Binomial distribution Questions in English

(C) Given that in $3$ trials of a binomial distribution,the probability of $2$ successes is $9$ times the probability of $3$ successes.
Let $p$ be the probability of success and $q$ be the probability of failure,where $p + q = 1$.
The probability of $r$ successes in $n$ trials is given by $P(X = r) = {}^nC_r p^r q^{n-r}$.
For $n = 3$:
$P(X = 2) = 9 \times P(X = 3)$
${}^3C_2 p^2 q^1 = 9 \times {}^3C_3 p^3 q^0$
$3 p^2 q = 9 p^3$
Since $p \neq 0$,we can divide by $3p^2$:
$q = 3p$
Substitute $q = 1 - p$:
$1 - p = 3p$
$1 = 4p$
$p = \frac{1}{4}$
Thus,the probability of success in each trial is $\frac{1}{4}$.
Hence,option $C$ is correct.

(B) Let $X$ be the number of times the man hits the target. This follows a binomial distribution $B(n, p)$ with $n=8$ and $p=\frac{1}{3}$.
Probability of success $p = \frac{1}{3}$,probability of failure $q = 1 - \frac{1}{3} = \frac{2}{3}$.
We need to find the probability of hitting the target at least twice,which is $P(X \ge 2)$.
$P(X \ge 2) = 1 - [P(X=0) + P(X=1)]$.
Using the binomial formula $P(X=k) = \binom{n}{k} p^k q^{n-k}$:
$P(X=0) = \binom{8}{0} (\frac{1}{3})^0 (\frac{2}{3})^8 = (\frac{2}{3})^8$.
$P(X=1) = \binom{8}{1} (\frac{1}{3})^1 (\frac{2}{3})^7 = 8 \cdot \frac{1}{3} \cdot (\frac{2}{3})^7$.
$P(X \ge 2) = 1 - [(\frac{2}{3})^8 + 8 \cdot \frac{1}{3} \cdot (\frac{2}{3})^7]$.
$P(X \ge 2) = 1 - [(\frac{2}{3})^7 \cdot (\frac{2}{3} + \frac{8}{3})] = 1 - [(\frac{2}{3})^7 \cdot \frac{10}{3}] = 1 - 5 \cdot (\frac{2}{3})^8$.

(D) Let $n = 5$ be the number of tests and $p = \frac{2}{3}$ be the probability of getting a distinction. Then $q = 1 - p = 1 - \frac{2}{3} = \frac{1}{3}$.
Using the binomial distribution formula $P(X = k) = \binom{n}{k} p^k q^{n-k}$,we need to find the probability of getting a distinction in at least $3$ tests,which is $P(X \ge 3) = P(X = 3) + P(X = 4) + P(X = 5)$.
$P(X = 3) = \binom{5}{3} (\frac{2}{3})^3 (\frac{1}{3})^2 = 10 \times \frac{8}{27} \times \frac{1}{9} = \frac{80}{243}$.
$P(X = 4) = \binom{5}{4} (\frac{2}{3})^4 (\frac{1}{3})^1 = 5 \times \frac{16}{81} \times \frac{1}{3} = \frac{80}{243}$.
$P(X = 5) = \binom{5}{5} (\frac{2}{3})^5 (\frac{1}{3})^0 = 1 \times \frac{32}{243} \times 1 = \frac{32}{243}$.
Summing these probabilities: $P(X \ge 3) = \frac{80 + 80 + 32}{243} = \frac{192}{243}$.
Dividing both numerator and denominator by $3$,we get $\frac{64}{81}$.

(B) Let $p$ be the probability that $A$ wins a single game,so $p = 0.6 = \frac{3}{5}$.
Let $n = 3$ be the number of games played.
We want to find the probability that $A$ wins at least two games,which is $P(X \ge 2) = P(X = 2) + P(X = 3)$,where $X$ follows a binomial distribution $B(n, p)$.
The probability mass function is $P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}$.
For $k = 2$: $P(X = 2) = \binom{3}{2} (0.6)^2 (0.4)^1 = 3 \times 0.36 \times 0.4 = 0.432$.
For $k = 3$: $P(X = 3) = \binom{3}{3} (0.6)^3 (0.4)^0 = 1 \times 0.216 \times 1 = 0.216$.
Summing these probabilities: $P(X \ge 2) = 0.432 + 0.216 = 0.648$.
Converting to a fraction: $0.648 = \frac{648}{1000} = \frac{81}{125}$.

(A) The probability mass function of a binomial distribution $X \sim B(n, p)$ is given by $P(X=k) = \binom{n}{k} p^k (1-p)^{n-k}$.
Given $n=9$,the equation $P(X=3)=P(X=6)$ implies:
$\binom{9}{3} p^3 (1-p)^{9-3} = \binom{9}{6} p^6 (1-p)^{9-6}$
Since $\binom{9}{3} = \binom{9}{6}$,we have:
$p^3 (1-p)^6 = p^6 (1-p)^3$
Dividing both sides by $p^3 (1-p)^3$ (assuming $p \neq 0, 1$):
$(1-p)^3 = p^3$
$1-p = p \implies 2p = 1 \implies p = \frac{1}{2}$.
Now,we need to find $P(X < 3) = P(X=0) + P(X=1) + P(X=2)$.
$P(X=0) = \binom{9}{0} (\frac{1}{2})^0 (\frac{1}{2})^9 = 1 \cdot 1 \cdot \frac{1}{512} = \frac{1}{512}$.
$P(X=1) = \binom{9}{1} (\frac{1}{2})^1 (\frac{1}{2})^8 = 9 \cdot \frac{1}{512} = \frac{9}{512}$.
$P(X=2) = \binom{9}{2} (\frac{1}{2})^2 (\frac{1}{2})^7 = 36 \cdot \frac{1}{512} = \frac{36}{512}$.
$P(X < 3) = \frac{1+9+36}{512} = \frac{46}{512} = \frac{23}{256}$.

