ = & P(A)+P(B)+P(C)-P(A) P(B) \\ & -P(A) P(C)-P(B) P(C)+P(A) P(B) P(C) \end{aligned} $

$R$ is also true.

$\therefore \mathrm{A}$ is true and R is true and R is the correct explanation for $A$.

$ P(X=r)=r / k $

$ \begin{array}{|c|c|c|c|c|c|} \hline X & 1 & 2 & 3 & 4 & 5 \\ \hline P(X) & 1 / k & 2 / k & 3 / k & 4 / k & 5 / k \\ \hline \end{array} $

$ \begin{array}{rlrl} &P\left(X_i\right) =1 \\ \Rightarrow & \frac{1+2+3+4+5}{k} =1 \\ \Rightarrow & k =15 \end{array} $

$ \begin{aligned} P(X= & \left.2 \text { or } X=\frac{k}{3}\right) \\ & =P(x=2)+P\left(X=\frac{15}{3}\right) \\ & =\frac{2}{15}+\frac{5}{15}=\frac{7}{15} \\ P(X= & \left.4 \text { or } X=\frac{k}{5}\right) \\ & =P(X=4)+P\left(X=\frac{15}{5}\right) \\ & =\frac{4}{15}+\frac{3}{15}=\frac{7}{15} \end{aligned} $

$ \begin{aligned} &\text { Here, }\\ &\begin{aligned} n & =2000 \\ P & =0.001 \\ \lambda & =n p=2 \\ P(X & >2) \\ & =1-[P(X=0)+P(X=1)+P(X=2)] \\ & =1-\left[\frac{2^0 e^{-2}}{0!}+\frac{2^1 e^{-2}}{1!}+\frac{2^2 e^{-2}}{2!}\right] \\ & =1-\left[e^{-2}+2 e^{-2}+2 e^{-2}\right] \\ & =1-5 e^{-2}=1-\frac{5}{e^2} \end{aligned} \end{aligned} $

According to Booley's Inequalities. If $A, B$ and $C$ are the events of a random experiment that then $P(A \cap B \cap C) \geq P(A)+P(B)+P(C)-2$ Similarly, If $A_1, A_2, A_3, \ldots, A_{15}$ are the events of a random experiment, then $P\left(\bigcap_{i=1}^{15} A \hat{\mathbf{i}}\right) \geq \sum_{i=1}^{15} P(A \hat{\mathbf{i}})-14$.

Consider the events

$E_1=$ Answer by student without guessing

$E_2=$ Answer by student guessing

$A=$ At least three question correctly

$ \begin{aligned} & P\left(E_1\right)=\frac{3}{7}, p\left(E_2\right)=\frac{4}{7} \\ & p\left(\frac{A}{E_1}\right)={ }^4 C_3\left(\frac{1}{2}\right)^4+{ }^4 C_4\left(\frac{1}{2}\right)^4 \\ & =\left(\frac{1}{2}\right)^4(4+1)=\frac{5}{16} \\ & p\left(\frac{A}{E_2}\right)={ }^4 C_3\left(\frac{2}{3}\right)^3 \times \frac{1}{3}+{ }^4 C_4\left(\frac{2}{3}\right)^4 \\ & =\left(\frac{2}{3}\right)^3\left(\frac{4}{3}+\frac{2}{3}\right) \\ & p\left(\frac{A}{E_2}\right)=\frac{16}{27} \end{aligned} $

$ \begin{aligned} & \text { Required probability }=p\left(\frac{E_2}{A}\right) \\ & =\frac{p\left(E_2\right) p\left(\frac{A}{E_2}\right)}{p\left(E_1\right) p\left(\frac{A}{E_1}\right)+p\left(E_2\right) p\left(\frac{A}{E_2}\right)} \\ & =\frac{\frac{4}{7} \times \frac{64}{27}}{\frac{3}{7} \times \frac{5}{16}+\frac{4}{7} \times \frac{64}{27}}=\frac{1024}{1429} \end{aligned} $

