MCQ 1011 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?
- 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. $\Big(\frac{9}{10}\Big)^5+\frac{1}{2}\Big(\frac{9}{10}\Big)^4$
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→MCQ 1021 Mark
In each of the following, choose the correct answer:
In a box containing 100 bulbs, 10 are defective. The probability that out of a sample of 5 bulbs, none is defective is
AnswerCorrect option: C. $\Big(\frac{9}{10}\Big)^5$
The repeated selections of defective bulbs from a box are Bernoulli trials. Let X denote the number of defective bulbs out of a sample of 5 bulbs.
Probability of getting a defective bulb, p $=\frac{10}{100}=\frac{1}{10}$
$\therefore\text{q}=1-\text{p}=1-\frac{1}{10}=\frac{9}{10}$
Clearly, X has a binomial distribution with n = 5 and $\text{p}=\frac{1}{10}$
$\therefore\ \text{P}(\text{X=x})=\ ^\text{n}\text{C}_\text{x}\text{q}^\text{n-x}\text{p}^\text{x}=\ ^5\text{C}_\text{x}\bigg(\frac{9}{10}\bigg)^{5-\text{x}}\bigg(\frac{1}{10}\bigg)^\text{x}$
P(none of the bulbs is defective) = P(X = 0)
$=\ ^5\text{C}_0\cdot\Big(\frac{9}{10}\Big)^5$
$=1\cdot\Big(\frac{9}{10}\Big)^5$
$=\Big(\frac{9}{10}\Big)^5$
The correct answer is C.
View full question & answer→MCQ 1031 Mark
A and B are two events such that P(A) = 0.25 and P(B) = 0.50. The probability pf both happening together is 0.14. The probability of both A and B hot happening is.
MCQ 1041 Mark
If one ball is drawn ar random from each of three boxes containing $3$ white and $1$ black$, 2$ white and $2$ black$, 1$ white and $3$ black balls, then the probability that $2$ white and $1$ black balls will be drawn is.
- ✓
$\frac{13}{32}$
- B
$\frac{1}{4}$
- C
$\frac{1}{32}$
- D
$\frac{3}{16}$
AnswerCorrect option: A. $\frac{13}{32}$
Total balls in first box $= 3$ white $+ 1$ black $= 4$
Total balls in second box $= 2$ white $+ 2$ black $= 4$
Total balls in third box $= 1$ white $+ 3$ black $= 4$
Probability of $2$ white and $1$ black
$= \text{P(WWB) + P(WBW) + P(BWW)}$
$=\frac{3}{4}\times\frac{2}{4}\times\frac{3}{4}+\frac{3}{4}\times\frac{2}{4}\times\frac{1}{4}+\frac{1}{4}\times\frac{2}{4}\times\frac{1}{4}$
$=\frac{18+6+2}{64}=\frac{13}{32}$
View full question & answer→MCQ 1051 Mark
From a set of $100$ cards numbered $1$ to $100,$ one card is drawn at randow. The probability number obtained on the card is divisible by $6$ or $8$ but not by $24$ is
- ✓
$\frac{6}{25}$
- B
$\frac{1}{4}$
- C
$\frac{1}{6}$
- D
$\frac{2}{6}$
AnswerCorrect option: A. $\frac{6}{25}$
Number of cards divisible by $6 = 16$
$\Rightarrow\ \text{P(A)}=\frac{16}{100}$
Number of cards divisible by $8 = 12$
$\Rightarrow\ \text{P(B)}=\frac{12}{100}$
Number of cards divisible by $24 = 4$
$\Rightarrow\ \text{P}(\text{A}\cap\text{B})=\frac{4}{100}$
$\text{P}(\text{A}\cup\text{B})=\frac{16}{100}+\frac{12}{100}-\frac{4}{100}$
$\text{P}(\text{A}\cup\text{B})=\frac{6}{25}$
View full question & answer→MCQ 1061 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→MCQ 1071 Mark
A bag containe $5$ black$, 4$ white balls and $3$ red balls. if a ball is selected randomwise, the probability that it is black or red ball is,
- A
$\frac{1}{3}$
- B
$\frac{1}{4}$
- ✓
$\frac{5}{12}$
- D
$\frac{2}{3}$
AnswerCorrect option: C. $\frac{5}{12}$
We know that the bag contains $5B ($black)$, 4W($white$)$ and $3R($red$)$ balls.
