Binomial distribution Questions in English

(A) When two coins are tossed $5$ times,the total number of tosses is $2 \times 5 = 10$ individual coin tosses.
Let $n = 10$ be the total number of trials.
The probability of getting a head in a single toss is $p = \frac{1}{2}$.
The probability of getting a tail in a single toss is $q = 1 - p = \frac{1}{2}$.
We want the probability of getting $5$ heads and $5$ tails in $10$ tosses.
Using the binomial distribution formula $P(X = r) = ^nC_r p^r q^{n-r}$,where $r = 5$:
$P(X = 5) = ^{10}C_5 \left(\frac{1}{2}\right)^5 \left(\frac{1}{2}\right)^5$
$P(X = 5) = \frac{10 \times 9 \times 8 \times 7 \times 6}{5 \times 4 \times 3 \times 2 \times 1} \times \left(\frac{1}{2}\right)^{10}$
$P(X = 5) = 252 \times \frac{1}{1024} = \frac{252}{1024} = \frac{63}{256}$.

(C) Given that the probability of success $p = \frac{1}{4}$.
Since $q = 1 - p$,we have $q = 1 - \frac{1}{4} = \frac{3}{4}$.
The standard deviation $\sigma = 3$,so the variance $\sigma^2 = 9$.
For a binomial distribution,the variance is given by $npq = 9$.
Substituting the values of $p$ and $q$: $n \times \frac{1}{4} \times \frac{3}{4} = 9$.
$n \times \frac{3}{16} = 9 \Rightarrow n = 9 \times \frac{16}{3} = 48$.
The mean of the binomial distribution is $\mu = np$.
Therefore,$\mu = 48 \times \frac{1}{4} = 12$.

(B) For a fair coin,the probability of getting a head in a single toss is $p = \frac{1}{2}$ and the probability of getting a tail is $q = 1 - p = \frac{1}{2}$.
Since the coin is tossed $n = 10$ times,this follows a binomial distribution $B(n, p) = B(10, \frac{1}{2})$.
The probability of getting exactly $k = 6$ heads is given by the formula $P(X = k) = {}^{n}C_k \cdot p^k \cdot q^{n-k}$.
Substituting the values: $P(X = 6) = {}^{10}C_6 \cdot (\frac{1}{2})^6 \cdot (\frac{1}{2})^{10-6}$.
$P(X = 6) = {}^{10}C_6 \cdot (\frac{1}{2})^{10}$.
Calculating ${}^{10}C_6 = \frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1} = 210$.
$P(X = 6) = 210 \times \frac{1}{1024} = \frac{210}{1024} = \frac{105}{512}$.

(A) When a dice is thrown once,the probability of getting a $4$ is $p = \frac{1}{6}$.
The probability of not getting a $4$ is $q = 1 - p = 1 - \frac{1}{6} = \frac{5}{6}$.
For two throws,the probability of getting $4$ at least once is given by $1 - P(\text{getting no } 4 \text{ in two throws})$.
The probability of not getting a $4$ in two throws is $q^2 = (\frac{5}{6})^2 = \frac{25}{36}$.
Therefore,the required probability is $1 - \frac{25}{36} = \frac{36 - 25}{36} = \frac{11}{36}$.

(A) For a binomial distribution,the mean is given by $np = 4$ and the variance is given by $npq = 2$.
Dividing the variance by the mean,we get $\frac{npq}{np} = \frac{2}{4}$,which simplifies to $q = \frac{1}{2}$.
Since $p + q = 1$,we have $p = 1 - \frac{1}{2} = \frac{1}{2}$.
Substituting $p = \frac{1}{2}$ into $np = 4$,we get $n(\frac{1}{2}) = 4$,so $n = 8$.
The probability mass function for a binomial distribution is $P(X = k) = \binom{n}{k} p^k q^{n-k}$.
For $X = 1$,we have $P(X = 1) = \binom{8}{1} (\frac{1}{2})^1 (\frac{1}{2})^{8-1} = 8 \times (\frac{1}{2})^8 = 8 \times \frac{1}{256} = \frac{1}{32}$.

(B) Let $X$ be the number of heads obtained in $n$ tosses of a coin.
$X$ follows a binomial distribution with parameters $n$ and $p = 1/2$.
The probability of getting at least one head is given by $P(X \ge 1) > 0.8$.
We know that $P(X \ge 1) = 1 - P(X = 0)$.
So,$1 - P(X = 0) > 0.8$,which implies $P(X = 0) < 0.2$.
The probability of getting zero heads in $n$ tosses is $P(X = 0) = ^nC_0 \times (1/2)^n = (1/2)^n$.
Thus,$(1/2)^n < 0.2$,which means $1/2^n < 1/5$.
This simplifies to $2^n > 5$.
For $n = 1$,$2^1 = 2 < 5$.
For $n = 2$,$2^2 = 4 < 5$.
For $n = 3$,$2^3 = 8 > 5$.
Therefore,the least integer value of $n$ is $3$.