(A) For a binomial distribution,mean $\mu = np = \frac{16}{5}$ and variance $\sigma^2 = npq = \frac{48}{25}$.
Dividing variance by mean,we get $q = \frac{npq}{np} = \frac{48/25}{16/5} = \frac{48}{25} \times \frac{5}{16} = \frac{3}{5}$.
Since $p + q = 1$,we have $p = 1 - \frac{3}{5} = \frac{2}{5}$.
Substituting $p$ in $np = \frac{16}{5}$,we get $n \times \frac{2}{5} = \frac{16}{5}$,so $n = 8$.
The probability mass function is $P(X = k) = \binom{n}{k} p^k q^{n-k} = \binom{8}{k} (\frac{2}{5})^k (\frac{3}{5})^{8-k}$.
We need $P(X \leq 2) = P(X=0) + P(X=1) + P(X=2)$.
$P(X=0) = \binom{8}{0} (\frac{2}{5})^0 (\frac{3}{5})^8 = \frac{3^8}{5^8}$.
$P(X=1) = \binom{8}{1} (\frac{2}{5})^1 (\frac{3}{5})^7 = 8 \times \frac{2}{5} \times \frac{3^7}{5^7} = \frac{16 \times 3^7}{5^8} = \frac{16 \times 3 \times 3^6}{5^8} = \frac{48 \times 3^6}{5^8}$.
$P(X=2) = \binom{8}{2} (\frac{2}{5})^2 (\frac{3}{5})^6 = 28 \times \frac{4}{25} \times \frac{3^6}{5^6} = \frac{112 \times 3^6}{5^8}$.
Summing these: $P(X \leq 2) = \frac{3^8 + 48 \times 3^6 + 112 \times 3^6}{5^8} = \frac{9 \times 3^6 + 160 \times 3^6}{5^8} = \frac{169 \times 3^6}{5^8}$.

(B) For a fair coin,the probability of heads is $p = 1/2$ and tails is $q = 1/2$. The random variable $X$ follows a binomial distribution $B(n, 1/2)$.
$P(X=k) = \binom{n}{k} (1/2)^n$.
Given that $P(X=4)$,$P(X=5)$,and $P(X=6)$ are in arithmetic progression,we have $2P(X=5) = P(X=4) + P(X=6)$.
Substituting the binomial probabilities: $2 \binom{n}{5} (1/2)^n = \binom{n}{4} (1/2)^n + \binom{n}{6} (1/2)^n$.
Dividing by $(1/2)^n$,we get $2 \binom{n}{5} = \binom{n}{4} + \binom{n}{6}$.
Using the formula $\binom{n}{k} = \frac{n!}{k!(n-k)!}$,we have $2 \frac{n!}{5!(n-5)!} = \frac{n!}{4!(n-4)!} + \frac{n!}{6!(n-6)!}$.
Dividing by $n!$ and multiplying by $6!(n-4)!$: $2 \times 6(n-4) = 6 \times 5 + (n-4)(n-5)$.
$12n - 48 = 30 + n^2 - 9n + 20$.
$n^2 - 21n + 98 = 0$.
$(n-7)(n-14) = 0$.
Thus,$n = 7$ or $n = 14$. The largest value of $n$ is $14$.

(D) For a binomial distribution,the mean is $\mu = np$ and the variance is $\sigma^2 = npq$,where $q = 1 - p$.
Given $n = 6$ and $\mu - \sigma^2 = \frac{27}{8}$.
Substituting the formulas: $np - npq = \frac{27}{8} \implies np(1 - q) = \frac{27}{8}$.
Since $1 - q = p$,we have $np^2 = \frac{27}{8}$.
Substituting $n = 6$: $6p^2 = \frac{27}{8} \implies p^2 = \frac{27}{48} = \frac{9}{16}$.
Thus,$p = \frac{3}{4}$ and $q = 1 - \frac{3}{4} = \frac{1}{4}$.
The probability of getting $X$ successes is given by $P(X = k) = \binom{n}{k} p^k q^{n-k}$.
We need to find $P(X \le 2) = P(X = 0) + P(X = 1) + P(X = 2)$.
$P(X = 0) = \binom{6}{0} (\frac{3}{4})^0 (\frac{1}{4})^6 = 1 \times 1 \times \frac{1}{4^6} = \frac{1}{4^6}$.
$P(X = 1) = \binom{6}{1} (\frac{3}{4})^1 (\frac{1}{4})^5 = 6 \times \frac{3}{4^6} = \frac{18}{4^6}$.
$P(X = 2) = \binom{6}{2} (\frac{3}{4})^2 (\frac{1}{4})^4 = 15 \times \frac{9}{4^6} = \frac{135}{4^6}$.
Summing these: $P(X \le 2) = \frac{1 + 18 + 135}{4^6} = \frac{154}{4^6}$.

(D) For a binomial distribution $X \sim B(n, p)$,the mean $\mu = np$ and variance $\sigma^2 = npq$,where $q = 1-p$.
Given $\mu = 2\sigma^2$,we have $np = 2npq$,which implies $1 = 2q$,so $q = \frac{1}{2}$ and $p = 1 - q = \frac{1}{2}$.
Given $\mu + \sigma^2 = 3$,we substitute $\mu = 2\sigma^2$ to get $3\sigma^2 = 3$,so $\sigma^2 = 1$.
Since $\sigma^2 = npq = n(\frac{1}{2})(\frac{1}{2}) = \frac{n}{4} = 1$,we find $n = 4$.
Thus,$X \sim B(4, \frac{1}{2})$.
We need to calculate $P(X \leq 3) = 1 - P(X = 4)$.
$P(X = 4) = \binom{4}{4} p^4 q^0 = 1 \times (\frac{1}{2})^4 \times 1 = \frac{1}{16}$.
Therefore,$P(X \leq 3) = 1 - \frac{1}{16} = \frac{15}{16}$.

(B) Let $n = 4$ be the number of scans and $p = 0.1$ be the probability of detecting the plane in a single scan. The probability of not detecting the plane is $q = 1 - p = 0.9$.
Using the binomial distribution,the probability of detecting the plane $X$ times in $n$ scans is given by $P(X = k) = \binom{n}{k} p^k q^{n-k}$.
We need to find the probability of detecting the plane at least twice,which is $P(X \ge 2) = 1 - [P(X = 0) + P(X = 1)]$.
$P(X = 0) = \binom{4}{0} (0.1)^0 (0.9)^4 = 1 \times 1 \times 0.6561 = 0.6561$.
$P(X = 1) = \binom{4}{1} (0.1)^1 (0.9)^3 = 4 \times 0.1 \times 0.729 = 0.2916$.
Therefore,$P(X \ge 2) = 1 - (0.6561 + 0.2916) = 1 - 0.9477 = 0.0523$.

