Probability

$ \begin{aligned} & \quad S=\{H, T H, T T H, T T T\} \\ & P(H)=\frac{1}{2}, P(T H)=\frac{1}{4} \\ & P(T T H)=\frac{1}{8}, P(T T T)=\frac{1}{8} \end{aligned}$

Let $E$ be the event "no head on first toss"

$\begin{aligned} & \therefore E=\{T H, T T H, T T T\} \\ & \therefore P(E)=\frac{1}{4}+\frac{1}{8}+\frac{1}{8}=\frac{1}{2} \end{aligned}$

Let $E^{\prime}$ be the event coin is tossed 3 times

$\begin{aligned} & \therefore E^{\prime}=\{T T H, T T T\} \\ & \therefore P\left(E^{\prime}\right)=\frac{1}{8}+\frac{1}{8}=\frac{1}{4} \\ & \therefore P\left(E^{\prime} / E\right)=\frac{P\left(E^{\prime} \cap E\right)}{P(E)}=\frac{P\left(E^{\prime}\right)}{P(E)}=\frac{1 / 4}{1 / 2}=\frac{1}{2} \end{aligned}$

$\Rightarrow$ No option is correct.

$x_1+x_2+x_3$ is odd if

(i) all three are odd

(ii) two even and one odd Now, total number of outcomes $=3 \times 5 \times 7=105$

Possible outcomes-

(i) (odd odd odd), then $P(O+O+O)$

$\text { Possibility }=\frac{2}{3} \times \frac{3}{5} \times \frac{4}{7}=\frac{24}{105}$

(ii) (odd even even), $P(O+E+E)$

$\text { Possibility }=\frac{2}{3} \times \frac{2}{5} \times \frac{3}{7}=\frac{12}{105}$

(iii) (even odd even), $p(E+O+E)$

$\text { Possibility }=\frac{1}{3} \times \frac{3}{5} \times \frac{3}{7}=\frac{9}{105}$

(iv) (even even odd), $P(E+E+O$ )$

Possibility $=\frac{1}{2} \times \frac{2}{5} \times \frac{4}{7}=\frac{8}{105}$

Probability that $x_1+x_2+x_3$ is odd is given as

$\frac{24}{105}+\frac{12}{105}+\frac{9}{105}+\frac{8}{105}=\frac{53}{105}$

We know $\Sigma p(x)=1$

$\Rightarrow \quad \frac{C}{1 !}+\frac{C^2}{2 !}+\frac{C^3}{3 !}+\ldots+=1$

Add 1 on both side

$\begin{aligned} 1+C+\frac{C^2}{2 !}+\frac{C^3}{3 !}+\ldots & =2 \\ \Rightarrow \quad e^C =2 \Rightarrow C=\log 2 \end{aligned}$

Let probability of getting head $=x$

Then, probability of getting tail $=1-x$

Tom wins if he gets head

H, TTH, TTTTH, ...(alternative through)

Probability of Tom winning game $=62.5 \%=\frac{625}{1000}$

$\mathrm{P}(\mathrm{H})+\mathrm{P}(\mathrm{TTH})+\mathrm{P}(\mathrm{TTTTH})+\ldots=\frac{625}{1000}$

$\begin{aligned} x+(1-x)(1-x) x+(1-x)^4 x+\ldots & =\frac{625}{1000} \\ x\left[1+(1-x)^2+(1-x)^4+\ldots\right] & =\frac{625}{1000} \\ x \cdot \frac{1}{1-(1-x)^2} & =\frac{625}{100} \\ \frac{x}{2 x-x^2} & =\frac{625}{1000} \\ \frac{1}{2-x} & =\frac{625}{1000} \end{aligned}$

or $\frac{1}{2-x}=\frac{625}{1000}$

or $x=\frac{2}{5}(\text { probability of head })$

or $1-x=\frac{3}{5} \text { (probability of tails) }$

$\therefore$ Probability of getting exactly 3 heads on tossing a coin or 5 times

$\begin{aligned} & ={ }^5 C_3 x^3(1-x)^2 \\ & =10 \times\left(\frac{2}{5}\right)^3\left(\frac{3}{5}\right)^2=\frac{144}{625} \end{aligned}$

Even or divisible by 5

$\begin{aligned} & A:\{2,4,5,6,8,10,12,14,15,16,18,20,22,24,25,26\} \\ & n(A)=16 \\ & \therefore \text { Probability }=\frac{n(A)}{n(S)}=\frac{16}{27} \end{aligned}$

$\begin{aligned} \text { Probability } & =\frac{{ }^5 C_3 \times{ }^7 C_6}{{ }^{12} C_9} \\ & =\frac{7 \times 10}{{ }^{12} C_9}=\frac{70}{\frac{12 \cdot 11 \cdot 10}{6}}=\frac{7}{22}\end{aligned}$

