MCQ 11 Mark
If $X$ is a binomial variate with parameters $n$ and $p$, where $0 < p < 1$ such that $\frac{\text{P(X = r)}}{\text{P(X = n - r})}$ is independent of $n$ and $r,$ then $p$ equals:
- ✓
$\frac{1}{2}$
- B
$\frac{1}{3}$
- C
$\frac{1}{4}$
- D
$\text{None of these}$
AnswerCorrect option: A. $\frac{1}{2}$
$\frac{1}{2}$
View full question & answer→MCQ 21 Mark
In a box containing $100$ bulbs, $10$ are defective. What is the probability that out of a sample of $5$ bulbs, none is defective?
AnswerCorrect option: A. $\big(\frac{9}{10}\big)^5$
$\big(\frac{9}{10}\big)^5$
View full question & answer→MCQ 31 Mark
A five$-$digit number is written down at raddom. The probability that the number is divisible by $5$, and no two consecutive digits are identical, is:
AnswerCorrect option: B. $\frac{1}{5}\big(\frac{9}{10}\big)^3$
$\frac{1}{5}\big(\frac{9}{10}\big)^3$
View full question & answer→MCQ 41 Mark
If the mean and variance of a binomial distribution are $4$ and $3$, respectively, the probability of getting exactly six successes in this distribution is:
- A
$\text{ }^{16}\text{C}_6\big(\frac{1}{4}\big)^{10}\big(\frac{3}{4}\big)^6$
- ✓
$\text{ }^{16}\text{C}_6\big(\frac{1}{4}\big)^{6}\big(\frac{3}{4}\big)^{10}$
- C
$\text{ }^{12}\text{C}_6\big(\frac{1}{20}\big)\big(\frac{3}{4}\big)^6$
- D
$\text{ }^{12}\text{C}_6\big(\frac{1}{20}\big)^6\big(\frac{3}{4}\big)^6$
AnswerCorrect option: B. $\text{ }^{16}\text{C}_6\big(\frac{1}{4}\big)^{6}\big(\frac{3}{4}\big)^{10}$
$\text{ }^{16}\text{C}_6\big(\frac{1}{4}\big)^{6}\big(\frac{3}{4}\big)^{10}$
View full question & answer→MCQ 51 Mark
If the mean and variance of a binomial variate $X$ are $2$ and $1$ respectively, then the probability that $X$ takes a value greater than $1$ is:
- A
$\frac{2}{3}$
- B
$\frac{4}{5}$
- C
$\frac{7}{8}$
- ✓
$\frac{15}{16}$
AnswerCorrect option: D. $\frac{15}{16}$
$\frac{15}{16}$
View full question & answer→MCQ 61 Mark
The least number of times a fair coin must be tossed so that the probability of getting at least one head is at least $0.8,$ is:
View full question & answer→MCQ 71 Mark
A fair coin is tossed $100$ times. The probability of getting tails an odd nimber of times is:
- ✓
$\frac{1}{2}$
- B
$\frac{1}{8}$
- C
$\frac{3}{8}$
- D
$\text{None of these}$
AnswerCorrect option: A. $\frac{1}{2}$
$\frac{1}{2}$
View full question & answer→MCQ 81 Mark
A coin is tossed $4$ times. The probability that at least one head turns up is:
- A
$\frac{1}{16}$
- B
$\frac{2}{16}$
- C
$\frac{14}{16}$
- ✓
$\frac{15}{16}$
AnswerCorrect option: D. $\frac{15}{16}$
$\frac{15}{16}$
View full question & answer→MCQ 91 Mark
Mark the correct alternative in the following question:
Suppose a random variable $X$ follows the binomial distribution with parameters $n$ and $p$, where $0 < p < 1.$ If $\frac{\text{P(X = r})}{\text{P(X = n} -\text{r})}$ is independent of $n$ and $r$, then $p$ equals:
- ✓
$\frac{1}{2}$
- B
$\frac{1}{3}$
- C
$\frac{1}{5}$
- D
$\frac{1}{7}$
AnswerCorrect option: A. $\frac{1}{2}$
$\frac{1}{2}$
View full question & answer→MCQ 101 Mark
If $X$ follows a binomial distribution with parameter $\text{n}=100$ and $\text{p}=\frac{1}{3},$ then $P(X = r)$ is maximum when $r =$
View full question & answer→MCQ 111 Mark
A rifleman is firing at a distant target and has only $10\%$ chance of hiting it. the least number of round he must fire in order to have more than $50\%$ chance of hitting it at least once is:
View full question & answer→MCQ 121 Mark
A coin is tossed $n$ times. The probability of geting at least once is greater than $0.8$. Then, the least value of $n$, is:
View full question & answer→MCQ 131 Mark
One hundred idential coins, each with probability $p$ of showing heads are tossed once. If $0 < p < 1$ and the probability of heads showing on $50$ coins is equal to that of heads showing on $51$ coins, the value of $p$ is:
- A
$\frac{1}{2}$
- ✓
$\frac{51}{101}$
- C
$\frac{49}{101}$
- D
$\text{None of these}$
AnswerCorrect option: B. $\frac{51}{101}$
$\frac{51}{101}$
View full question & answer→MCQ 141 Mark
Mark the correct alternative in the following question : Which one is not a requirement of a binomial dstribution?