(A) Let $X$ be the number of defective bulbs chosen. This follows a binomial distribution $B(n, p)$ where $n = 5$ and $p = 20\% = \frac{1}{5}$.
Then $q = 1 - p = \frac{4}{5}$.
The probability of getting exactly $k$ defective bulbs is given by $P(X = k) = { }^nC_k \cdot p^k \cdot q^{n-k}$.
For $k = 3$,we have:
$P(X = 3) = { }^5C_3 \cdot (\frac{1}{5})^3 \cdot (\frac{4}{5})^{5-3}$
$P(X = 3) = \frac{5 \times 4}{2 \times 1} \cdot (\frac{1}{125}) \cdot (\frac{16}{25})$
$P(X = 3) = 10 \cdot \frac{16}{3125}$
$P(X = 3) = \frac{160}{3125} = \frac{32}{625}$.

(B) Given that the difference between the mean and variance is $1$:
$np - npq = 1 \Rightarrow np(1-q) = 1 \Rightarrow np^2 = 1$ $\qquad (i)$
Also,the difference between their squares is $11$:
$(np)^2 - (npq)^2 = 11$ $\qquad (ii)$
$n^2p^2 - n^2p^2q^2 = 11 \Rightarrow n^2p^2(1-q^2) = 11$
Dividing $(ii)$ by $(i)$,we get:
$\frac{n^2p^2(1-q^2)}{np^2} = 11 \Rightarrow n(1-q^2) = 11$
Since $n=36$,$36(1-q^2) = 11 \Rightarrow 1-q^2 = \frac{11}{36} \Rightarrow q^2 = 1 - \frac{11}{36} = \frac{25}{36} \Rightarrow q = \frac{5}{6}$
Thus,$p = 1 - q = 1 - \frac{5}{6} = \frac{1}{6}$
Now,$P(X=2) = {}^{36}C_2 \left(\frac{1}{6}\right)^2 \left(\frac{5}{6}\right)^{34} = m\left(\frac{5}{6}\right)^{36}$
$\frac{36 \times 35}{2} \times \frac{1}{36} \times \left(\frac{5}{6}\right)^{34} = m\left(\frac{5}{6}\right)^{36}$
$\frac{35}{2} \times \left(\frac{5}{6}\right)^{34} = m \times \left(\frac{5}{6}\right)^{34} \times \left(\frac{5}{6}\right)^2$
$m = \frac{35}{2} \times \left(\frac{6}{5}\right)^2 = \frac{35}{2} \times \frac{36}{25} = \frac{7 \times 18}{5} = \frac{126}{5} = 25.2$
Wait,re-evaluating $m$: $m = \frac{35}{2} \times \frac{36}{25} = \frac{7 \times 18}{5} = 25.2$.
Given $m : n = 25.2 : 36 = 252 : 360 = 7 : 10$.

(A) Let $p$ be the probability of hitting the target and $q$ be the probability of failing to hit the target.
Given $q = \frac{1}{3}$,so $p = 1 - q = 1 - \frac{1}{3} = \frac{2}{3}$.
The number of trials is $n = 4$.
We need to find the probability of hitting the target at least thrice,which is $P(X \geq 3) = P(X = 3) + P(X = 4)$.
Using the binomial distribution formula $P(X = k) = {}^nC_k p^k q^{n-k}$:
$P(X = 3) = {}^4C_3 \left(\frac{2}{3}\right)^3 \left(\frac{1}{3}\right)^1 = 4 \times \frac{8}{27} \times \frac{1}{3} = \frac{32}{81}$.
$P(X = 4) = {}^4C_4 \left(\frac{2}{3}\right)^4 \left(\frac{1}{3}\right)^0 = 1 \times \frac{16}{81} \times 1 = \frac{16}{81}$.
Therefore,$P(X \geq 3) = \frac{32}{81} + \frac{16}{81} = \frac{48}{81} = \frac{16}{27}$.

(C) For a binomial distribution with $n=7$ and $p=q=\frac{1}{2}$:
$\mu = np = 7 \times \frac{1}{2} = \frac{7}{2}$
$\sigma^2 = npq = 7 \times \frac{1}{2} \times \frac{1}{2} = \frac{7}{4}$
$P(X=3) = {}^{7}C_{3} \times (\frac{1}{2})^{3} \times (\frac{1}{2})^{4} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} \times (\frac{1}{2})^{7} = 35 \times \frac{1}{128} = \frac{35}{128}$
$\frac{\mu \sigma^2}{P(X=3)} = \frac{(\frac{7}{2}) \times (\frac{7}{4})}{\frac{35}{128}} = \frac{49}{8} \times \frac{128}{35} = \frac{7}{1} \times \frac{16}{5} = \frac{112}{5}$

(C) Given $X \sim B(5, p)$.
$P(X=3) = P(X=4)$
$\Rightarrow { }^5 C_3 p^3(1-p)^2 = { }^5 C_4 p^4(1-p)$
$\Rightarrow 10(1-p) = 5p$
$\Rightarrow 10 - 10p = 5p \Rightarrow 15p = 10 \Rightarrow p = \frac{2}{3}$.
Now,we need to find $P(|X-3| < 2)$.
$P(|X-3| < 2) = P(-2 < X-3 < 2) = P(1 < X < 5) = P(X=2) + P(X=3) + P(X=4)$.
Since $P(X=3) = P(X=4)$,we have $P(X=2) + 2P(X=3)$.
$P(X=2) = { }^5 C_2 (\frac{2}{3})^2 (\frac{1}{3})^3 = 10 \times \frac{4}{9} \times \frac{1}{27} = \frac{40}{243}$.
$P(X=3) = { }^5 C_3 (\frac{2}{3})^3 (\frac{1}{3})^2 = 10 \times \frac{8}{27} \times \frac{1}{9} = \frac{80}{243}$.
$P(|X-3| < 2) = \frac{40}{243} + 2(\frac{80}{243}) = \frac{40 + 160}{243} = \frac{200}{243}$.

