Question 11 Mark
If A and B are two independent events with $\text{P(A)}=\frac{1}{3}$ and $\text{P(B)}=\frac{1}{4},$ then P(B'|A) is equal to:
Answer
- $\frac{3}{4}$
Solution:
$\text{P(A)}=\frac{1}{3},\text{ P(B)}=\frac{1}{4},\text{ P}\Big(\frac{\text{B}'}{\text{A}}\Big)=?$
$\text{P}(\text{A}\cap\text{B})=\text{P(A)}\times\text{P(B)}$
$\text{P}\Big(\frac{\text{B}'}{\text{A}}\Big)=\text{P}\frac{(\text{A}\cap\text{B}')}{\text{P(A)}}$
$\text{P}(\text{B}')=1-\frac{1}{4}=\frac{3}{4}$
$\therefore\text{P}\Big(\frac{\text{B}'}{\text{A}}\Big)=\frac{\text{P(A)}\times\text{P(B})'}{\text{P(A)}}$
$=\text{P(B}')=\frac{3}{4}$ View full question & answer→Question 21 Mark
The probability distribution of a random variable $X$ is:
| $X$ |
$ 0$ |
$1$ |
$2$ |
$3$ |
$4$ |
| $P(X)$ |
$ 0.1$ |
$ k$ |
$ 2k$ |
$ k$ |
$0.1$ |
where $k$ is some unknown constant.
The probability that the random variable $X$ takes the value $2$ is: Answer$\text {As } \sum_{i=1}^n p_i=1 \Rightarrow 0.1+k+2 k+k+0.1=1$
$\Rightarrow 4 k=1-0.2=0.8 \Rightarrow k=\frac{0.8}{4}=0.2$
$\therefore P(x=2)=2 k=2 \times 0.2=0.4=\frac{4}{10}=\frac{2}{5}$
View full question & answer→Question 31 Mark
Let $E$ be an event of a sample space $S$ of an experiment then $P(S \mid E)=$
Answer$P(S \mid E)=\frac{P(S \cap E)}{P(E)}=\frac{P(E)}{P(E)}=1$
View full question & answer→Question 41 Mark
If $A$ and $B$ are events such that $P(A / B)=P(B / A) \neq 0$, then
AnswerGiven, $P(A / B)=P(B / A)$
$\Rightarrow \frac{P(A \cap B)}{P(B)}=\frac{P(B \cap A)}{P(A)}$
$\Rightarrow P(A)=P(A) \quad[\because P(A \cap B)=P(B \cap A)]$
View full question & answer→Question 51 Mark
A problem in Mathematics is given to three students whose chances of solving it are $\frac{1}{2}, \frac{1}{3}, \frac{1}{4}$ respectively. If the events of their solving the problem are independent then the probability that the problem will be solved, is
AnswerLet $A, B, C$ be the respective events of solving the problem. Then, $P(A)=\frac{1}{2}, P(B)=\frac{1}{3}$ and $P(C)=\frac{1}{4}$. Here, $A, B$, $C$ are independent events. Problem is solved if at least one of them solves the problem.
Required probability is $=P(A \cup B \cup C)=1-P(\bar{A}) P(\bar{B}) P(\bar{C})$
$
=1-\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)=1-\frac{1}{2} \cdot \frac{2}{3} \cdot \frac{3}{4}=1-\frac{1}{4}=\frac{3}{4} \text {. }$
View full question & answer→Question 61 Mark
$M$ and $N$ are two events such that $P(M \cap N)=0$. Which of the following is equal to $P(M \mid(M \cup N))$ ?