Consider the event

$\mathrm{A}=$ Box A is selected

$\mathrm{B}=$ Box B is selected

$\mathrm{C}=$ Box C is selected

$\mathrm{D}=$ Box D is selected

E = Defective fuses

$ \begin{aligned} p(A) & =\frac{5000}{11000}=\frac{5}{11} \\ p(B) & =\frac{3}{11^{\prime}}, p(C)=\frac{2}{11} \\ p(D) & =\frac{1}{11^{\prime}}, p\left(\frac{E}{A}\right)=\frac{3}{100} \\ p\left(\frac{E}{B}\right) & =\frac{2}{100^{\prime}}, p\left(\frac{E}{C}\right)=\frac{1}{100^{\prime}} \\ p\left(\frac{E}{D}\right) & =\frac{1}{200} \end{aligned} $

∴ Required probability

$ \begin{array}{r} =p\left(\frac{D}{E}\right)=\frac{p(D) \times p\left(\frac{E}{D}\right)}{p(A) \times p\left(\frac{E}{A}\right)+p(B) \times p\left(\frac{E}{B}\right)} \\ +p(C) \times p\left(\frac{E}{C}\right)+p(D) \times p\left(\frac{E}{D}\right) \\ =\frac{\frac{1}{11} \times \frac{1}{200}}{\frac{5}{11} \times \frac{2}{100}+\frac{3}{11} \times \frac{2}{100}+\frac{2}{11} \times \frac{1}{100}+\frac{1}{11} \times \frac{1}{200}} \\ =\frac{1}{47} \end{array} $

$P($ getting 1 or 6$)=p=\frac{2}{6}=\frac{1}{3}$

$ \therefore \quad q=1-\frac{1}{3}=\frac{2}{3} $

Let $X$ be the random variable showing the number of success as

$ \begin{aligned} \therefore P(X=0) & ={ }^3 C_0\left(\frac{1}{3}\right)^0\left(\frac{2}{3}\right)^3=\frac{8}{27} \\ P(X=0) & ={ }^3 C_1\left(\frac{1}{3}\right)^1\left(\frac{2}{3}\right)^2 \\ & =3 \times \frac{1}{3} \times \frac{4}{9}=\frac{4}{9} \\ P(X=2) & ={ }^3 C_2\left(\frac{1}{3}\right)^2\left(\frac{2}{3}\right)^1 \\ & =3 \times \frac{1}{9} \times \frac{2}{3}=\frac{2}{9} \\ P(X=3) & ={ }^3 C_3\left(\frac{1}{3}\right)^3\left(\frac{2}{3}\right)^0 \\ & =1 \times \frac{1}{27} \times 1=\frac{1}{27} \end{aligned} $

∴ Hence, the probability distribution of the number of six is given

$ \begin{array}{c|c|c|c|c} \hline X & 0 & 1 & 2 & 3 \\ \hline P(X) & 8 / 27 & 4 / 9 & 2 / 9 & 1 / 27 \\ \hline \end{array} $

$ \begin{aligned} & \therefore \text { Mean }=\Sigma X P(X) \\ & \quad=0 \times \frac{8}{27}+1 \times \frac{4}{9}+2 \times \frac{2}{9}+3 \times \frac{1}{27} \\ & \quad=\frac{4}{9}+\frac{4}{9}+\frac{1}{9}=1 \\ & \therefore \text { Variance }=\Sigma X^2 P(x)-(\text { mean })^2 \end{aligned} $

$ \begin{aligned} =0^2 \times \frac{8}{27}+(1)^2 & \times \frac{4}{9}+(2)^2 \times \frac{2}{9} +(3)^2 \times \frac{1}{27}-(1)^2 \end{aligned} $

$ =\left(\frac{4}{9}+\frac{8}{9}+\frac{3}{9}\right)-1=\frac{15}{9}-1=\frac{6}{9}=\frac{2}{3} $

Given,

Average birth in a week $=35$

∴ Average birth in a day $=\frac{35}{7}=5$

$ \begin{aligned} \therefore \lambda=5 & \\ P(X<3)= & P(X=0)+P(X=1) \quad+P(X \equiv 2) \\ = & \lambda^0 e^{-\lambda}+\frac{\lambda^1 e^{-\lambda}}{1!}+\frac{\lambda^2 e^{-\lambda}}{2!} \\ = & e^{-5}+5 e^{-5}+\frac{25}{2} e^{-5} \\ = & e^{-5}\left(1+5+\frac{25}{2}\right)=\frac{37}{2 e^5} \end{aligned} $