AnswerCorrect option: C. The outcomes must be dependent on each other.
The outcomes must be dependent on each other.
View full question & answer→MCQ 151 Mark
A box contain $100$ pens of which $10$ are defective. What is the probability that out of a sample of $5$ pens draws one by one with replacement at most one is defective?
- A
$\big(\frac{9}{10}\big)^5$
- B
$\frac{1}{2}\big(\frac{9}{10}\big)^4$
- C
$\frac{1}{2}\big(\frac{9}{10}\big)^5$
- ✓
$\big(\frac{9}{10}\big)^5+\frac{1}{2}\big(\frac{9}{10}\big)^4$
AnswerCorrect option: D. $\big(\frac{9}{10}\big)^5+\frac{1}{2}\big(\frac{9}{10}\big)^4$
$\big(\frac{9}{10}\big)^5+\frac{1}{2}\big(\frac{9}{10}\big)^4$
View full question & answer→MCQ 161 Mark
A biased coin with probabilty $p, 0 < p < 1$, of heads is tossed until a head appears for the first time. If the probability that the number of tosses required is even is $\frac{2}{5},$ then $p$ equals:
- ✓
$\frac{1}{3}$
- B
$\frac{2}{3}$
- C
$\frac{2}{5}$
- D
$\frac{3}{5}$
AnswerCorrect option: A. $\frac{1}{3}$
$\frac{1}{3}$
View full question & answer→MCQ 171 Mark
Fifteen coupons are numbered $1$ to $15.$ Seven coupons are selected at random one at a time with replacement. The probability that the largest number appearing on a selected coupon is $9$ is:
- A
$\big(\frac{3}{7}\big)^7$
- B
$\big(\frac{1}{15}\big)^7$
- C
$\big(\frac{8}{15}\big)^7$
- ✓
$\text{None of these}$
AnswerCorrect option: D. $\text{None of these}$
$\text{None of these}$
View full question & answer→MCQ 181 Mark
Mark the correct alternative in the following question:The probability of guessing correctly at least $8$ out of $10$ answers of a true false type examination is:
- A
$\frac{7}{64}$
- ✓
$\frac{7}{128}$
- C
$\frac{45}{1024}$
- D
$\frac{7}{41}$
AnswerCorrect option: B. $\frac{7}{128}$
$\frac{7}{128}$
View full question & answer→MCQ 191 Mark
A fair die is tossed eight times. The probability that a third six is observed in the eight throw is:
- ✓
$\frac{\text{ }^7\text{C}_2\times5^5}{6^7}$
- B
$\frac{\text{ }^7\text{C}_2\times5^5}{6^8}$
- C
$\frac{\text{ }^7\text{C}_2\times5^5}{6^6}$
- D
$\text{None of these}$
AnswerCorrect option: A. $\frac{\text{ }^7\text{C}_2\times5^5}{6^7}$
$\frac{\text{ }^7\text{C}_2\times5^5}{6^7}$
View full question & answer→MCQ 201 Mark
If in a binomial distribution $\text{n}=4,\text{P(X}=0)=\frac{16}{81},$ then $\text{P(X}=4)$ equals:
- A
$\frac{1}{16}$
- ✓
$\frac{1}{81}$
- C
$\frac{1}{27}$
- D
$\frac{1}{8}$
AnswerCorrect option: B. $\frac{1}{81}$
$\frac{1}{81}$
View full question & answer→MCQ 211 Mark
Mark the correct alternative in the following question:
The probability that a person is not a swimmer is $0.3$. The probability that out of $5$ persons $4$ are swimmers is:
- ✓
$\text{ }^5\text{C}_4(0.7)^4(0.3)$
- B
$\text{ }^5\text{C}_1(0.7)(0.3)^4$
- C
$\text{ }^5\text{C}_4(0.7)(0.3)^4$
- D
$(0.7)^4(0.3)$
AnswerCorrect option: A. $\text{ }^5\text{C}_4(0.7)^4(0.3)$
$\text{ }^5\text{C}_4(0.7)^4(0.3)$
View full question & answer→MCQ 221 Mark
A fair coin is tossed a fixed number of times. If the probability of getting seven heads is equal to that of getting nine heads, the probability of getting two heads is:
- A
$\frac{15}{2^8}$
- B
$\frac{2}{15}$
- ✓
$\frac{15}{2^{13}}$
- D
$\text{None of these}$
AnswerCorrect option: C. $\frac{15}{2^{13}}$
$\frac{15}{2^{13}}$