Answer$\frac{P(M)}{P(M)+P(N)}$
View full question & answer→Question 71 Mark
For two events $A$ and $B$, if $P(A)=0.4, P(B)=0.8$ and $P(B / A)=0.6$, then $P(A \cup B)$ is
AnswerWe have, $P(A)=0.4, P(B)=0.8$ and $P(B / A)=0.6$
We know that $P(B / A)=\frac{P(A \cap B)}{P(A)}$
$
\Rightarrow 0.6=\frac{P(A \cap B)}{0.4} \Rightarrow P(A \cap B)=0.24
$
Now: $P(A \cup B)=P(A)+P(B)-P(A \cap B)=0.4+0.8-0.24=0.96$
Hence. $P(A \cup B)=0.96$
View full question & answer→Question 81 Mark
If $P\left(\frac{A}{B}\right)=0.3, P(A)=0.4$ and $P(B)=0.8$, then $P\left(\frac{B}{A}\right)$ is equal to:
AnswerWe have, $P\left(\frac{A}{B}\right)=0.3, P(A)=0.4$ and $P(B)=0.8$
Now, $P\left(\frac{A}{B}\right)=\frac{P(A \cap B)}{P(B)}$
$
\Rightarrow \quad 0.3=\frac{P(A \cap B)}{0.8} \Rightarrow P(A \cap B)=0.3 \times 0.8=0.24
$
Now, $P\left(\frac{B}{A}\right)=\frac{P(B \cap A)}{P(A)}=\frac{0.24}{0.4}=0.6$
View full question & answer→Question 91 Mark
A family has 2 children and the elder child is a girl. The probability that both children are girls is:
AnswerLet $G_i(i=1,2)$ and $B_i(i=1,2)$ denote the $i^{\text {th }}$ child is a girl or a boy respectively.
Then sample space, $S=\left\{G_1 G_2, G_1 B_2, B_1 G_2, B_1 B_2\right\}$
Let $A$ be the event that both children are girls,
$B$ be the event that the elder child is a girl.
Thus, $A=\left\{G_1 G_2\right\}$ and $B=\left\{G_1 G_2, G_1 B_2\right\}$
$\Rightarrow A \cap B=\left\{G_1 G_2\right\}$
$\therefore \quad$ Required probability $=P(A / B)=\frac{P(A \cap B)}{P(B)}=\frac{1 / 4}{2 / 4}=\frac{1}{2}$
View full question & answer→Question 101 Mark
If for any two events $A$ and $B$, $P(A)=\frac{4}{5}$ and $P(A \cap B)=\frac{7}{10}$, then $P(B / A)$ is
AnswerWe know that, $P(B / A)=\frac{P(B \cap A)}{P(A)}=\frac{7 / 10}{4 / 5}=\frac{7}{8}$
View full question & answer→Question 111 Mark
Five fair coins are tossed simultaneously. The probability of the events that atleast one head comes up is
AnswerSince each coin turns up on either a head or tail.
$\therefore \quad$ Total possible outcomes $=2^5=32$
Let $A$ be the event that all tails comes up.
$\therefore \quad n(A)=1\{$ i.e., $(T, T, T, T, T)$
So, required probability $=1-P(A)=1-\frac{1}{32}=\frac{31}{32}$
View full question & answer→Question 121 Mark
A card is picked at random from a pack of 52 playing cards. Given that the picked card is a queen, the probability of this card to be a card of spade is
AnswerLet $A$ be the event that the card is a spade and $B$ be the event that the picked card is a queen.
We have a total of 13 spades and 4 queen cards.
Also only one queen is from spade.
$
\therefore \quad P(A \mid B)=\frac{P(A \cap B)}{P(B)}=\frac{\frac{1}{52}}{\frac{4}{52}}=\frac{1}{4}
$
View full question & answer→Question 131 Mark
A die is thrown once. Let $A$ be the event that the number obtained is greater than $3$ . Let $B$ be the event that the number obtained is less than $5$ . Then $P(A \cup B)$ is
AnswerHere, $A=\{4,5,6\}, B=\{1,2,3,4\}$
$A \cap B=\{4\}$
Now, $P(A \cup B)=P(A)+P(B)-P(A \cap B)=\frac{3}{6}+\frac{4}{6}-\frac{1}{6}=1$
View full question & answer→Question 141 Mark
If $A$ and $B$ are two independent events with $P(A)=\frac{1}{3}$ and $P(B)=\frac{1}{4}$, then $P\left(B^{\prime} \mid A\right)$ is equal to
AnswerGiven, $A$ and $B$ are independent events.
Also, $P(A)=\frac{1}{3}$ and $P(B)=\frac{1}{4}$
$\text { Now, } P\left(B^{\prime} \mid A\right)=\frac{P\left(B^{\prime} \cap A\right)}{P(A)}$
$=\frac{P\left(B^{\prime}\right) P(A)}{P(A)} \quad[\because A, B \text { are independent events }]$
$=P\left(B^{\prime}\right)=1-P(B)=1-\frac{1}{4}=\frac{3}{4}$
View full question & answer→Question 151 Mark
Choose the correct answer from the given four options.