Now,
$\text{P(B)}=\frac{5}{12}$
$\text{P(R)}=\frac{3}{12}$
$\text{P}(\text{B}$ or $R)=\text{P(B)}+\text{P(R)}$
$=\frac{5}{12}+\frac{3}{12}$
$=\frac{8}{12}=\frac{2}{3}$
View full question & answer→MCQ 1081 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. $\text{ }^5\text{C}_4(0.7)^4(0.3)$
Given that a person is not a swimmer
$\Rightarrow\text{q}=0.3$
$\Rightarrow\text{p}=0.7$
$\text{n = 5, X = 4}$
$\text{P(X}=4)=\text{ }^5\text{C}_4\times0.7^{4}\times0.3$
View full question & answer→MCQ 1091 Mark
The probability of obtaining an even prime number on each die, when a pair of dice is rolled is
- A
- B
$\frac{1}{3}$
- C
$\frac{1}{12}$
- ✓
$\frac{1}{36}$
AnswerCorrect option: D. $\frac{1}{36}$
When 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→MCQ 1101 Mark
An urn contains $5$ red and $2$ black balls. Two balls are randomly drawn. Let $X$ represent the number of black balls. What are the possible values of $X?$ Is $X$ a random variable?
- ✓
$0, 1, 2$
- B
$3, 5, 7$
- C
$7, 7, 8$
- D
$1, 5, 7$
AnswerCorrect option: A. $0, 1, 2$
The two balls selected can be represented as $\text{BB, BR, RB, RR,}$
where $B$ represents a black ball and $R$ represents a red ball.
$X$ represents the number of black balls.
$\therefore X(BB) = 2$
$\text{X(BR) = 1}$
$\text{X(RB) = 1}$
$\text{X(RR) = 0}$
Therefore, the possible values of $X$ are $0, 1$ and $2.$
Yes$, X$ is a random variable.
View full question & answer→MCQ 1111 Mark
A special lottery is to be held to select a student who will live in the only deluxe room in a hostel. There are $100$ Year$\text{-III,} 150$ Year$-II$ and $200$ Year$-I$ students who applied. Each Year$-III\ 's$ name is placed in the lottery $3$ times; each Year$-II\ 's$ name$, 2$ times and Year$-I\ 's$ name$, 1$ time. What is the probability that a Year$-III\ 's$ name will be chosen?
- A
$\frac18$
- B
$\frac28$
- ✓
$\frac38$
- D
$\frac12$
AnswerCorrect option: C. $\frac38$
Total names in the lottery
$= 3 \times 100 + 2 \times 150 + 200 = 800$
Number of Year$-III\ 's$ names $= 3 \times 100 = 300$
Required probability $=\frac{300}{800}=\frac38$
View full question & answer→MCQ 1121 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$\text{S = [(HHH), (HHT), (HTH), (HTT), (THH), (THT), (TTH), (TTT)]}$
$\text{P(A)}=\text{P}(2$ heads$)=\frac{3}{8}$
$\text{P(B)}=\text{P}($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→MCQ 1131 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. $\frac{15}{2^{13}}$
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 $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→MCQ 1141 Mark
If $A$ and $B$ are two independent events with $\text{P(A)}=\frac{3}{5}$ and $\text{P(B)}=\frac{4}{9},$ then $\text{P}(\overline{\text{A}}\cap\overline{\text{B}})$ equals,
- A
$\frac{4}{15}$
- B
$\frac{8}{45}$
- C
$\frac{1}{3}$
- ✓
$\frac{2}{9}$
AnswerCorrect option: D. $\frac{2}{9}$
$\text{P(A)}=\frac{3}{5},\text{P(B)}=\frac{4}{9}$
$\text{P}(\overline{\text{A}}\cap\overline{\text{B}})=\text{P}(\overline{\text{A}\cup\text{B}})$
$\text{P}(\overline{\text{A}}\cap\overline{\text{B}})=1-\text{P}(\text{A}\cup\text{B})$
$\text{P}(\overline{\text{A}}\cap\overline{\text{B}})=1-\big[\text{P(A)}+\text{P(B)}-\text{P}(\text{A}\cap\text{B})\big]$
$\text{P}(\overline{\text{A}}\cap\overline{\text{B}})=1-\Big[\frac{3}{5}+\frac{4}{9}-\frac{3}{5}\times\frac{4}{9}\Big]$
$(\because A$ and $B$ are independent$)$
$\text{P}(\overline{\text{A}}\cap\overline{\text{B}})=1-\frac{7}{9}$
$\text{P}(\overline{\text{A}}\cap\overline{\text{B}})=\frac{2}{9}$
View full question & answer→MCQ 1151 Mark
The probability that a leap year will have $53$ fridays or $53$ Saturdays is.
- A
$\frac{2}{7}$
- ✓
$\frac{3}{7}$
- C
$\frac{4}{7}$
- D
$\frac{1}{7}$
AnswerCorrect option: B. $\frac{3}{7}$
Non$-$leap year has $365$ days $= 52$ weeks $+ 1$
$366$ days in leap year.