(A) Let $p$ be the probability of getting a tail,so $p = \frac{1}{2}$.
Let $q$ be the probability of getting a head,so $q = \frac{1}{2}$.
Here,$n = 100$ trials are performed.
The probability of getting a tail $k$ times is given by the binomial distribution: $P(X = k) = ^{100}C_k p^k q^{100-k}$.
We need the probability of getting an odd number of tails,which is $P(X = 1) + P(X = 3) + \dots + P(X = 99)$.
Using the binomial expansion, we know that $(q + p)^n = \sum_{k=0}^{n} {^{n}C_k} p^k q^{n-k}$ and $(q - p)^n = \sum_{k=0}^{n} {^{n}C_k} (-p)^k q^{n-k}$.
Subtracting these two equations: $(q + p)^n - (q - p)^n = 2 \times [{^{n}C_1} p^1 q^{n-1} + {^{n}C_3} p^3 q^{n-3} + \dots]$.
Since $p = q = \frac{1}{2}$,we have $q + p = 1$ and $q - p = 0$.
Therefore,the sum of probabilities for odd $k$ is $\frac{(q + p)^n - (q - p)^n}{2} = \frac{1^n - 0^n}{2} = \frac{1}{2}$.

(D) Let $p$ be the probability of getting an odd number in a single throw of a dice.
The possible outcomes on a dice are $\{1, 2, 3, 4, 5, 6\}$.
The odd numbers are $\{1, 3, 5\}$.
So,$p = \frac{3}{6} = \frac{1}{2}$.
The probability of failure $q = 1 - p = 1 - \frac{1}{2} = \frac{1}{2}$.
Since the dice is thrown $n = 2$ times,we use the binomial distribution formula $P(X = k) = {}^nC_k p^k q^{n-k}$.
For two successes,$k = 2$.
$P(X = 2) = {}^2C_2 \left(\frac{1}{2}\right)^2 \left(\frac{1}{2}\right)^{2-2} = 1 \times \frac{1}{4} \times 1 = \frac{1}{4}$.

(A) For a binomial distribution,the mean is given by $\mu = np = 4$ and the variance is given by $\sigma^2 = npq = 2$.
Dividing the variance by the mean,we get $\frac{npq}{np} = \frac{2}{4}$,which implies $q = \frac{1}{2}$.
Since $p + q = 1$,we have $p = 1 - \frac{1}{2} = \frac{1}{2}$.
Substituting $p = \frac{1}{2}$ into $np = 4$,we get $n(\frac{1}{2}) = 4$,so $n = 8$.
The probability of $X$ successes is given by $P(X = k) = \binom{n}{k} p^k q^{n-k}$.
For $k = 2$,we have $P(X = 2) = \binom{8}{2} (\frac{1}{2})^2 (\frac{1}{2})^{8-2} = \binom{8}{2} (\frac{1}{2})^8$.
Calculating the combination,$\binom{8}{2} = \frac{8 \times 7}{2 \times 1} = 28$.
Thus,$P(X = 2) = 28 \times \frac{1}{256} = \frac{28}{256}$.

(B) The probability mass function for a binomial distribution is given by $P(X = k) = ^nC_k p^k q^{n-k}$.
We need to evaluate the ratio $\frac{P(X = k)}{P(X = k - 1)}$.
$\frac{P(X = k)}{P(X = k - 1)} = \frac{^nC_k p^k q^{n-k}}{^nC_{k-1} p^{k-1} q^{n-k+1}}$
Using the formula for combinations,$^nC_k = \frac{n!}{k!(n-k)!}$,we have:
$\frac{^nC_k}{^nC_{k-1}} = \frac{n!}{k!(n-k)!} \cdot \frac{(k-1)!(n-k+1)!}{n!} = \frac{n-k+1}{k}$
Substituting this back into the ratio:
$\frac{P(X = k)}{P(X = k - 1)} = \left( \frac{n-k+1}{k} \right) \cdot \frac{p^k q^{n-k}}{p^{k-1} q^{n-k+1}} = \frac{n-k+1}{k} \cdot \frac{p}{q}$.

(C) Let $X$ be the random variable representing the number of aces obtained in $2$ draws.
Since the cards are drawn with replacement,this follows a Binomial distribution $B(n, p)$ where $n = 2$ and $p = \frac{4}{52} = \frac{1}{13}$.
The probability of success (getting an ace) is $p = \frac{1}{13}$ and the probability of failure is $q = 1 - p = \frac{12}{13}$.
The mean of a Binomial distribution is given by $E(X) = np$.
Therefore,$E(X) = 2 \times \frac{1}{13} = \frac{2}{13}$.

(A) Let $p$ be the probability of hitting the target in one shot,so $p = \frac{10}{100} = 0.1$.
Let $q$ be the probability of missing the target in one shot,so $q = 1 - p = 0.9$.
Let $n$ be the number of rounds fired.
The probability of hitting the target at least once in $n$ shots is given by $P(\text{at least one hit}) = 1 - P(\text{no hits}) = 1 - q^n$.
We want this probability to be at least $50\%$,so $1 - (0.9)^n \ge 0.5$.
This simplifies to $(0.9)^n \le 0.5$.
Taking the logarithm on both sides: $n \log(0.9) \le \log(0.5)$.
Since $\log(0.9)$ is negative,the inequality sign flips: $n \ge \frac{\log(0.5)}{\log(0.9)}$.
Using $\log(0.5) \approx -0.3010$ and $\log(0.9) \approx -0.04576$,we get $n \ge \frac{-0.3010}{-0.04576} \approx 6.57$.
Since $n$ must be an integer,the minimum number of rounds is $7$.