The probability that a person is not a swimmer is 0.3. The probability that out of 5 persons 4 are swimmers is:
Answer
- ${^5}\text{C}_4(0.7)^4(0.3)$
Solution:
Here, $\bar{\text{p}}=0.3\Rightarrow\text{p}=0.7$
and $\text{q}=0.3,\text{n}=5$ and $\text{r}=4$
$\therefore$ Required probability $={^5}\text{C}_4(0.7)^4(0.3)$ View full question & answer→Question 161 Mark
A fair die is tossed eight times. The probability that a third six is observed in the eight throw is:
Answer
- $\frac{\text{ }^7\text{C}_2\times5^5}{6^8}$
Solution:
probability of getting $6=\text{p}=\frac{1}{6},\text{q}=\frac{5}{6}$
probability of getting third six in eight throw.
= probability of getting 2 sixes in first seven throw + probability of getting six in eight throw
$=\Big(\text{ }^7\text{C}_2\big(\frac{1}{6}\big)^2\big(\frac{5}{6}\big)^5\Big)\big(\frac{1}{6}\big)$
$=\frac{\text{ }^7\text{C}_2\times5^5}{6^8}$ View full question & answer→Question 171 Mark
A fair die is thrown twenty times. The probability that on the tenth throw the fourth six appears is:
AnswerA 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→Question 181 Mark
A coin is tossed three times. If events A and B are defined as A = Two heads come, B = Last should be head, Then, A and B are
Answer
- Dependent.
Solution:
S = [(HHH), (HHT), (HTH), (HTT), (THH), (THT), (TTH), (TTT)]
$\text{P(A)}=\text{P}(2\text{heads})=\frac{3}{8}$
$\text{P(B)}=\text{P}(\text{last one is heads})=\frac{4}{8}$
$\text{P}(\text{A}\cap\text{B})=\frac{2}{8}=\frac{1}{4}\neq\text{P(A) P(B)}$
Thus, A and B are dependent. View full question & answer→Question 191 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:
Answer
- $\frac{119}{128}$
Solution:
$\text{n = 8,}\text{p}=\frac{1}{2}=\text{q}$
$\text{P(|X}-4|)\leq2$
$\Rightarrow-2\leq\text{x}-4\leq2$
$\Rightarrow4-2\leq\text{x}\leq2+4$
$\Rightarrow2\leq\text{x}\leq6$
$\text{P}(2\leq\text{x}\leq6)=\text{P(2)+P(3)+P(4)+P(5)+P(6)}$
$\text{P(2}\leq\text{x}\leq6)=\text{ }^8\text{C}_2\Big(\frac{1}{2^8}\Big)+\text{ }^8\text{C}_3\Big(\frac{1}{2^8}\Big)\\+\text{ }^8\text{C}_4\Big(\frac{1}{2^8}\Big)\text{ }^8\text{C}_5\Big(\frac{1}{2^8}\Big)+\text{ }^8\text{C}_6\Big(\frac{1}{2^8}\Big)$
$=\frac{119}{128}$ View full question & answer→Question 201 Mark
If P(A) + P(B) = 1; then which of the following option explains the event A and B correctly?
Answer
- Event A and B are mutually exclusive, exhaustive and complementary events.
Solution:
Since P(A) + P(B) = 1
$\therefore\text{A}\cap\text{B}=0.$
Thus, event A and B are mutually exclusive, exhaustive and complementary events. View full question & answer→Question 211 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:
Answer
- $\frac{15}{2^{13}}$
Solution:
Let X be the number of heads.