(C) Let the mean be $\mu = np$ and standard deviation be $\sigma = \sqrt{npq}$,where $q = 1-p$.
Given,$\mu - \sigma = 3 \Rightarrow \mu - 3 = \sigma$.
Squaring both sides,$(\mu - 3)^2 = \sigma^2 = npq$.
Also given,$\mu^2 - \sigma^2 = 21$.
Substituting $\sigma^2 = \mu^2 - 21$ into the first equation:
$(\mu - 3)^2 = \mu^2 - 21$
$\mu^2 - 6\mu + 9 = \mu^2 - 21$
$-6\mu = -30 \Rightarrow \mu = 5$.
Since $\mu = np = 5$,then $\sigma^2 = 5^2 - 21 = 25 - 21 = 4$.
We know $\sigma^2 = npq = 5q = 4$,so $q = \frac{4}{5}$ and $p = 1 - \frac{4}{5} = \frac{1}{5}$.
Since $np = 5$,$n(\frac{1}{5}) = 5 \Rightarrow n = 25$.
The probability mass function is $P(X=k) = {}^{n}C_k p^k q^{n-k}$.
$\frac{P(X=1)}{P(X=2)} = \frac{{}^{25}C_1 p^1 q^{24}}{{}^{25}C_2 p^2 q^{23}} = \frac{25 \cdot q}{300 \cdot p} = \frac{1}{12} \cdot \frac{4/5}{1/5} = \frac{1}{12} \cdot 4 = \frac{1}{3}$.

(A) Let $X$ be the number of times the radar detects the plane in $n = 4$ scans. This follows a binomial distribution $B(n, p)$ where $n = 4$ and $p = 0.1$ (probability of detection).
The probability of not detecting the plane is $q = 1 - p = 0.9$.
We want to find the probability that it cannot detect the plane at least two times,which is equivalent to $1 - P(\text{detecting the plane } 2, 3, \text{ or } 4 \text{ times})$.
$P(X \ge 2) = P(X=2) + P(X=3) + P(X=4)$.
$P(X=2) = {}^{4}C_{2} (0.1)^{2} (0.9)^{2} = 6 \times 0.01 \times 0.81 = 0.0486$.
$P(X=3) = {}^{4}C_{3} (0.1)^{3} (0.9)^{1} = 4 \times 0.001 \times 0.9 = 0.0036$.
$P(X=4) = {}^{4}C_{4} (0.1)^{4} (0.9)^{0} = 1 \times 0.0001 \times 1 = 0.0001$.
$P(X \ge 2) = 0.0486 + 0.0036 + 0.0001 = 0.0523$.
The required probability is $1 - P(X \ge 2) = 1 - 0.0523 = 0.9477$.

(C) For a Binomial distribution $B(n, p)$,Mean $\mu = np$ and Variance $\sigma^2 = npq$,where $q = 1-p$.
Given:
Sum: $np + npq = 5 \Rightarrow np(1+q) = 5$ ...$(i)$
Product: $(np)(npq) = n^2p^2q = 6$ ...(ii)
From $(i)$,$np = \frac{5}{1+q}$. Substituting this into (ii):
$\left(\frac{5}{1+q}\right)^2 q = 6 \Rightarrow 25q = 6(1+q)^2 \Rightarrow 25q = 6(1+2q+q^2) \Rightarrow 6q^2 - 13q + 6 = 0$.
Solving the quadratic equation: $6q^2 - 9q - 4q + 6 = 0 \Rightarrow 3q(2q-3) - 2(2q-3) = 0 \Rightarrow (3q-2)(2q-3) = 0$.
Since $q < 1$,we have $q = \frac{2}{3}$. Then $p = 1 - q = \frac{1}{3}$.
Substituting $q = \frac{2}{3}$ into $(i)$: $np(1 + \frac{2}{3}) = 5 \Rightarrow np(\frac{5}{3}) = 5 \Rightarrow np = 3$.
Since $p = \frac{1}{3}$,$n(\frac{1}{3}) = 3 \Rightarrow n = 9$.
Finally,$6(n+p-q) = 6(9 + \frac{1}{3} - \frac{2}{3}) = 6(9 - \frac{1}{3}) = 54 - 2 = 52$.

(C) Given mean $= np = 6$ ...$(i)$
Variance $= npq = 2$ ...(ii)
Dividing (ii) by $(i)$,we get $q = \frac{2}{6} = \frac{1}{3}$.
Since $p + q = 1$,we have $p = 1 - \frac{1}{3} = \frac{2}{3}$.
Substituting $p$ in $(i)$,$n \times \frac{2}{3} = 6 \Rightarrow n = 9$.
We need to find $P(X \geq 2) = 1 - [P(X=0) + P(X=1)]$.
$P(X=0) = {}^9C_0 \times (\frac{2}{3})^0 \times (\frac{1}{3})^9 = \frac{1}{3^9}$.
$P(X=1) = {}^9C_1 \times (\frac{2}{3})^1 \times (\frac{1}{3})^8 = 9 \times \frac{2}{3} \times \frac{1}{3^8} = \frac{18}{3^9}$.
$P(X \geq 2) = 1 - [\frac{1}{3^9} + \frac{18}{3^9}] = 1 - \frac{19}{3^9}$.

(A) For a binomial distribution,Mean $= np$ and Variance $= npq$,where $n=5$ is the number of trials,$p$ is the probability of success,and $q=1-p$ is the probability of failure.
Given that the difference between mean and variance is $\frac{5}{9}$:
$np - npq = \frac{5}{9}$
$np(1-q) = \frac{5}{9}$
Since $1-q = p$,we have $np^2 = \frac{5}{9}$.
Substituting $n=5$:
$5p^2 = \frac{5}{9} \Rightarrow p^2 = \frac{1}{9} \Rightarrow p = \frac{1}{3}$.
Then $q = 1 - \frac{1}{3} = \frac{2}{3}$.
The probability of getting $2$ successes in $5$ trials is given by the binomial formula $P(X=k) = {^nC_k} p^k q^{n-k}$:
$P(X=2) = {^5C_2} \times (\frac{1}{3})^2 \times (\frac{2}{3})^{5-2}$
$P(X=2) = 10 \times \frac{1}{9} \times \frac{8}{27} = \frac{80}{243}$.

(A) For a binomial distribution,the mean is given by $\mu = np$ and the variance is given by $\sigma^2 = np(1-p)$.
Given,$np = 2.4$ and $np(1-p) = 1.44$.
Substituting the value of $np$ in the variance equation:
$2.4(1-p) = 1.44$
$1-p = \frac{1.44}{2.4} = 0.6$
$p = 1 - 0.6 = 0.4 = \frac{2}{5}$.
Now,substituting $p = 0.4$ into $np = 2.4$:
$n(0.4) = 2.4$
$n = \frac{2.4}{0.4} = 6$.
Thus,the parameters are $n = 6$ and $p = \frac{2}{5}$.