(A) For a binomial distribution,the mean is given by $\mu = np = 3$ and the standard deviation is $\sigma = \sqrt{npq} = \frac{3}{2}$.
Squaring the standard deviation,we get $npq = \frac{9}{4}$.
Dividing $npq$ by $np$,we find $q = \frac{npq}{np} = \frac{9/4}{3} = \frac{3}{4}$.
Since $p + q = 1$,we have $p = 1 - q = 1 - \frac{3}{4} = \frac{1}{4}$.
Now,substituting $p = \frac{1}{4}$ into $np = 3$,we get $n(\frac{1}{4}) = 3$,which implies $n = 12$.
The binomial distribution is given by $(q + p)^n = (\frac{3}{4} + \frac{1}{4})^{12}$.

(A) Let $X$ be the number of times $1, 3$ or $4$ occur on the die.
Then $X$ follows a binomial distribution with parameters $N = 2n + 1$ and $p = \frac{3}{6} = \frac{1}{2}$.
We have $q = 1 - p = \frac{1}{2}$.
The probability of getting $1, 3$ or $4$ at most $n$ times is $P(X \le n) = \sum_{k=0}^{n} {}^{2n+1}C_k p^k q^{2n+1-k}$.
Since $p = q = \frac{1}{2}$,this becomes $P(X \le n) = \sum_{k=0}^{n} {}^{2n+1}C_k (\frac{1}{2})^{2n+1}$.
Let $S = \sum_{k=0}^{n} {}^{2n+1}C_k$. We know that $\sum_{k=0}^{2n+1} {}^{2n+1}C_k = 2^{2n+1}$.
Since ${}^{2n+1}C_k = {}^{2n+1}C_{2n+1-k}$,we have $2S = \sum_{k=0}^{n} {}^{2n+1}C_k + \sum_{k=n+1}^{2n+1} {}^{2n+1}C_k = 2^{2n+1}$.
Thus,$S = 2^{2n}$.
Therefore,the required probability is $S \times (\frac{1}{2})^{2n+1} = 2^{2n} \times \frac{1}{2^{2n+1}} = \frac{1}{2}$.

(C) The probability of drawing a white ball in a single draw is $p = \frac{12}{24} = \frac{1}{2}$.
For the $4^{th}$ white ball to be drawn on the $7^{th}$ draw,there must be exactly $3$ white balls in the first $6$ draws,and the $7^{th}$ draw must be a white ball.
This follows a binomial distribution scenario.
The probability is given by: $P = \binom{6}{3} \times p^3 \times (1-p)^3 \times p$.
Substituting $p = \frac{1}{2}$:
$P = 20 \times (\frac{1}{2})^3 \times (\frac{1}{2})^3 \times \frac{1}{2} = 20 \times (\frac{1}{2})^7 = \frac{20}{128} = \frac{5}{32}$.

(B) The probability of getting $k$ heads in $n$ tosses of a fair coin is given by the binomial distribution formula: $P(X = k) = \binom{n}{k} (\frac{1}{2})^n$.
Given that $P(X = 6) = P(X = 8)$,we have:
$\binom{n}{6} (\frac{1}{2})^n = \binom{n}{8} (\frac{1}{2})^n$.
This simplifies to $\binom{n}{6} = \binom{n}{8}$.
Using the property $\binom{n}{r} = \binom{n}{n-r}$,we know that if $\binom{n}{a} = \binom{n}{b}$,then either $a = b$ or $a + b = n$.
Since $6 \neq 8$,we must have $6 + 8 = n$.
Therefore,$n = 14$.

(A) When two coins are tossed $5$ times,the total number of tosses is $2 \times 5 = 10$.
The total number of possible outcomes is $2^{10} = 1024$.
We want to find the probability of getting exactly $5$ heads and $5$ tails in $10$ tosses.
The number of ways to choose $5$ positions for heads out of $10$ total positions is given by the combination formula $\binom{n}{r} = \frac{n!}{r!(n-r)!}$.
Here,$n = 10$ and $r = 5$,so the number of favorable outcomes is $\binom{10}{5} = \frac{10!}{5! \times 5!} = \frac{10 \times 9 \times 8 \times 7 \times 6}{5 \times 4 \times 3 \times 2 \times 1} = 252$.
The probability is $\frac{\text{favorable outcomes}}{\text{total outcomes}} = \frac{252}{1024}$.
Simplifying the fraction by dividing both numerator and denominator by $4$,we get $\frac{63}{256}$.