$\text{p}=\frac{1}{2}\Rightarrow\text{q}=\frac{1}{2}\dots(1)$
$\text{P(X}=7)=\text{P(X}=9)$
$\text{ }^{\text{n}}\text{C}_7\text{p}^7\text{q}^{\text{n}-7}=\text{ }^{\text{n}}\text{C}_9\text{p}^9\text{q}^{\text{n}-9}$
$\frac{\text{ }^{\text{n}}\text{C}_7}{\text{ }^{\text{n}}\text{C}_9}=\frac{\text{q}^{\text{n}-9}}{\text{q}^{\text{n}-7}}\times\frac{\text{p}^9}{\text{p}^7}$
$\frac{\frac{\text{n}!}{7!(\text{n}-7)!}}{\frac{\text{n}!}{9!(\text{n}-9)!}}=\text{q}^{-2}\text{p}^2$
$\frac{9!(\text{n}-9)!}{7!(\text{n}-7)!}=\frac{\text{p}^2}{\text{q}^2}$
$\frac{9\times8\times7!(\text{n}-9)!}{7!(\text{n}-7)(\text{n}-8)(\text{n}-9)!}=1\dots\big[\because\text{from (1)}\big]$
$9\times8=(\text{n}-7)(\text{n}-8)$
Comparing both sides,
$\text{n}-7=9\Rightarrow\text{n}=16$
$\Rightarrow\text{P(X}=2)=\text{ }^{16}\text{C}_2\times0.5^2\times0.5^{14}$
$\Rightarrow\text{P(X}=2)=\frac{15}{2^{13}}$ View full question & answer→Question 221 Mark
Choose the correct answer from the given four options.
If the events A and B are independent, then $\text{P}(\text{A}\cap\text{B})$ is equal to:
Answer
- $\text{P}(\text{A})-\text{P}(\text{B})$
Solution:
If A and B are independent, then $\text{P}(\text{A}\cap\text{B})=\text{P}(\text{A})\cdot\text{P}(\text{B})$ View full question & answer→Question 231 Mark
A fair coin is tossed 100 times. The probability of getting tails an odd nimber of times is:
Answer
- $\frac{1}{2}$
Solution:
Here, $\text{n}=100$
Let X denote the number of times a tail is obtained.
Here, $\text{p = q}=\frac{1}{2}$
$\text{P(X = odd})=\text{P(X}=1,3,5,\dots99)$
$=\big(\text{ }^{100}\text{C}_1+\text{ }^{100}\text{C}_3+\dots+\text{ }^{100}{\text{C}}_{99}\big)\big(\frac{1}{2}\big)^{100}$
= Sum of odd coefficients in binomial expansion in $(1+\text{x})^{100}\big(\frac{1}{2}\big)^{100}$
$=\frac{2^{(100-1)}}{2^{100}}$
$=\frac{1}{2}$ View full question & answer→Question 241 Mark
Difference between sample space and subset of sample space is considered as:
Answer
- Complementary events.
Solution:
The set of all the possible outcomes is called the sample space of the experiment and is usually denoted by S.
Any subset E of the sample space S
Difference between sample space and subset of sample space is considered as complementary events.
View full question & answer→Question 251 Mark
If A and B are two events such that $\text{P(A)}\neq0$ and $\text{P(B)}\neq1,$ then $\text{P}(\overline{\text{A}}|\overline{\text{B}})=$
Answer
- $\frac{1-\text{P}(\text{A}\cup\text{B})}{\text{P}(\overline{\text{B}})}$
Solution:
We have,
$\text{P(A)}\neq0$ and $\text{P(B)}\neq1$
Now,
$\text{P}(\overline{\text{A}}|\overline{\text{B}})=\frac{\text{P}(\overline{\text{A}}\cap\overline{\text{B}})}{\text{P}(\overline{\text{B}})}$
$=\frac{\text{P}(\overline{\text{A}\cap\text{B}})}{\text{P}(\overline{\text{B}})}$
$=\frac{1-\text{P}(\text{A}\cup\text{B})}{\text{P}(\overline{\text{B}})}$
Hence, the correct alternative is option (C). View full question & answer→Question 261 Mark
Choose the correct answer from the given four options.If two events are independent, then:
Answer
- None of the above is correct.
Solution:
If two events A and B are independent, then we know that
$\text{P}(\text{A}\cap\text{B})=\text{P}(\text{A})\cdot\text{P}(\text{B}),\text{P}(\text{A})\neq0,\text{P}(\text{B})\neq0$
Since, A and B have a common outcome.
Further, mutually exclusive events never have a common outcome.