(A) The Poisson distribution is a limiting case of the Binomial distribution under specific conditions: $n$ is very large $(n \to \infty)$ and $p$ is very small $(p \to 0)$ such that $np = \lambda$ remains constant.
Statement $(i)$ describes the basic requirement for a Bernoulli trial.
Statement (ii) is a necessary condition for the Poisson approximation.
Statement (iii) is a requirement for both Binomial and Poisson distributions.
Statement (iv) states that the probability $p$ of success is very large. This contradicts the fundamental assumption of the Poisson distribution,which models rare events where $p$ is small. Therefore,(iv) is not suitable.

(D) For a binomial distribution,Mean $= np = 2$ and Variance $= npq = 1$.
Since $q = 1 - p$,we have $np(1 - p) = 1$.
Substituting $np = 2$,we get $2(1 - p) = 1$,which implies $1 - p = 1/2$,so $p = 1/2$.
Then $n(1/2) = 2$,so $n = 4$.
We need to find $P(X > 1) = 1 - P(X \leq 1) = 1 - [P(X = 0) + P(X = 1)]$.
Using the formula $P(X = k) = ^nC_k p^k q^{n-k}$:
$P(X = 0) = ^4C_0 (1/2)^0 (1/2)^4 = 1 \times 1 \times 1/16 = 1/16$.
$P(X = 1) = ^4C_1 (1/2)^1 (1/2)^3 = 4 \times 1/2 \times 1/8 = 4/16$.
Therefore,$P(X > 1) = 1 - (1/16 + 4/16) = 1 - 5/16 = 11/16$.

(A) Let $p$ be the probability that a person is left-handed,so $p = 0.1$.
Let $q$ be the probability that a person is not left-handed,so $q = 1 - p = 0.9$.
We use the binomial distribution formula for $n = 10$ trials and $k = 1$ success:
$P(X = k) = {}^{n}C_{k} p^{k} q^{n-k}$
Substituting the values:
$P(X = 1) = {}^{10}C_{1} (0.1)^{1} (0.9)^{10-1}$
$P(X = 1) = 10 \times 0.1 \times (0.9)^9$
$P(X = 1) = 1 \times (0.9)^9 = (0.9)^9$

(C) The properties of a Binomial Distribution are as follows:
- There are a fixed number of $n$ independent trials.
- Each trial has exactly two possible outcomes: success or failure.
- The probability of success $(p)$ and failure $(q = 1 - p)$ remains constant for every trial.
- Each trial is independent,meaning the outcome of one trial does not affect the outcome of any other trial.
Therefore,the statement that 'the probabilities of the two outcomes can change from one trial to the next' is incorrect and is not a property of a Binomial distribution.

(C) For a binomial distribution,the mean is given by $\mu = np = 15$.
The variance is given by $\sigma^2 = np(1-p) = 10$.
Substituting the value of $np$ into the variance equation:
$15(1-p) = 10$.
$1-p = \frac{10}{15} = \frac{2}{3}$.
Therefore,$p = 1 - \frac{2}{3} = \frac{1}{3}$.
Now,substitute $p$ back into the mean equation:
$n \times \frac{1}{3} = 15$.
$n = 15 \times 3 = 45$.
Thus,the parameter $n$ is $45$.

(B) Given that $X \sim B(n, p)$,the probability mass function is $P(X=k) = \binom{n}{k} p^k q^{n-k}$,where $q = 1-p$.
Given $P(X=2) = P(X=3)$,we have:
$\binom{n}{2} p^2 q^{n-2} = \binom{n}{3} p^3 q^{n-3}$
$\frac{n!}{2!(n-2)!} p^2 q^{n-2} = \frac{n!}{3!(n-3)!} p^3 q^{n-3}$
Dividing both sides by $n! p^2 q^{n-3}$:
$\frac{1}{2(n-2)!} q = \frac{1}{6(n-3)!} p$
$\frac{q}{2} = \frac{p}{6(n-2)}$
$3q = p(n-2)$
Since $q = 1-p$,we substitute:
$3(1-p) = np - 2p$
$3 - 3p = np - 2p$
$np = 3 - p$
The mean of a binomial distribution is given by $E(X) = np$.
Therefore,the mean is $3-p$.

(B) Given $X$ is a random variable with $B(n=20, p=0.4)$.
So,$n=20$,$p=0.4 = \frac{2}{5}$,and $q = 1 - p = \frac{3}{5}$.
We need to calculate $5 - 5 P(X \geq 2)$.
$P(X \geq 2) = 1 - (P(X=0) + P(X=1))$.
Substituting this into the expression:
$5 - 5(1 - (P(X=0) + P(X=1))) = 5 - 5 + 5(P(X=0) + P(X=1)) = 5(P(X=0) + P(X=1))$.
Using the Binomial probability formula $P(X=k) = {}^{n}C_{k} p^k q^{n-k}$:
$P(X=0) = {}^{20}C_{0} (\frac{2}{5})^0 (\frac{3}{5})^{20} = (\frac{3}{5})^{20}$.
$P(X=1) = {}^{20}C_{1} (\frac{2}{5})^1 (\frac{3}{5})^{19} = 20 \times \frac{2}{5} \times (\frac{3}{5})^{19} = 8 \times (\frac{3}{5})^{19}$.
Now,$5(P(X=0) + P(X=1)) = 5 [(\frac{3}{5})^{20} + 8(\frac{3}{5})^{19}]$
$= 5 [(\frac{3}{5}) \times (\frac{3}{5})^{19} + 8(\frac{3}{5})^{19}]$
$= 5 [(\frac{3}{5} + 8) \times (\frac{3}{5})^{19}]$
$= 5 [(\frac{3+40}{5}) \times (\frac{3}{5})^{19}]$
$= 5 [\frac{43}{5} \times (\frac{3}{5})^{19}] = 43 \times (\frac{3}{5})^{19}$.