(C) The probability of at least one success is given by $P(X \geq 1) = 1 - P(X = 0)$.
For a binomial distribution,$P(X = 0) = {}^nC_0 p^0 q^n = q^n$,where $q = 1 - p = 1 - 1/4 = 3/4$.
We are given $P(X \geq 1) \geq 9/10$,which implies $1 - (3/4)^n \geq 9/10$.
Rearranging the inequality,we get $1 - 9/10 \geq (3/4)^n$,so $(3/4)^n \leq 1/10$.
Taking the logarithm with base $10$ on both sides,we get $n \log_{10}(3/4) \leq \log_{10}(1/10)$.
Since $\log_{10}(1/10) = -1$,we have $n(\log_{10} 3 - \log_{10} 4) \leq -1$.
Multiplying by $-1$ reverses the inequality sign: $n(\log_{10} 4 - \log_{10} 3) \geq 1$.
Thus,$n \geq \frac{1}{\log_{10} 4 - \log_{10} 3}$.

(D) For a binomial distribution,the mean is given by $\mu = np = 6$ and the variance is given by $\sigma^2 = npq = 2$.
Dividing the variance by the mean,we get $\frac{npq}{np} = \frac{2}{6}$,which simplifies to $q = \frac{1}{3}$.
Since $p + q = 1$,we have $p = 1 - \frac{1}{3} = \frac{2}{3}$.
Substituting $p = \frac{2}{3}$ into $np = 6$,we get $n(\frac{2}{3}) = 6$,which implies $n = 6 \times \frac{3}{2} = 9$.
The binomial distribution is given by $(q + p)^n$,which is $(\frac{1}{3} + \frac{2}{3})^9$.

(D) For a binomial distribution,the mean is given by $E(X) = np = 2$ and the variance is given by $Var(X) = npq = 1$.
Dividing the variance by the mean,we get $q = \frac{npq}{np} = \frac{1}{2}$.
Since $p + q = 1$,we have $p = 1 - \frac{1}{2} = \frac{1}{2}$.
Substituting $p = \frac{1}{2}$ into $np = 2$,we get $n(\frac{1}{2}) = 2$,which implies $n = 4$.
The probability $P(X \geq 1)$ is given by $1 - P(X = 0)$.
Using the binomial probability formula $P(X = k) = \binom{n}{k} p^k q^{n-k}$,we have $P(X = 0) = \binom{4}{0} (\frac{1}{2})^0 (\frac{1}{2})^4 = 1 \times 1 \times \frac{1}{16} = \frac{1}{16}$.
Therefore,$P(X \geq 1) = 1 - \frac{1}{16} = \frac{15}{16}$.

(A) Let $p$ be the probability that a person is cured by the medicine. Given $p = 75\% = 0.75 = 3/4$.
Let $n = 3$ be the number of people.
Since the events are independent,the probability that all three people are cured is given by $P(X = 3) = p^3$.
$P(X = 3) = (3/4)^3 = 27/64$.
Therefore,the correct option is $A$.

(D) Let $n$ be the number of trials,where $n = 6$.
Let $X$ be the random variable representing the number of times a $4$ is obtained.
Since we are throwing a pair of dice $6$ times,the maximum number of times we can get a $4$ is $6$.
We are asked to find the probability of getting a $4$ exactly $7$ times,i.e.,$P(X = 7)$.
Since the number of trials $n = 6$ is less than the required number of successes $k = 7$,it is impossible to get $7$ successes in $6$ trials.
Therefore,$P(X = 7) = 0$.
Since $0$ is not among the given options,the correct answer is $D$ (None of these).

(C) The total number of outcomes when a coin is tossed $2n$ times is $2^{2n} = 4^n$.
Let $A$ be the event that the number of heads and tails are not equal.
Let $A'$ be the event that the number of heads and tails are equal.
For $A'$,we must have exactly $n$ heads and $n$ tails.
The number of ways to arrange $n$ heads and $n$ tails in $2n$ tosses is given by the binomial coefficient $\binom{2n}{n} = \frac{(2n)!}{n!n!} = \frac{(2n)!}{(n!)^2}$.
The probability of event $A'$ is $P(A') = \frac{\binom{2n}{n}}{2^{2n}} = \frac{(2n)!}{(n!)^2 \times 4^n}$.
The probability of event $A$ is $P(A) = 1 - P(A') = 1 - \frac{(2n)!}{(n!)^2 \times 4^n}$.

(B) Let $X$ be the number of heads in $10$ tosses of a coin.
This follows a binomial distribution with parameters $n = 10$,$p = 1/2$,and $q = 1 - p = 1/2$.
The probability of getting exactly $x$ heads is given by $P(X = x) = {}^{n}C_{x} p^{x} q^{n-x}$.
For $x = 6$:
$P(X = 6) = {}^{10}C_{6} \left(\frac{1}{2}\right)^{6} \left(\frac{1}{2}\right)^{10-6}$
$P(X = 6) = {}^{10}C_{6} \left(\frac{1}{2}\right)^{10}$
Since ${}^{10}C_{6} = \frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1} = 210$,
$P(X = 6) = \frac{210}{2^{10}} = \frac{210}{1024} = \frac{105}{512}$.