In other words, two independents events having non-zero probabilities of occurrence cannot be mutually exclusive and conversely, i.e., two mutually exclusive events having non-zero probabilities of outcome cannot be independent. View full question & answer→Question 271 Mark
If A and B are independent events such that P(A) > 0 and P(B) > 0, then.
Answer
- $\text{P}(\text{A}\cap\text{B})=\text{P}(\text{A}).\text{P}(\text{B})$
Solution:
Since, A and B are independent events
$\therefore\text{P}(\text{A}\cap\text{B})=\text{P}(\text{A}).\text{P}(\text{B})$
$\text{P}\big(\frac{\text{A}}{\text{B}}\big)=\frac{\text{P}(\text{A}\cap\text{B})}{\text{P}(\text{B})}=\text{P}(\text{A})$
and $\text{P}\big(\frac{\text{B}}{\text{A}}\big)=\frac{\text{P}(\text{B}\cap\text{A})}{\text{P}(\text{A})}=\text{P}(\text{B})$ View full question & answer→Question 281 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:
Answer
- $\frac{51}{101}$
Solution:
Let X denote the number of coins showing head.
Therefore, X follows a binomial distribution with p and n as parameters.
Given that $\text{P(X}=50)=\text{P(X}=51)$
$\Rightarrow\text{ }^{100}\text{C}_{50}\text{p}^{50}\text{q}^{50}=\text{ }^{100}\text{C}_{51}\text{p}^{51}\text{q}^{49}$
on simplifying we get,
$\frac{51}{50}=\frac{\text{p}}{\text{q}}$
$\Rightarrow\frac{51}{50}=\frac{\text{p}}{1-\text{p}}$ (Since p + q = 1)
$\Rightarrow\text{p}=\frac{51}{101}$ View full question & answer→Question 291 Mark
Choose the correct answer from the given four options.Which one is not a requirement of a binomial distribution?
Answer
- The outcomes must be dependent on each othere.
Solution:
We know that, in a Binomial distribution:
- There are 2 outcomes of each trail.
- There is a fixed number of trails.
- The probability of success must be the same for all the trails.
View full question & answer→Question 301 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?
Answer
- $\big(\frac{9}{10}\big)^5+\frac{1}{2}\big(\frac{9}{10}\big)^4$
Solution:
$\text{p}=\frac{10}{100}=\frac{1}{10},\text{q}=\frac{90}{100}=\frac{9}{10},\text{n}=5$
$\text{P(X}\leq1)=\text{P(0)}+\text{P(1)}$
$\text{P(X}\leq1)=\big(\frac{9}{10}\big)^{5}+\text{ }^5\text{C}_1\big(\frac{1}{10}\big)\big(\frac{9}{10}\big)^{4}$
$\text{P(X}\leq1)=\big(\frac{9}{10}\big)^{5}+\big(\frac{1}{2}\big)\big(\frac{9}{10}\big)^4$ View full question & answer→Question 311 Mark
Choose the correct answer from the given four options.
If A and B are two events and $\text{A}\neq\phi,\text{B}\neq\phi,$ then:
Answer
- $\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{\text{P}(\text{A}\cap\text{B})}{\text{P}(\text{B})}$
Solution:
If $\text{A}\neq\phi,\text{B}\neq\phi,$ then $\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{\text{P}(\text{A}\cap\text{B})}{\text{P}(\text{B})}$ View full question & answer→Question 321 Mark
Choose the correct answer from the given four options.
A box has 100 pens of which 10 are defective. What is the probability that out of a sample of 5 pens drawn one by one with replacement at most one is defective?
Answer
- $\Big(\frac{9}{10}\Big)^5+\frac{1}{2}\Big(\frac{9}{10}\Big)^4$
Solution:
We have, $\text{n}=5,\text{P}=\frac{10}{100}=\frac{1}{10}$ and $\text{q}=\frac{9}{10}$
$\text{r}<1\Rightarrow\text{r}=0,1 $
Also, $\text{P}(\text{X}=\text{r})={^\text{n}}\text{C}_\text{r}\text{P}^\text{r}\text{q}^{\text{n}-\text{r}}$
$\therefore\text{P}(\text{X}=\text{r})=\text{P}(\text{r}=0)+\text{P}(\text{r}=1)$
$={^5}\text{C}_0\Big(\frac{1}{10}\Big)^0\Big(\frac{9}{10}\Big)^5+{^5}\text{C}_1\Big(\frac{1}{10}\Big)^1\Big(\frac{9}{10}\Big)^4$
$=\Big(\frac{9}{10}\Big)^5+5\cdot\frac{1}{10}\cdot\Big(\frac{9}{10}\Big)^4$
$=\Big(\frac{9}{10}\Big)^5+\frac{1}{2}\Big(\frac{9}{10}\Big)^4$ View full question & answer→Question 331 Mark
If P(A) + P(B) = 1; then which of the following option explains the event A and B correctly?