View full question & answer→MCQ 231 Mark
In a binomial distribution, the probability of getting success is $\frac{1}{4}$ and standard deviation is $3$. Then, its mean is:
View full question & answer→MCQ 241 Mark
A fair coin is tossed $99$ times. If $X$ is the number of times head appears, then $P(X = r)$ is maximum when $r$ is:
- ✓
$49, 50$
- B
$50, 51$
- C
$51,52$
- D
AnswerCorrect option: A. $49, 50$
$49, 50$
View full question & answer→MCQ 251 Mark
For a binomial variate $X$, if $\text{n}=3$ and $\text{P(X}=1)=8\text{ P(X = 3}),$ then $p =$
- A
$\frac{4}{5}$
- B
$\frac{1}{5}$
- C
$\frac{1}{3}$
- ✓
$\frac{2}{3}$
AnswerCorrect option: D. $\frac{2}{3}$
$\frac{2}{3}$
View full question & answer→MCQ 261 Mark
A fair die is thrown twenty times. The probability that on the tenth throw the fourth six appears is:
- A
$\frac{\text{ }^{20}\text{C}_{10}\times5^6}{6^{20}}$
- B
$\frac{120\times5^7}{6^{10}}$
- ✓
$\frac{84\times5^6}{6^{10}}$
- D
$\text{None of these}$
AnswerCorrect option: C. $\frac{84\times5^6}{6^{10}}$
A fair die is thrown then probebility of getting 6 isp $=\frac{1}{6}.$
$\Rightarrow\text{q}=\frac{5}{6}$
To find probability that on tenth throw $4^{th}$ six appears, in the first nine throw $3$ six should appear.
Required probability $= P(3$ six in first $9$ throw$) \times P($a six in tenth throw$)$
Required probability $=\text{ }^9\text{C}_3\big(\frac{1}{6}\big)^3\big(\frac{5}{6}\big)^6\times\frac{1}{6}$
Required probability $=\frac{84\times5^6}{6^{10}}$
View full question & answer→MCQ 271 Mark
If $X$ follows a binomial distribution with parameter $\text{n}=8$ and $\text{p}=\frac{1}{2},$ then $\text{P(|X}-4|\leq2)$ equals:
- A
$\frac{118}{128}$
- ✓
$\frac{119}{128}$
- C
$\frac{117}{128}$
- D
$\text{None of these}$
AnswerCorrect option: B. $\frac{119}{128}$
$\text{n = 8,}\text{p}=\frac{1}{2}=\text{q}$
$\text{P(|X}-4|)\leq2$
$\Rightarrow-2\leq\text{x}-4\leq2$
$\Rightarrow4-2\leq\text{x}\leq2+4$
$\Rightarrow2\leq\text{x}\leq6$
$\text{P}(2\leq\text{x}\leq6)=\text{P(2)+P(3)+P(4)+P(5)+P(6)}$
$\text{P(2}\leq\text{x}\leq6)=\text{ }^8\text{C}_2\Big(\frac{1}{2^8}\Big)+\text{ }^8\text{C}_3\Big(\frac{1}{2^8}\Big)+\text{ }^8\text{C}_4\Big(\frac{1}{2^8}\Big)\text{ }^8\text{C}_5\Big(\frac{1}{2^8}\Big)+\text{ }^8\text{C}_6\Big(\frac{1}{2^8}\Big)$
$=\frac{119}{128}$
View full question & answer→MCQ 281 Mark
A coin is tossed $10$ times. The probability of getting exactly six heads is:
AnswerCorrect option: B. $\frac{105}{512}$
$\text{n}=10,\text{x}=6,\text{p = q}=\frac{1}{2}$
$\text{P(X}=6)=\text{ }^{10}\text{C}_6\big(\frac{1}{2}\big)^{10}$
$=\frac{105}{512}$
View full question & answer→MCQ 291 Mark
The probablity of selecting a male or a female is same. If the probability that in an office of $n$ persons $(n - 1)$ males being selected is $\frac{3}{2^{10}},$ the value of $n$ is:
Answer$X$ represents number of males.
$\text{p = q}=\frac{1}{2}$
$\text{p(n}-1)=\frac{3}{2^{10}}$
$\text{ }^{\text{n}}\text{C}_{\text{n}-1}\text{p}^{\text{n}-1}\text{q}=\frac{3}{2^{10}}$
$\text{n}\big(\frac{1}{2}\big)^{\text{n}-1}\big(\frac{1}{2}\big)^{\text{n}}=\frac{3}{2^{10}}$
$\text{n}\big(\frac{1}{2}\big)^{\text{n}}=\frac{1}{4}\times\frac{3\times4}{2^{10}}$
$\text{n}\big(\frac{1}{2}\big)^{\text{n}}=12\big(\frac{1}{2}\big)^{12}$
$\Rightarrow\text{n}=12$
View full question & answer→