We want to find probability of $53$ Fridays or $53$ Saturday.
Favourable cases $=\{($Thursday, Friday$), ($Friday, Saturday$), ($Saturday, Sunday$)\}$
Required probability $=\frac{3}{7}$
View full question & answer→MCQ 1161 Mark
Choose the correct answer from the given four options:If $A$ and $B$ are such events that $\text{P}(\text{A})>0$ and $\text{P}(\text{B})\neq1,$ then $\text{P}\Big(\frac{\text{A}'}{\text{B}'}\Big)$ equals to:
- A
$1-\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)$
- B
$1-\text{P}\Big(\frac{\text{A}'}{\text{B}}\Big)$
- ✓
$\frac{1-\text{P}(\text{A}\cup\text{B})}{\text{P}(\text{B}')}$
- D
$\frac{\text{P}(\text{A}')}{\text{P}(\text{B}')}$
AnswerCorrect option: C. $\frac{1-\text{P}(\text{A}\cup\text{B})}{\text{P}(\text{B}')}$
We have, $\text{P}(\text{A})>0$ and $\text{P}(\text{B})\neq1$
$\text{P}\Big(\frac{\text{A}'}{\text{B}'}\Big)=\frac{\text{P}(\text{A}'\cap\text{B}')}{\text{P}(\text{B}')}$
$=\frac{1-\text{P}(\text{A}\cup\text{B})}{\text{P}(\text{B}')}$
View full question & answer→MCQ 1171 Mark
Choose the correct answer from the given four options.Let $A$ and $B$ be two events such that $\text{P}(\text{A})=\frac{3}{8},\text{P}({\text{B}})=\frac{5}{8}$ and $\text{P}(\text{A}\cup\text{B})=\frac{3}{4}.$Then $\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)\cdot\text{P}\Big(\frac{\text{A'}}{\text{B}}\Big)$ is equal to:
- A
$\frac{2}{5}$
- B
$\frac{3}{8}$
- C
$\frac{3}{20}$
- ✓
$\frac{6}{25}$
AnswerCorrect option: D. $\frac{6}{25}$
We have, $\text{P}(\text{A})=\frac{3}{8},\text{P}({\text{B}})=\frac{5}{8}$ and $\text{P}(\text{A}\cup\text{B})=\frac{3}{4}$
Now $\text{P}(\text{A}\cup\text{B})=\text{P}(\text{A})+\text{P}(\text{B})-\text{P}(\text{A}\cap\text{B})$
$\Rightarrow\text{P}(\text{A}\cap\text{B})=\frac{3}{8}+\frac{5}{8}-\frac{3}{4}=\frac{1}{4}$
$\because\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{\text{P}(\text{A}\cap\text{B})}{\text{P}(\text{B})}$
$=\frac{\frac{1}{4}}{\frac{5}{8}}=\frac{2}{5}$
and $\text{P}\Big(\frac{\text{A}'}{\text{B}}\Big)=\frac{\text{P}(\text{A}'\cap\text{B})}{\text{P}(\text{B})}=\frac{\text{P}(\text{B})-\text{P}(\text{A}\cap\text{B})}{\text{P}(\text{B})}$
$=\frac{\frac{5}{8}-\frac{1}{4}}{\frac{5}{8}}=\frac{3}{5}$
$\therefore\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)\cdot\text{P}\Big(\frac{\text{A}'}{\text{B}}\Big)$
$=\frac{2}{5}\cdot\frac{3}{5}=\frac{6}{25}$
View full question & answer→MCQ 1181 Mark
Choose the correct answer from the given four options.A bag contains $5$ red and $3$ blue balls. If $3$ balls are drawn at random without replacement the probability of getting exactly one red ball is:
- A
$\frac{45}{196}$
- B
$\frac{135}{392}$
- ✓
- D
$\frac{15}{29}$
AnswerProbability of getting exactly one red $(R)$ ball
$=\text{P}_{\text{R}}\cdot\text{P}_{\bar{\text{R}}}\cdot\text{P}_{\bar{\text{R}}}+\text{P}_{\bar{\text{R}}}\cdot\text{P}_{\text{R}}\cdot\text{P}_{\bar{\text{R}}}+\text{P}_{\bar{\text{R}}}\cdot\text{P}_{\bar{\text{R}}}\cdot\text{P}_{\text{R}}$
$=\frac{5}{8}\cdot\frac{3}{7}\cdot\frac{2}{6}+\frac{3}{8}\cdot\frac{5}{7}\cdot\frac{2}{7}+\frac{3}{8}\cdot\frac{2}{7}\cdot\frac{5}{6}$