Answer
- Event A and B are mutually exclusive, exhaustive and complementary events.
Solution:
Since P(A) + P(B) = 1
$∴ \text{A ∩ B} =0.$
Thus, event A and B are mutually exclusive, exhaustive and complementary events. View full question & answer→Question 341 Mark
If A and B are two events such that $\text{A}\neq\phi,\text{B}=\phi,$ then,
Answer
- $\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{\text{P}(\text{A}\cap\text{B})}{\text{P(B)}}$
Solution:
If A and B are two events such that $\text{A}\neq\phi, \text{B}=\phi$ then,
$\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{\text{P}(\text{A}\cap\text{B})}{\text{P(B)}}$ View full question & answer→Question 351 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?
Answer
- $\big(\frac{9}{10}\big)^5$
Solution:
Let X denote the number of defective bulbs.
Hence, the binomial distribution is given by
$\text{n}=5,\text{p}=\frac{10}{100}=\frac{1}{10}$
$\& \text{ q}=\frac{90}{100}=\frac{9}{10}$
Hence, the distribution is given by
$\text{P(X = r})=\text{ }^5\text{C}_{\text{r}}\big(\frac{1}{10}\big)^{\text{r}}\big(\frac{9}{10}\big)^{5-\text{r}}$
$\therefore\text{P(X}=0)=\big(\frac{9}{10}\big)^5$ View full question & answer→Question 361 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:
Answer
- $\text{ }^{16}\text{C}_6\big(\frac{1}{4}\big)^6\big(\frac{3}{4}\big)^{10}$
Solution:
$\text{np}=4,\text{npq}=3$
$\Rightarrow\text{q}=\frac{3}4{},\text{p}=\frac{1}{4},\text{n}=16$
$\text{P(X}=6)=\text{ }^{16}\text{C}_6\big(\frac{1}{4}\big)^6\big(\frac{3}{4}\big)^{10}$ View full question & answer→Question 371 Mark
A coin is tossed 4 times. The probability that at least one head turns up is:
Answer
- $\frac{15}{16}$
Solution:
$\text{n}=4,\text{p = q}=\frac{1}{2}$
$\text{P(X}\geq1)=1-\text{P(X}=0)$
$\text{P(X}\geq1)=1-\big(\frac{1}{2}\big)^4$
$\text{P(X}\geq1)=\frac{15}{16}$ View full question & answer→Question 381 Mark
The probability distribution of a discrete random variable $X$ is given below$:$
|
$\text{X}:$
|
$1$
|
$2$
|
$3$
|
$4$
|
|
$\text{P}(\text{X}):$
|
$\frac{1}{10}$
|
$\frac{1}{5}$
|
$\frac{3}{10}$
|
$\frac{2}{5}$
|
The value of $E(X^2)$ is$:$ Answer
| $\text{X}$ |
$1$ |
$2$ |
$3$ |
$4$ |
|
| $\text{P}(\text{X})$ |
$\frac{1}{10}$ |
$\frac{1}{5}$ |
$\frac{3}{10}$ |
$\frac{2}{5}$ |
|
| $\text{X}^2\text{P(X)}$ |
$\frac{1}{10}$ |
$\frac{4}{5}$ |
$\frac{27}{10}$ |
$\frac{32}{5}$ |
$\text{E}(\text{X}^2)=10$ |
View full question & answer→Question 391 Mark
The probability distribution of a discrete random variable X is given below:
| $\text{X}:$ |
$2$ |
$3$ |
$4$ |
$5$ |
| $\text{P}(\text{X}):$ |
$\frac{5}{\text{k}}$ |
$\frac{7}{\text{k}}$ |
$\frac{9}{\text{k}}$ |
$\frac{11}{\text{k}}$ |
The value of k is: Answer
- 32
Solution:
$\sum\limits_2^5\text{P}(\text{x})=1$
$\frac{5}{\text{k}}+\frac{7}{\text{k}}+\frac{9}{\text{k}}+\frac{11}{\text{k}}=1$
$\text{k}=32$
NOTE: Question is modified. View full question & answer→Question 401 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:
Answer
- 3
Solution:
A fair coin is tossed $\Rightarrow\text{p = q}=\frac{1}{2}$
$\text{P(X}\geq1)\geq0.8$
$\Rightarrow1-\text{P}(0)\geq0.8$
$\Rightarrow\text{P(0)}=0.2$
$\Rightarrow\big(\frac{1}{2}\big)^{\text{n}}=0.2$
$\Rightarrow2^{-\text{n}}=0.2$
$\Rightarrow2^{\text{n}}\geq5$
$\Rightarrow\text{n}\geq3$ View full question & answer→Question 411 Mark
Mark the correct alternative in the following question:Which one is not a requirement of a binomial dstribution?