(C) Step $- 1$: Identify the parameters for the binomial distribution.
Each question has $3$ choices with $1$ correct answer.
Probability of success $p = \frac{1}{3}$.
Probability of failure $q = 1 - p = \frac{2}{3}$.
Number of trials $n = 5$.
Step $- 2$: Calculate the probability for $4$ or more correct answers.
We need to find $P(X \ge 4) = P(X = 4) + P(X = 5)$.
Using the binomial formula $P(X = k) = ^nC_k \cdot p^k \cdot q^{n-k}$:
For $k = 4$:
$P(X = 4) = ^5C_4 \cdot (\frac{1}{3})^4 \cdot (\frac{2}{3})^1 = 5 \cdot \frac{1}{81} \cdot \frac{2}{3} = \frac{10}{3^5}$.
For $k = 5$:
$P(X = 5) = ^5C_5 \cdot (\frac{1}{3})^5 \cdot (\frac{2}{3})^0 = 1 \cdot \frac{1}{3^5} \cdot 1 = \frac{1}{3^5}$.
Step $- 3$: Sum the probabilities.
$P(X \ge 4) = \frac{10}{3^5} + \frac{1}{3^5} = \frac{11}{3^5}$.
Thus,the probability is $\frac{11}{3^5}$.

(C) Let $p$ be the probability of hitting the bird,so $p = 3/4$.
Let $q$ be the probability of not hitting the bird,so $q = 1 - p = 1 - 3/4 = 1/4$.
The person tries $n = 5$ times.
The probability of not hitting the bird in any of the $5$ attempts is given by $q^n$.
Therefore,the required probability is $(1/4)^5 = 1/1024$.

(A) Let $X$ be the number of successes in $5$ independent Bernoulli trials. $X$ follows a binomial distribution $B(n, p)$ with $n = 5$.
The probability of success is $p$ and the probability of failure is $q = 1 - p$.
The probability of at least one failure is $P(X < 5) = 1 - P(X = 5)$.
Since $P(X = 5) = p^5$,the probability of at least one failure is $1 - p^5$.
We are given that $1 - p^5 \ge 31/32$.
Rearranging the inequality: $1 - 31/32 \ge p^5$,which simplifies to $1/32 \ge p^5$.
Taking the fifth root on both sides: $(1/2)^5 \ge p^5$,which implies $p \le 1/2$.
Since $p$ is a probability,$p \ge 0$.
Thus,$p \in [0, 1/2]$.

(D) The probability of getting $x$ heads in $n=8$ tosses is given by the binomial distribution formula: $P(X=x) = {^nC_x} (p)^x (q)^{n-x}$,where $p = 1/2$ and $q = 1/2$.
For at least $6$ heads,we need to calculate $P(X \ge 6) = P(X=6) + P(X=7) + P(X=8)$.
$P(X=6) = {^8C_6} (1/2)^8 = 28 \times (1/256) = 28/256$.
$P(X=7) = {^8C_7} (1/2)^8 = 8 \times (1/256) = 8/256$.
$P(X=8) = {^8C_8} (1/2)^8 = 1 \times (1/256) = 1/256$.
Summing these probabilities: $P(X \ge 6) = (28 + 8 + 1) / 256 = 37/256$.

(D) For a binomial distribution,the mean is given by $np = 4$ and the variance is given by $npq = 2$.
Dividing the variance by the mean,we get $q = \frac{npq}{np} = \frac{2}{4} = \frac{1}{2}$.
Since $p + q = 1$,we have $p = 1 - \frac{1}{2} = \frac{1}{2}$.
Substituting $p = \frac{1}{2}$ into $np = 4$,we get $n \times \frac{1}{2} = 4$,which implies $n = 8$.
The probability mass function for a binomial distribution is $P(X = k) = \binom{n}{k} p^k q^{n-k}$.
For $X = 2$,$P(X = 2) = \binom{8}{2} (\frac{1}{2})^2 (\frac{1}{2})^{8-2} = \binom{8}{2} (\frac{1}{2})^8$.
Calculating the combination,$\binom{8}{2} = \frac{8 \times 7}{2 \times 1} = 28$.
Thus,$P(X = 2) = 28 \times \frac{1}{256} = \frac{28}{256}$.

(D) This is a binomial distribution problem where $n = 7$ (number of trials),$k = 4$ (number of successes),$p = \frac{1}{6}$ (probability of getting a $5$ in a single throw),and $q = 1 - p = \frac{5}{6}$ (probability of not getting a $5$).
Using the binomial probability formula $P(X = k) = ^n{C_k} \cdot p^k \cdot q^{n-k}$:
$P(X = 4) = ^7{C_4} \cdot \left( \frac{1}{6} \right)^4 \cdot \left( \frac{5}{6} \right)^{7-4}$
$P(X = 4) = ^7{C_4} \cdot \left( \frac{1}{6} \right)^4 \cdot \left( \frac{5}{6} \right)^3$
Thus,the correct option is $D$.

(C) Let $X$ be the number of heads obtained in $7$ tosses. Since the coin is fair,the probability of getting a head is $p = 1/2$ and the probability of getting a tail is $q = 1/2$.
The person wins if they get more heads than tails. Since there are $7$ tosses,the person wins if they get $4, 5, 6,$ or $7$ heads.
The probability is given by the sum of binomial probabilities:
$P(X \ge 4) = \sum_{k=4}^{7} \binom{7}{k} (1/2)^k (1/2)^{7-k} = \frac{1}{2^7} \sum_{k=4}^{7} \binom{7}{k}$
We know that $\sum_{k=0}^{7} \binom{7}{k} = 2^7 = 128$.
Also,by symmetry,$\sum_{k=0}^{3} \binom{7}{k} = \sum_{k=4}^{7} \binom{7}{k} = \frac{128}{2} = 64$.
Therefore,the required probability is $\frac{64}{128} = \frac{1}{2}$.