(B) Let $n = 5$ be the total number of questions.
Let $p$ be the probability of answering a question incorrectly. Since there are $3$ incorrect responses out of $4$,$p = \frac{3}{4}$.
Let $q$ be the probability of answering a question correctly,so $q = 1 - p = \frac{1}{4}$.
We need to find the probability of answering at least $3$ questions incorrectly,which is $P(X \ge 3)$,where $X$ follows a binomial distribution $B(n, p)$.
$P(X \ge 3) = P(X=3) + P(X=4) + P(X=5)$
$P(X=k) = \binom{n}{k} p^k q^{n-k}$
$P(X=3) = \binom{5}{3} (\frac{3}{4})^3 (\frac{1}{4})^2 = 10 \times \frac{27}{64} \times \frac{1}{16} = \frac{270}{1024}$
$P(X=4) = \binom{5}{4} (\frac{3}{4})^4 (\frac{1}{4})^1 = 5 \times \frac{81}{256} \times \frac{1}{4} = \frac{405}{1024}$
$P(X=5) = \binom{5}{5} (\frac{3}{4})^5 (\frac{1}{4})^0 = 1 \times \frac{243}{1024} \times 1 = \frac{243}{1024}$
$P(X \ge 3) = \frac{270 + 405 + 243}{1024} = \frac{918}{1024} = \frac{459}{512}$

(A) For a Binomial distribution with parameters $n$ and $p$,let $q = 1 - p$ be the probability of failure.
Mean $= np$
Variance $= npq$
Since success is not an impossible event,$p > 0$. Since failure is not an impossible event (implied by the context of a standard Binomial distribution where $0 < p < 1$),we have $0 < q < 1$.
Since $q < 1$,it follows that $npq < np$.
Therefore,the Mean is always greater than the Variance.

(B) The variance of a Binomial distribution is given by $\sigma^2 = n p q$,where $q = 1 - p$.
To find the maximum value of the variance,we express it as a function of $p$: $f(p) = n p(1 - p) = n(p - p^2)$.
Taking the derivative with respect to $p$ and setting it to zero: $f'(p) = n(1 - 2p) = 0$.
This gives $1 - 2p = 0$,or $p = \frac{1}{2}$.
Since $q = 1 - p$,we have $q = 1 - \frac{1}{2} = \frac{1}{2}$.
Substituting these values into the variance formula: $\sigma^2_{\max} = n \times \frac{1}{2} \times \frac{1}{2} = \frac{n}{4}$.

(D) Given,$p=q$.
Since $p+q=1$,we have $p=q=\frac{1}{2}$.
Now,the probability mass function for a Binomial distribution is $P(X=k) = { }^n C_k p^k q^{n-k}$.
Substituting $k=5$ and $p=q=\frac{1}{2}$:
$P(X=5) = { }^n C_5 \left(\frac{1}{2}\right)^5 \left(\frac{1}{2}\right)^{n-5} = { }^n C_5 \left(\frac{1}{2}\right)^n$.
Therefore,$2^n P(X=5) = 2^n \cdot { }^n C_5 \left(\frac{1}{2}\right)^n = { }^n C_5$.

(D) Let $X$ denote the number of successes in $3$ trials. Since the die is tossed $3$ times,$n=3$.
The probability of getting an even number (success) in a single toss is $p = \frac{3}{6} = \frac{1}{2}$.
The probability of failure is $q = 1 - p = 1 - \frac{1}{2} = \frac{1}{2}$.
$X$ follows a binomial distribution $B(n, p)$,where $P(X=r) = { }^n C_r p^r q^{n-r}$.
We need to find the probability of getting at least $2$ successes,which is $P(X \geq 2) = P(X=2) + P(X=3)$.
$P(X=2) = { }^3 C_2 \left(\frac{1}{2}\right)^2 \left(\frac{1}{2}\right)^{3-2} = 3 \times \left(\frac{1}{2}\right)^3 = \frac{3}{8}$.
$P(X=3) = { }^3 C_3 \left(\frac{1}{2}\right)^3 \left(\frac{1}{2}\right)^{3-3} = 1 \times \left(\frac{1}{2}\right)^3 = \frac{1}{8}$.
Therefore,$P(X \geq 2) = \frac{3}{8} + \frac{1}{8} = \frac{4}{8} = \frac{1}{2}$.

(A) For a binomial distribution,the mean is given by $np = 5$ and the variance is given by $npq = 4$.
Substituting $np = 5$ into the variance formula,we get $5q = 4$,which implies $q = \frac{4}{5}$.
Since $p + q = 1$,we have $p = 1 - q = 1 - \frac{4}{5} = \frac{1}{5}$.
Using $np = 5$ and $p = \frac{1}{5}$,we find $n = \frac{5}{p} = 5 \times 5 = 25$.
The probability mass function for a binomial distribution is $P(X=k) = {}^{n}C_{k} p^{k} q^{n-k}$.
For $X=1$,we have $P(X=1) = {}^{25}C_{1} \left(\frac{1}{5}\right)^{1} \left(\frac{4}{5}\right)^{25-1}$.
$P(X=1) = 25 \times \frac{1}{5} \times \left(\frac{4}{5}\right)^{24} = 5 \times \frac{4^{24}}{5^{24}} = \frac{4^{24}}{5^{23}}$.

(A) Let success be selecting a smart phone used for office purposes.
Here,the probability of success $p = 50 \% = \frac{1}{2}$.
The probability of failure $q = 1 - p = 1 - \frac{1}{2} = \frac{1}{2}$.
We have $n = 10$ trials and we want exactly $r = 2$ successes.
Using the Binomial distribution formula $P(X = r) = { }^n C_r p^r q^{n-r}$:
$P(X = 2) = { }^{10} C_2 \cdot (\frac{1}{2})^2 \cdot (\frac{1}{2})^{10-2}$
$P(X = 2) = { }^{10} C_2 \cdot (\frac{1}{2})^2 \cdot (\frac{1}{2})^8$
$P(X = 2) = { }^{10} C_2 \cdot (\frac{1}{2})^{10} = { }^{10} C_2 \frac{1}{2^{10}}$.

(C) In a binomial distribution $B(n, p)$,the mean is given by $\mu = np$ and the standard deviation is given by $\sigma = \sqrt{npq}$.
Given,$np = 200$ and $\sqrt{npq} = 10$.
Squaring the standard deviation equation,we get $npq = 100$.
Substituting $np = 200$ into $npq = 100$,we get $200q = 100$,which implies $q = \frac{1}{2}$.
Since $p + q = 1$,we have $p = 1 - \frac{1}{2} = \frac{1}{2}$.
Now,$np = 200 \implies n(\frac{1}{2}) = 200 \implies n = 400$.
We need to calculate $n^2 + \frac{1}{p^2} + \frac{1}{q^2}$.
Substituting the values: $(400)^2 + \frac{1}{(1/2)^2} + \frac{1}{(1/2)^2} = 160000 + 4 + 4 = 160008$.