Answer
- The outcomes must be dependent on each other.
Solution:
In binomial distribution trails are independent.
View full question & answer→Question 421 Mark
For the following probability distribution:
| X: |
-4 |
-3 |
-2 |
-1 |
0 |
| P(X): |
0.1 |
0.2 |
0.3 |
0.2 |
0.2 |
The value of E(X) is:
Answer
- -1.8
Solution:
The probability distribution of X is given below:
| X: |
-4 |
-3 |
-2 |
-1 |
0 |
| P(X): |
0.1 |
0.2 |
0.3 |
0.2 |
0.2 |
E(X) = (-4) × 0.1 + (-3) × 0.2 + (-2) × 0.3 + (-1) × 0.2 + 0 × 0.2
= -0.4 - 0.6 - 0.6 - 0.2
= -1.8
Hence, the correct alternative is option (d).
View full question & answer→Question 431 Mark
If $X$ is a random variable with probability distribution as given below$:$
| $X = x_i$ |
$0$ |
$1$ |
$2$ |
$3$ |
| $P(X = X_i)$ |
$k$ |
$3k$ |
$3k$ |
$k$ |
The value of $k$ and its variance are$:$ Answer$\sum\limits_0^3\text{P}(\text{x})=1$
$\text{k}+3\text{k}+3\text{k}+\text{k}=1$
$\text{k}=\frac{1}{8}$
| $\text{x}$ |
$\text{P}(\text{x})$ |
$\text{x}\text{P}(\text{x})$ |
$\text{x}^2\text{P}(\text{x})$ |
| $0$ |
$\frac{1}{8}$ |
$0$ |
$0$ |
| $1$ |
$\frac{3}{8}$ |
$\frac{3}{8}$ |
$\frac{3}{8}$ |
| $2$ |
$\frac{3}{8}$ |
$\frac{6}{8}$ |
$\frac{12}{8}$ |
| $3$ |
$\frac{1}{8}$ |
$\frac{3}{8}$ |
$\frac{9}{8}$ |
| $\text{Total}$ |
|
$\text{E(x)}=\frac{12}{8}=1.5$ |
$\text{E}(\text{x}^2)=3$ |
$\text{V(x)}=\text{E}(\text{x}^2)-[\text{E}(\text{x})^2]$
$\text{V(x)}=3-(1.5)^2$
$\text{V(x)}=0.75$
$=\frac{3}{4}$ View full question & answer→Question 441 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:
Answer
- $\frac{7}{128}$
Solution:
$\text{n}=10,\text{p = q}=\frac{1}{2}$
$\text{P(X}\geq8)=\text{P(8) + P(9) + P(10)}$
$\text{P(X}\geq8)=\text{ }^{10}\text{C}_8\big(\frac{1}{2}\big)^{10}+\text{ }^{10}\text{C}_{9}\big(\frac{1}{2}\big)^{10}+\text{ }^{10}\text{C}_{10}\big(\frac{1}{2}\big)^{10}$
$\text{P(X}\geq8)=\frac{45+10+1}{2^8}$
$\text{P(X}\geq8)=\frac{56}{256}=\frac{7}{128}$ View full question & answer→Question 451 Mark
The probability of obtaining an even prime number on each die, when a pair of dice is rolled is
AnswerWhen two dice are rolled, the number of outcomes is 36. The only even prime number is 2.Let E be the event of getting an even prime number on each die.