(A) Let $n = 5$ be the number of games and $p = 1/2$ be the probability of winning a single game.
India wins at least three games if they win $3, 4,$ or $5$ games.
Using the binomial probability formula $P(X = k) = {^nC_k} p^k q^{n-k}$,where $q = 1-p = 1/2$:
$P(X \ge 3) = P(X=3) + P(X=4) + P(X=5)$
$P(X \ge 3) = {^5C_3} (1/2)^5 + {^5C_4} (1/2)^5 + {^5C_5} (1/2)^5$
$P(X \ge 3) = (10 + 5 + 1) \times (1/2)^5$
$P(X \ge 3) = 16 \times (1/32) = 1/2$.

(D) When two coins are tossed,the number of heads in one trial can be $0, 1,$ or $2$.
In $5$ trials,the total number of tosses is $10$.
Let $X$ be the total number of heads in $10$ independent tosses of a single coin.
$X$ follows a binomial distribution $B(n, p)$ with $n = 10$ and $p = 1/2$.
We want the probability that $X$ is odd,i.e.,$P(X \in \{1, 3, 5, 7, 9\})$.
$P(X \text{ is odd}) = \sum_{k \in \{1, 3, 5, 7, 9\}} \binom{10}{k} (\frac{1}{2})^k (\frac{1}{2})^{10-k} = (\frac{1}{2})^{10} \sum_{k \in \{1, 3, 5, 7, 9\}} \binom{10}{k}$.
Using the property of binomial coefficients,$\sum_{k \text{ odd}} \binom{n}{k} = 2^{n-1}$.
Here,$n = 10$,so the sum is $2^{10-1} = 2^9$.
Therefore,$P(X \text{ is odd}) = \frac{2^9}{2^{10}} = \frac{1}{2}$.

(A) Let $X$ be the outcome of a single toss of the die.
We are given that the value must be $\ge 2$ and $\le 5$.
The favorable outcomes are $\{2, 3, 4, 5\}$.
The number of favorable outcomes is $4$.
The total number of outcomes on a die is $6$.
The probability of success in a single trial is $p = \frac{4}{6} = \frac{2}{3}$.
The probability of failure in a single trial is $q = 1 - p = 1 - \frac{2}{3} = \frac{1}{3}$.
Since the die is tossed $4$ times,we use the binomial distribution formula $P(X = k) = \binom{n}{k} p^k q^{n-k}$ where $n=4$ and $k=4$.
The required probability is $P(X = 4) = \binom{4}{4} \left(\frac{2}{3}\right)^4 \left(\frac{1}{3}\right)^0 = 1 \times \frac{16}{81} \times 1 = \frac{16}{81}$.

(D) Let the coin be tossed $n$ times. Let $X$ be the number of heads obtained.
$X$ follows a binomial distribution with parameters $n$ and $p = 1/2$.
We want $P(X \geqslant 1) \geqslant 0.8$.
This is equivalent to $1 - P(X = 0) \geqslant 0.8$.
$1 - (1/2)^n \geqslant 0.8$.
$(1/2)^n \leqslant 0.2$.
$(1/2)^n \leqslant 1/5$.
$2^n \geqslant 5$.
For $n = 1$,$2^1 = 2 < 5$.
For $n = 2$,$2^2 = 4 < 5$.
For $n = 3$,$2^3 = 8 \geqslant 5$.
Thus,the minimum value of $n$ is $3$.

(D) Given the probability of success $p = 3/4$,the probability of failure $q = 1 - 3/4 = 1/4$,and the number of trials $n = 5$.
We need to find the probability of at least $3$ successes,which is $P(X \ge 3) = P(X=3) + P(X=4) + P(X=5)$.
Using the binomial distribution formula $P(X=k) = \binom{n}{k} p^k q^{n-k}$:
$P(X=3) = \binom{5}{3} (3/4)^3 (1/4)^2 = 10 \times (27/64) \times (1/16) = 270/1024$.
$P(X=4) = \binom{5}{4} (3/4)^4 (1/4)^1 = 5 \times (81/256) \times (1/4) = 405/1024$.
$P(X=5) = \binom{5}{5} (3/4)^5 (1/4)^0 = 1 \times (243/1024) \times 1 = 243/1024$.
Summing these probabilities: $P(X \ge 3) = (270 + 405 + 243) / 1024 = 918 / 1024$.
Simplifying the fraction: $918 / 1024 = 459 / 512$.

(D) For a fair die,the probability of getting an odd number (success) is $p = \frac{3}{6} = \frac{1}{2}$.
The probability of failure is $q = 1 - p = 1 - \frac{1}{2} = \frac{1}{2}$.
The number of trials is $n = 5$.
The variance of a binomial distribution is given by the formula $\text{Variance} = npq$.
Substituting the values: $\text{Variance} = 5 \times \frac{1}{2} \times \frac{1}{2} = \frac{5}{4}$.