We have, $P(A \cup B)=\frac{3}{4}$,

$ P(A \cap B)=\frac{1}{4} \text { and } P(\bar{A})=\frac{2}{3} $

We know that,

$ \begin{aligned} & P(A \cup B)=P(A)+P(B)-P(A \cap B) \\ & \Rightarrow P(B)-P(A \cap B)=P(A \cup B)-P(A) \\ & \Rightarrow P(\bar{A} \cap B)=P(A \cup B)-P(A) \\ & \quad=\frac{3}{4}-\left(1-\frac{2}{3}\right)=\frac{3}{4}-\frac{1}{3}=\frac{5}{12} \end{aligned} $

We have, $P(X)=\frac{1}{3}, P(X / Y)=\frac{1}{2}$

Now, $\quad P(Y / X)=\frac{2}{5}$

$ \begin{aligned} & \Rightarrow \quad \frac{P(Y \cap X)}{P(X)}=\frac{2}{5} \\ & \Rightarrow P(Y \cap X)=\frac{2}{5} \times \frac{1}{3}=\frac{2}{15} \\ & \text { Again, } P(X / Y)=\frac{P(X \cap Y)}{P(Y)} \\ & \qquad \frac{1}{2}=\frac{\frac{2}{15}}{P(Y)} \Rightarrow P(Y)=\frac{4}{15} \end{aligned} $

$ \begin{aligned} & \text { Also, } P(X \cup Y)=P(X)+P(Y)-P(X \cap Y) \\ & \quad=\frac{1}{3}+\frac{4}{15}-\frac{2}{15}=\frac{5+4-2}{15}=\frac{7}{15} \\ & \therefore P(\bar{X} / Y)=\frac{P(\bar{X} \cap Y)}{P(Y)} \\ & \quad=\frac{P(Y)-P(X \cap Y)}{P(Y)} \\ & \quad=1-\frac{P(X \cap Y)}{P(Y)}=1-\frac{\frac{2}{15}}{\frac{4}{15}}=1-\frac{1}{2} \\ & =\frac{1}{2} \end{aligned} $

We have,

$ P(A)=\frac{4}{9}, P(A \cap \bar{B})=\frac{3}{7} $

Now, $P(A \cap \bar{B})=P(A)-P(A \cap B)$

$ \begin{aligned} & \Rightarrow \quad P(A \cap B)=P(A)-P(A \cap \bar{B}) \\ & \quad=\frac{4}{9}-\frac{3}{7}=\frac{28-27}{63}=\frac{1}{63} \\ & \therefore P(B / A)=\frac{P(B \cap A)}{P(A)}=\frac{\left(\frac{1}{63}\right)}{\left(\frac{4}{9}\right)}=\frac{1}{28} \end{aligned} $

We have,

$ n=4, p=\frac{20}{100}=\frac{1}{5}, q=1-p=1-\frac{1}{5}=\frac{4}{5} $

∴ Required probability $=P(x<2)$

$ \begin{aligned} & =P(x=0)+P(x=1) \\ & ={ }^4 C_0\left(\frac{1}{5}\right)^0\left(\frac{4}{5}\right)^4+{ }^4 C_1\left(\frac{1}{5}\right)^1\left(\frac{4}{5}\right)^3 \\ & =\left(\frac{4}{5}\right)^4+\left(\frac{4}{5}\right)^4 \\ & =2\left(\frac{4}{5}\right)^4=0.8192 \end{aligned} $

$ \begin{aligned} & \text { We have, } \lambda=\frac{60}{600}=0.1 \\ & \begin{aligned} \therefore \quad P(X & =x)=\frac{e^{-\lambda}(\lambda)^x}{x!} \\ & =\frac{e^{-0.1}(0.1)^x}{x!} \end{aligned} \end{aligned} $

$ \begin{aligned} &\text { Required probability }\\ &\begin{aligned} & =P(x=0)+P(x=1)+P(x=2) \\ & =e^{-0.1}\left[\frac{(0.1)^0}{0!}+\frac{(0.1)^1}{1!}+\frac{(0.1)^2}{2!}\right] \\ & =e^{-0.1}\left[1+\frac{1}{10}+\frac{1}{200}\right] \\ & =e^{-0.1}\left[\frac{200+20+1}{200}\right] \\ & =e^{-0.1}\left(\frac{221}{200}\right)=\frac{1}{e^{0.1}}\left(\frac{221}{200}\right) \end{aligned} \end{aligned} $