(B) Let $n$ be the number of times he fires at the target.
Let $X$ be the number of times he hits the target.
Since hitting the target is a Bernoulli trial,$X$ follows a binomial distribution.
Here,$p = \frac{2}{3}$ (probability of hitting) and $q = 1 - p = \frac{1}{3}$ (probability of missing).
The probability of hitting the target at least once is given by $P(X \geq 1) = 1 - P(X = 0)$.
We are given that $P(X \geq 1) > 90 \%$,which means $P(X \geq 1) > 0.9$.
Therefore,$1 - P(X = 0) > 0.9$.
Since $P(X = 0) = {}^nC_0 q^n p^0 = (\frac{1}{3})^n$,we have:
$1 - (\frac{1}{3})^n > 0.9$
$0.1 > (\frac{1}{3})^n$
$(\frac{1}{3})^n < \frac{1}{10}$
$3^n > 10$.
For $n = 1$,$3^1 = 3 < 10$.
For $n = 2$,$3^2 = 9 < 10$.
For $n = 3$,$3^3 = 27 > 10$.
Thus,the minimum number of times he must fire is $3$.

(B) Given Mean = $np = 6$ ... $(i)$
Variance = $npq = 2$ where $q = 1 - p$ ... $(ii)$
Substituting $(i)$ into $(ii)$:
$6q = 2 \implies q = \frac{1}{3}$
Since $p = 1 - q$,we have $p = 1 - \frac{1}{3} = \frac{2}{3}$
From $(i)$,$n = \frac{6}{p} = \frac{6}{2/3} = 9$
The probability mass function is $P(X = r) = {}^nC_r p^r q^{n-r}$
For $X = 8$:
$P(X = 8) = {}^9C_8 \left(\frac{2}{3}\right)^8 \left(\frac{1}{3}\right)^{9-8}$
$P(X = 8) = 9 \times \frac{2^8}{3^8} \times \frac{1}{3}$
$P(X = 8) = 9 \times \frac{2^8}{3^9} = \frac{3^2 \times 2^8}{3^9} = \frac{2^8}{3^7}$

(D) Given,mean of a binomial distribution $= 20$.
$\Rightarrow np = 20$ $(i)$.
Standard deviation of a binomial distribution $= 4$.
$\Rightarrow \sqrt{np(1-p)} = 4$.
Squaring both sides,we get $np(1-p) = 16$ $(ii)$.
Substitute $np = 20$ from equation $(i)$ into equation $(ii)$:
$20(1-p) = 16$.
$1-p = \frac{16}{20} = \frac{4}{5}$.
$p = 1 - \frac{4}{5} = \frac{1}{5}$.
Substitute the value of $p$ into equation $(i)$:
$n \times \frac{1}{5} = 20$.
$n = 20 \times 5 = 100$.
Therefore,the number of trials is $100$.

(B) Total toys = $30$. White toys = $10$. Blue toys = $30 - 10 = 20$.
Probability of drawing a white toy $(p)$ = $\frac{10}{30} = \frac{1}{3}$.
Probability of drawing a blue toy $(q)$ = $1 - p = 1 - \frac{1}{3} = \frac{2}{3}$.
Since the toys are replaced,this follows a binomial distribution $B(n, p)$ with $n = 5$ and $p = \frac{1}{3}$.
We need the probability of getting at most $2$ white toys,which is $P(X \le 2) = P(X=0) + P(X=1) + P(X=2)$.
$P(X=k) = {}^nC_k p^k q^{n-k}$.
$P(X=0) = {}^5C_0 (\frac{1}{3})^0 (\frac{2}{3})^5 = (\frac{2}{3})^5$.
$P(X=1) = {}^5C_1 (\frac{1}{3})^1 (\frac{2}{3})^4 = 5 \times \frac{1}{3} \times \frac{16}{81} = \frac{80}{243}$.
$P(X=2) = {}^5C_2 (\frac{1}{3})^2 (\frac{2}{3})^3 = 10 \times \frac{1}{9} \times \frac{8}{27} = \frac{80}{243}$.
$P(X \le 2) = \frac{32}{243} + \frac{80}{243} + \frac{80}{243} = \frac{192}{243}$.
Dividing numerator and denominator by $3$,we get $\frac{64}{81} = (\frac{8}{9})^2$.

(A) For a binomial distribution with $n$ trials and probability of success $p$,let $q = 1 - p$.
Mean $\mu = np$ and Variance $\sigma^2 = npq$.
Given:
$np + npq = 15$
$np(npq) = 54$
Let $X = np$ and $Y = npq$.
Then $X + Y = 15$ and $XY = 54$.
These are roots of the quadratic equation $t^2 - 15t + 54 = 0$.
$(t - 9)(t - 6) = 0$,so the roots are $9$ and $6$.
Since $np > npq$ (because $q < 1$),we have $np = 9$ and $npq = 6$.
Dividing the two equations: $\frac{npq}{np} = \frac{6}{9} \implies q = \frac{2}{3}$.
Then $p = 1 - q = 1 - \frac{2}{3} = \frac{1}{3}$.
Substituting $p$ into $np = 9$: $n(\frac{1}{3}) = 9 \implies n = 27$.
Thus,the number of trials is $27$.

(C) We know that the binomial distribution is given by $P(X=r) = { }^n C_r p^r q^{n-r}$ where $p+q=1$.
Given $n=5, p=\frac{3}{4}, q=\frac{1}{4}$.
First,calculate $P(X \geq 3) = P(X=3) + P(X=4) + P(X=5)$.
$P(X=3) = { }^5 C_3 (\frac{3}{4})^3 (\frac{1}{4})^2 = 10 \times \frac{27}{64} \times \frac{1}{16} = \frac{270}{1024}$.
$P(X=4) = { }^5 C_4 (\frac{3}{4})^4 (\frac{1}{4})^1 = 5 \times \frac{81}{256} \times \frac{1}{4} = \frac{405}{1024}$.
$P(X=5) = { }^5 C_5 (\frac{3}{4})^5 (\frac{1}{4})^0 = 1 \times \frac{243}{1024} \times 1 = \frac{243}{1024}$.
$P(X \geq 3) = \frac{270+405+243}{1024} = \frac{918}{1024}$.
Then $\alpha = \frac{1}{9} \times \frac{918}{1024} = \frac{102}{1024}$.
Next,calculate $\beta = P(X \leq 2) = 1 - P(X \geq 3) = 1 - \frac{918}{1024} = \frac{106}{1024}$.
Finally,$256(\beta - \alpha) = 256(\frac{106}{1024} - \frac{102}{1024}) = 256(\frac{4}{1024}) = 256(\frac{1}{256}) = 1$.