(D) The probability of getting an odd number on a single throw of a die is $p = \frac{3}{6} = \frac{1}{2}$.
The probability of failure is $q = 1 - p = 1 - \frac{1}{2} = \frac{1}{2}$.
Given the number of trials $n = 5$.
For a binomial distribution,the variance is given by the formula $\text{Var}(X) = npq$.
Substituting the values,we get $\text{Var}(X) = 5 \times \frac{1}{2} \times \frac{1}{2} = \frac{5}{4}$.

(A) For a binomial distribution,the mean is given by $\mu = np$ and the variance is given by $\sigma^2 = npq$.
Given that $np = 6$ and $npq = 4$.
Dividing the variance by the mean,we get:
$\frac{npq}{np} = \frac{4}{6}$
$q = \frac{2}{3}$
Since $p + q = 1$,we have $p = 1 - q = 1 - \frac{2}{3} = \frac{1}{3}$.
Substituting the value of $p$ into the mean equation:
$np = 6$
$n \times \frac{1}{3} = 6$
$n = 6 \times 3 = 18$.
Therefore,the value of $n$ is $18$.

(C) The probability of getting the same numbers on both dice in a single throw is $p = \frac{6}{36} = \frac{1}{6}$.
Here,we use the binomial distribution formula $P(X = r) = ^nC_r p^r q^{n-r}$,where $n = 4$,$r = 2$,$p = \frac{1}{6}$,and $q = 1 - p = \frac{5}{6}$.
The required probability is $P(X = 2) = ^4C_2 \left(\frac{1}{6}\right)^2 \left(\frac{5}{6}\right)^{4-2}$.
$P(X = 2) = 6 \times \left(\frac{1}{36}\right) \times \left(\frac{25}{36}\right) = \frac{150}{1296} = \frac{25}{216}$.

(A) The total number of outcomes when a coin is tossed $4$ times is $2^4 = 16$.
Let $E$ be the event of getting at least $1$ tail.
The complement event $E'$ is the event of getting no tails (i.e.,all heads).
The number of outcomes for $E'$ is $1$ (which is $HHHH$).
Therefore,the probability of getting no tails is $P(E') = \frac{1}{16}$.
The probability of getting at least $1$ tail is $P(E) = 1 - P(E') = 1 - \frac{1}{16} = \frac{15}{16}$.

(C) Since $A$ and $B$ are equally skilled,the probability of $A$ winning a single game is $p = 1/2$.
The probability of $A$ winning exactly $k = 3$ games out of $n = 4$ games follows the binomial distribution formula:
$P(X = k) = { }^n C_k p^k (1-p)^{n-k}$
Substituting the values:
$P(X = 3) = { }^4 C_3 \left(\frac{1}{2}\right)^3 \left(1-\frac{1}{2}\right)^{4-3}$
$P(X = 3) = 4 \times \left(\frac{1}{8}\right) \times \left(\frac{1}{2}\right)$
$P(X = 3) = 4 \times \frac{1}{16} = \frac{1}{4}$

(A) Let $n = 5$ be the number of trials and $X$ be the number of times an even number appears.
The probability of getting an even number in a single throw is $p = 3/6 = 1/2$.
The probability of not getting an even number is $q = 1 - p = 1/2$.
Using the binomial distribution formula $P(X = k) = \binom{n}{k} p^k q^{n-k}$,we need to find $P(X = 3)$:
$P(X = 3) = \binom{5}{3} \times (1/2)^3 \times (1/2)^{5-3}$
$P(X = 3) = 10 \times (1/8) \times (1/4)$
$P(X = 3) = 10/32 = 5/16$.

(A) Let $p$ be the probability that a student is a swimmer and $q$ be the probability that a student is not a swimmer.
Given $q = 1/5$,so $p = 1 - 1/5 = 4/5$.
Using the binomial distribution formula $P(X = k) = ^nC_k p^k q^{n-k}$,where $n = 5$ and $k = 4$:
$P(X = 4) = ^5C_4 \left( \frac{4}{5} \right)^4 \left( \frac{1}{5} \right)^{5-4}$
$P(X = 4) = ^5C_4 \left( \frac{4}{5} \right)^4 \left( \frac{1}{5} \right)$

(B) The total number of outcomes when a pair of dice is thrown is $6 \times 6 = 36$.
To get a score of $9$,the possible outcomes are $(3, 6), (4, 5), (5, 4), (6, 3)$.
Thus,the number of favorable outcomes is $4$.
The probability of getting a score of $9$ in a single throw is $p = \frac{4}{36} = \frac{1}{9}$.
The probability of not getting a score of $9$ is $q = 1 - p = 1 - \frac{1}{9} = \frac{8}{9}$.
Using the binomial distribution formula $P(X = k) = nC_k \times p^k \times q^{n-k}$,where $n = 3$ and $k = 2$:
$P(X = 2) = 3C_2 \times (\frac{1}{9})^2 \times (\frac{8}{9})^{3-2}$
$= 3 \times \frac{1}{81} \times \frac{8}{9}$
$= \frac{24}{729} = \frac{8}{243}$.