$\therefore\text{E}=\left\{\left(2,\ 2\right)\right\}$
$\Rightarrow\text{P}(\text{E})=\frac{1}{36}$
Therefore, the correct answer is D.
View full question & answer→Question 461 Mark
A random variable $X$ takes the values $0, 1, 2, 3$ and its mean is $1.3.$ If $P(X = 3) = 2P(X = 1)$ and $P(X = 2) = 0.3,$ then $P(X = 0)$ is$:$
AnswerLet:
$P(X = 0) = m$
$P(X = 1) = k$
Now,
$P(X = 3) = 2k$
| $x_i$ |
$p_i$ |
$p_ix_i$ |
| $0$ |
$m$ |
$0$ |
| $1$ |
$k$ |
$k$ |
| $2$ |
$0.3$ |
$0.6$ |
| $3$ |
$2k$ |
$6k$ |
Mean $=\sum\text{p}_\text{i}\text{x}_\text{i}$
$\Rightarrow 0 + k + 0.6 + 6k = 1.3$
$\Rightarrow 7k = 1.6 - 0.6$
$\Rightarrow\text{k}=\frac{0.7}{7}$
$\Rightarrow 0.1$
We know that the sum of probabilities in a probability distribution is always $1.$
$\therefore P(X = 0) + P(X = 1) + P(X = 3) = 1$
$\Rightarrow m + 0.1 + 0.3 + 0.2 = 1$
$\Rightarrow m + 0.6 = 1$
$\Rightarrow m = 0.4$ View full question & answer→Question 471 Mark
In each of the following choose the correct answer:$\text{If}\ \text{P}(\text{A})=\frac{1}{2},\ \text{P}(\text{B})=0,\ \text{then}\ \text{P}(\text{A}|\text{B})\ \text{is}:$
Answer$\text{P}(\text{A})=\frac{1}{2},\ \text{P}(\text{B})=0$ $\therefore\ \text{P}(\text{A}\cap\text{B})=0$$\therefore\ \text{P}(\text{A}|\text{B})=\frac{\text{P}(\text{A}\cap\text{B})}{\text{P}(\text{B})}=\frac{0}{0}=\text{not defined}$
Therefore, option (C) is correct.
View full question & answer→Question 481 Mark
A random variable has the following probability distribution:
| $X = x_i$ |
$0$ |
$1$ |
$2$ |
$3$ |
$4$ |
$5$ |
$6$ |
$7$ |
| $P(X = X_i)$ |
$0$ |
$2p$ |
$2p$ |
$3p$ |
$p^2$ |
$2p^2$ |
$7p^2$ |
$2p$ |
AnswerWe know that the sum of probabilities in a probability distribution is always $1.$
$\therefore P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) +P(X = 6) + P(X = 7) +P(X = 8) = 1$
$\Rightarrow 0 + 2p + 2p + 3p + p^2 + 2p^2 + 7p^2 + 2p = 1$
$\Rightarrow 10p^{2 }+ 9p - 1 = 0$
$\Rightarrow (10p - 1)(p + 1) = 0$
$\Rightarrow\text{p}=\frac{1}{10}\text{ or }-1 ($Negleting $-1$ as the value of the probability cannot be negative$)$
View full question & answer→Question 491 Mark
In a binomial distribution, the probability of getting success is $\frac{1}{4}$ and standard deviation is 3. Then, its mean is:
Answer
- 12
Solution:
$\text{p}=\frac{1}{4},\sqrt{\text{npq}}=3$
$\Rightarrow\text{q}=\frac{3}{4},\text{npq}=9$
$\Rightarrow\text{Mean = np}=\frac{9}{\text{q}}$
$\Rightarrow\text{Mean}=9\times\frac{4}{3}=12$ View full question & answer→Question 501 Mark
For a binomial variate X, if $\text{n}=3$ and $\text{P(X}=1)=8\text{ P(X = 3}),$ then p =
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