(D) For a binomial distribution $X \sim B(n, p)$,the probability mass function is given by $P(X=k) = \binom{n}{k} p^k q^{n-k}$,where $q = 1-p$.
Given $n=7$ and $P(X=3) = P(X=4)$:
$\binom{7}{3} p^3 q^4 = \binom{7}{4} p^4 q^3$
Since $\binom{7}{3} = \binom{7}{4} = 35$,we have $35 p^3 q^4 = 35 p^4 q^3$.
Dividing both sides by $35 p^3 q^3$ (assuming $p, q \neq 0$),we get $q = p$.
Since $p+q=1$,we have $p = q = \frac{1}{2}$.
Now,we calculate $P(X=5)$:
$P(X=5) = \binom{7}{5} (\frac{1}{2})^5 (\frac{1}{2})^{7-5} = \binom{7}{5} (\frac{1}{2})^7$.
$\binom{7}{5} = \binom{7}{2} = \frac{7 \times 6}{2} = 21$.
Thus,$P(X=5) = 21 \times \frac{1}{2^7} = \frac{21}{2^7}$.

(D) For a binomial distribution,the mean is given by $\mu = np = 4$ ... $(i)$
The variance is given by $\sigma^2 = npq = 2$ ... (ii)
Dividing equation (ii) by equation $(i)$,we get:
$\frac{npq}{np} = \frac{2}{4} \implies q = \frac{1}{2}$
Since $p + q = 1$,we have $p = 1 - \frac{1}{2} = \frac{1}{2}$
Substituting $p = \frac{1}{2}$ in equation $(i)$:
$n \times \frac{1}{2} = 4 \implies n = 8$
The probability of $X$ successes in a binomial distribution is given by $P(X=k) = { }^n C_k p^k q^{n-k}$
For $k = 2$:
$P(X=2) = { }^8 C_2 \left(\frac{1}{2}\right)^2 \left(\frac{1}{2}\right)^6 = 28 \times \left(\frac{1}{2}\right)^8 = \frac{28}{256} = \frac{7}{64}$

(A) Given,probability of failure $q = 0.8$,so probability of success $p = 1 - 0.8 = 0.2$. Number of trials $n = 3$.
Using the binomial distribution formula $P(X = r) = { }^n C_r p^r q^{n-r}$:
We need to find the probability that the event happens at most once,which is $P(X \leq 1) = P(X = 0) + P(X = 1)$.
$P(X = 0) = { }^3 C_0 (0.2)^0 (0.8)^3 = 1 \times 1 \times 0.512 = 0.512$.
$P(X = 1) = { }^3 C_1 (0.2)^1 (0.8)^2 = 3 \times 0.2 \times 0.64 = 0.384$.
Therefore,$P(X \leq 1) = 0.512 + 0.384 = 0.896$.

(C) Given,mean of binomial variable,$np = 8$ and variance of binomial variable,$npq = 4$.
$\therefore q = \frac{npq}{np} = \frac{4}{8} = \frac{1}{2}$.
Since $p = 1 - q$,we have $p = 1 - \frac{1}{2} = \frac{1}{2}$.
Substituting $p = \frac{1}{2}$ in $np = 8$,we get $n \times \frac{1}{2} = 8$,so $n = 16$.
We need to find $P(X < 3) = P(X=0) + P(X=1) + P(X=2)$.
Using the formula $P(X=k) = {}^{n}C_{k} p^{k} q^{n-k}$:
$P(X=0) = {}^{16}C_{0} (\frac{1}{2})^{0} (\frac{1}{2})^{16} = 1 \times \frac{1}{2^{16}} = \frac{1}{2^{16}}$.
$P(X=1) = {}^{16}C_{1} (\frac{1}{2})^{1} (\frac{1}{2})^{15} = 16 \times \frac{1}{2^{16}} = \frac{16}{2^{16}}$.
$P(X=2) = {}^{16}C_{2} (\frac{1}{2})^{2} (\frac{1}{2})^{14} = \frac{16 \times 15}{2} \times \frac{1}{2^{16}} = 120 \times \frac{1}{2^{16}} = \frac{120}{2^{16}}$.
Therefore,$P(X < 3) = \frac{1 + 16 + 120}{2^{16}} = \frac{137}{2^{16}}$.

(A) Given that $X$ follows a binomial distribution,the probability mass function is $P(X=r) = { }^n C_r p^r q^{n-r}$,where $q = 1-p$.
The ratio is given by:
$\frac{P(X=r)}{P(X=n-r)} = \frac{{ }^n C_r p^r q^{n-r}}{{ }^n C_{n-r} p^{n-r} q^r}$
Since ${ }^n C_r = { }^n C_{n-r}$,the expression simplifies to:
$\frac{P(X=r)}{P(X=n-r)} = \frac{p^r q^{n-r}}{p^{n-r} q^r} = \left(\frac{p}{q}\right)^r \left(\frac{q}{p}\right)^{n-r} = \left(\frac{q}{p}\right)^{n-2r}$
For this expression to be independent of $n$,the base of the exponent involving $n$ must be $1$.
Thus,$\frac{q}{p} = 1$,which implies $q = p$.
Since $p + q = 1$,we have $p + p = 1$,which gives $2p = 1$.
Therefore,$p = \frac{1}{2}$.

(A) Let $X$ be the number of tails obtained in $100$ tosses of a fair coin. $X$ follows a binomial distribution $B(n, p)$ where $n = 100$ and $p = \frac{1}{2}$.
The probability of getting tails $k$ times is given by $P(X=k) = {}^{100}C_k (\frac{1}{2})^k (\frac{1}{2})^{100-k} = {}^{100}C_k (\frac{1}{2})^{100}$.
We need to find the probability of getting an odd number of tails,which is $P(X=1) + P(X=3) + \dots + P(X=99)$.
This sum is equal to $(\frac{1}{2})^{100} \times ({}^{100}C_1 + {}^{100}C_3 + \dots + {}^{100}C_{99})$.
We know that the sum of odd-indexed binomial coefficients ${}^{n}C_1 + {}^{n}C_3 + \dots = 2^{n-1}$.
For $n=100$,the sum is $2^{100-1} = 2^{99}$.
Therefore,the required probability is $(\frac{1}{2})^{100} \times 2^{99} = \frac{2^{99}}{2^{100}} = \frac{1}{2}$.