(A) The probability of at least one success is given by $P(X \geq 1) = 1 - P(X = 0)$.
In a binomial distribution $B(n, p)$,$P(X = 0) = q^n$,where $q = 1 - p$.
Given $p = \frac{1}{4}$,we have $q = 1 - \frac{1}{4} = \frac{3}{4}$.
The condition is $1 - (\frac{3}{4})^n \geq \frac{9}{10}$.
Rearranging the inequality,we get $1 - \frac{9}{10} \geq (\frac{3}{4})^n$,which simplifies to $\frac{1}{10} \geq (\frac{3}{4})^n$.
Taking the logarithm with base $10$ on both sides: $\log_{10}(\frac{1}{10}) \geq \log_{10}((\frac{3}{4})^n)$.
$-1 \geq n(\log_{10} 3 - \log_{10} 4)$.
Multiplying by $-1$ reverses the inequality sign: $1 \leq n(\log_{10} 4 - \log_{10} 3)$.
Therefore,$n \geq \frac{1}{\log_{10} 4 - \log_{10} 3}$.

(B) Let $X$ be the number of successes in $5$ independent Bernoulli trials. The probability of success is $p$ and the probability of failure is $q = 1 - p$.
The probability of at least one failure is given by $1 - P(\text{no failure}) = 1 - P(X = 5)$.
Since the trials are independent,$P(X = 5) = p^5$.
According to the problem,the probability of at least one failure is at least $\frac{31}{32}$:
$1 - p^5 \geq \frac{31}{32}$
Rearranging the inequality:
$1 - \frac{31}{32} \geq p^5$
$\frac{1}{32} \geq p^5$
Taking the fifth root on both sides:
$p \leq (\frac{1}{32})^{1/5}$
$p \leq \frac{1}{2}$
Since $p$ is a probability,$p \geq 0$. Therefore,$p \in [0, \frac{1}{2}]$.

(C) This is a binomial distribution problem where $n = 5$,$p = \frac{1}{3}$ (probability of success),and $q = \frac{2}{3}$ (probability of failure).
We need to find the probability of getting $4$ or more correct answers,which is $P(X \ge 4) = P(X = 4) + P(X = 5)$.
Using the formula $P(X = k) = {^nC_k} \cdot p^k \cdot q^{n-k}$:
$P(X = 4) = {^5C_4} \cdot (\frac{1}{3})^4 \cdot (\frac{2}{3})^1 = 5 \cdot \frac{1}{81} \cdot \frac{2}{3} = \frac{10}{3^5}$.
$P(X = 5) = {^5C_5} \cdot (\frac{1}{3})^5 \cdot (\frac{2}{3})^0 = 1 \cdot \frac{1}{243} \cdot 1 = \frac{1}{3^5}$.
Total probability = $\frac{10}{3^5} + \frac{1}{3^5} = \frac{11}{3^5}$.

(B) The total number of ways to place $12$ identical balls into $3$ distinct boxes is given by the stars and bars formula: $\binom{12+3-1}{3-1} = \binom{14}{2} = 91$. However,in probability problems involving placing balls into boxes,we typically treat the balls as being placed independently into boxes with probability $p = \frac{1}{3}$ for each box.
Let $X$ be the number of balls in a specific box. $X$ follows a binomial distribution $B(n, p)$ where $n = 12$ and $p = \frac{1}{3}$.
The probability that a specific box contains exactly $3$ balls is $P(X = 3) = \binom{12}{3} \left(\frac{1}{3}\right)^3 \left(\frac{2}{3}\right)^9$.
Since there are $3$ boxes,the probability that at least one box contains exactly $3$ balls is calculated using the binomial distribution logic for the selection of boxes.
$P = \binom{12}{3} \left(\frac{1}{3}\right)^3 \left(\frac{2}{3}\right)^9 = \frac{12 \times 11 \times 10}{3 \times 2 \times 1} \times \frac{1}{27} \times \left(\frac{2}{3}\right)^9 = 220 \times \frac{1}{27} \times \left(\frac{2}{3}\right)^9 = \frac{220}{27} \times \left(\frac{2}{3}\right)^9$.
Simplifying,$\frac{220}{27} = \frac{55 \times 4}{27} = \frac{55}{3} \times \frac{4}{9} = \frac{55}{3} \times \left(\frac{2}{3}\right)^2$.
Thus,$P = \frac{55}{3} \times \left(\frac{2}{3}\right)^2 \times \left(\frac{2}{3}\right)^9 = \frac{55}{3} \left(\frac{2}{3}\right)^{11}$.

(B) The problem follows a binomial distribution because the balls are drawn with replacement,making each trial independent.
Here,the total number of trials is $n = 10$.
The probability of drawing a green ball in a single trial is $p = \frac{15}{15 + 10} = \frac{15}{25} = \frac{3}{5}$.
The probability of not drawing a green ball (drawing a yellow ball) is $q = 1 - p = 1 - \frac{3}{5} = \frac{2}{5}$.
The variance of a binomial distribution is given by the formula $\text{Variance} = npq$.
Substituting the values: $\text{Variance} = 10 \times \frac{3}{5} \times \frac{2}{5} = 10 \times \frac{6}{25} = \frac{60}{25} = \frac{12}{5}$.