They will contradict each other in two case

Case I $A \rightarrow$ True, $B \rightarrow$ False

Case II $A \rightarrow$ False, $B \rightarrow$ True

$\text { Probability }=\frac{4}{5} \times \frac{1}{4}+\frac{1}{5} \times \frac{3}{4}=\frac{7}{20}$

$\begin{aligned} & \text { Mean }=n p=2 \\ & \text { Variance }=n p q=1 \\ & \Rightarrow \quad q=\frac{1}{2} \text { and } p=\frac{1}{2} \Rightarrow n\left(\frac{1}{2}\right)=2 \\ & \Rightarrow \quad n=4 \\ & P(x>1)=P(x=2)+P(x=3)+P(x=4) \\ & =\left({ }^4 C_2+{ }^4 C_3+{ }^4 C_4\right) \frac{1}{2^4}=\frac{11}{16}\end{aligned}$

$P$ speaks truth $=70 \%=7 / 10$

$Q$ speaks truth $=80 \%=8 / 10$

They agree when both speaks truth or both does not speaks truth

$\begin{aligned} & =\frac{7}{10} \times \frac{8}{10}+\left(1-\frac{7}{10}\right)\left(1-\frac{8}{10}\right) \\ & =\frac{56}{100}+\left(\frac{3}{10}\right)\left(\frac{2}{10}\right)=\frac{56+6}{100}=\frac{62}{100}=62 \% \end{aligned}$

Given,

$\begin{array}{ll} & P(A \cap B)=\frac{1}{3}, P(A \cup B)=\frac{5}{6}, P\left(A^c\right)=\frac{1}{2} \\ \because & P(A)+P(B)=P(A \cup B)+P(A \cap B) \\ \because & P(A)+P(B)=P(A \cup B)+P(A \cap B) \\ \text { or } & 1-P\left(A^c\right)+1-P\left(B^c\right)=\frac{5}{6}+\frac{1}{3}=\frac{7}{6} \\ \text { or } & P\left(B^c\right)=2-\frac{7}{6}-\frac{1}{2}=\frac{1}{3} \end{array}$

Every time we toss, probability of getting head is $1 / 2$.

So, on tossing $n$ times, probability of getting head at $r$th toss $=1 / 2$

$\therefore$ Probability of getting head on 1947th toss $=1 / 2$

Expected value of $X=E(X)=\Sigma px$

$=10\times0.3+20\times0.3+30\times0.2+40\times0.2$

$=3+6+6+8=23$

$\therefore \Sigma P=1$

$\Rightarrow K+8K+27K+64K=1$

$\Rightarrow K=\frac{1}{100}$

$\begin{aligned} P\left(\frac{1}{2} < X < \frac{5}{2} X>1\right) & =\frac{P\left(\left(\frac{1}{2} < X < \frac{5}{2}\right) \cap(X >1)\right)}{P(X > 1)} \\ & =\frac{P\left(1 < X < \frac{5}{2}\right)}{1-P(X \leq 1)}=\frac{P(X=2)}{1-P(X=1)} \\ & =\frac{\frac{8}{100}}{1-\frac{1}{100}}=\frac{8}{99}\end{aligned}$

Total distribution = 3$^{12}$

Number of distribution when first box contains three balls $ = {}^{12}{C_3} \times {2^9}$

Required probability $ = {{{}^{12}{C_3} \times {2^9}} \over {{3^{12}}}}$

$\begin{aligned} & \text { } P(E)=P(X=2)+P(X=3)+P(X=5) \\ & +P(X=7) \\ & =0.23+0.12+0.20+0.07=0.62 \\ & P(F)=P(X=1)+P(X=2)+P(X=3) \\ & =0.15+0.23+0.12=0.50 \\ & P(E \cap F)=P(X=2)+P(X=3) \\ & =0.23+0.12=0.35 \\ & P(E \cup F)=P(E)+P(F)-P(E \cap F) \\ & =0.62+0.50-0.35 \\ & =0.77 \\ & \end{aligned}$

Sample space of a dice $S=\{1,2,3,4,5,6\}$

Let $E=$ getting even number $\Rightarrow E=\{2,4,6\}$

$P(E)=\frac{n(E)}{n(S)}=\frac{3}{6}=\frac{1}{2}$

If dice is thrown three times

Let $X=$ success, then

Probability of getting at least 2 successes

$\begin{aligned} & =P(X \geq 2)=P(X=2)+P(X=3) \\ & ={ }^3 C_2\left(\frac{1}{2}\right)^3+{ }^3 C_3\left(\frac{1}{2}\right)^3=\frac{3}{8}+\frac{1}{8}=\frac{1}{2} \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} $