$=\frac{15}{4\cdot7\cdot6}+\frac{15}{4\cdot7\cdot6}+\frac{15}{4\cdot7\cdot6}$
$=\frac{5}{56}+\frac{5}{56}+\frac{5}{56}=\frac{15}{56}$
View full question & answer→MCQ 1191 Mark
Choose the correct answer from the given four options.If $\text{P}(\text{B})=\frac{3}{5},\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{1}{2}$ and $\text{P}(\text{A}\cup\text{B})=\frac{4}{5},$ then $\text{P}(\text{A}\cup\text{B})\ '+\text{P}(\text{A}\ '\cup\text{B})=$
- A
$\frac{1}{5}$
- B
$\frac{4}{5}$
- C
$\frac{1}{2}$
- ✓
$1.$
AnswerWe have, $\text{P}(\text{B})=\frac{3}{5},\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{1}{2}$
$\therefore\text{P}(\text{A}\cap\text{B})=\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)\cdot\text{P}(\text{B})$
$=\frac{1}{2}\cdot\frac{3}{5}=\frac{3}{10}$
Now, $\text{P}(\text{A}\cup\text{B})=\text{P}(\text{A})+\text{P}(\text{B})-\text{P}(\text{A}\cap\text{B})$
$\Rightarrow\text{P}(\text{A})=\frac{4}{5}-\frac{3}{5}+\frac{3}{10}=\frac{1}{2}$
$\therefore\text{P}(\text{A}\cup\text{B})\ '=1-\text{P}(\text{A}\cup\text{B})$
$=1-\frac{4}{5}=\frac{1}{5}$
And $\text{P}(\text{A}\ '\cap\text{B})=1-\text{P}(\text{A}-\text{B})$
$=1-\big[\text{P}(\text{A})-\text{P}(\text{A}\cap\text{B})\big]$
$=1-\Big(\frac{1}{2}-\frac{3}{10}\Big)=\frac{4}{5}$
$\Rightarrow\text{P}(\text{A}\cup\text{B})\ '+\text{P}(\text{A}\ '\cup\text{B})$
$=\frac{1}{5}+\frac{4}{5}=1$
View full question & answer→MCQ 1201 Mark
Two cards are drawn from a well shuffled deck of $52$ playing cards with replacement. The probability that both cards are queen is
- ✓
$\frac{1}{13}\times\frac{1}{13}$
- B
$\frac{1}{13}+\frac{1}{13}$
- C
$\frac{1}{13}\times\frac{1}{17}$
- D
$\frac{1}{13}\times\frac{4}{5}$
AnswerCorrect option: A. $\frac{1}{13}\times\frac{1}{13}$
Two cards are drawn from $52$ cards.
Let$, E_1$ be the event that getting queen in first draw and $E_2$ be the event that getting queen in second draw,
$\text{P}(\text{E}_1\cap\text{E}_2)=\frac{4}{52}\times\frac{4}{52}=\frac{1}{13}\times\frac{1}{13}$
View full question & answer→MCQ 1211 Mark
Choose the correct answer from the given four options. The probability that exactly two of the three balls were red, the first ball being red, is:
- A
$\frac{1}{3}$
- ✓
$\frac{4}{7}$
- C
$\frac{15}{28}$
- D
$\frac{5}{28}$
AnswerCorrect option: B. $\frac{4}{7}$
Let $E_1=$ Event that first ball being red
And $E_2 =$ Event that exactly two of three balls being red
$\therefore\text{P}(\text{E}_1)=\text{P}_{\text{R}}\cdot\text{P}_{\text{R}}\cdot\text{P}_{\text{R}}+\text{P}_{\text{R}}\cdot\text{P}_{\text{R}}\cdot\text{P}{_\bar{\text{R}}}+\text{P}_{\text{R}}\cdot\text{P}{_\bar{\text{R}}}\cdot\text{P}_{\text{R}}+\text{P}_{\text{R}}\cdot\text{P}{_\bar{\text{R}}}\cdot\text{P}{_\bar{\text{R}}}$
$=\frac{5}{8}\cdot\frac{4}{7}\cdot\frac{3}{6}+\frac{5}{8}\cdot\frac{4}{7}\cdot\frac{3}{6}+\frac{5}{8}\cdot\frac{3}{7}\cdot\frac{4}{6}+\frac{5}{8}\cdot\frac{3}{7}\cdot\frac{2}{6}$
$=\frac{60+60+60+30}{336}=\frac{210}{336}$
$\text{P}(\text{E}_1\cap\text{E}_2)=\text{P}{_\text{R}}\cdot\text{P}{_\bar{\text{R}}}\cdot\text{P}{_\text{R}}+\text{P}{_\text{R}}\cdot\text{P}{_\text{R}}\cdot\text{P}{_\bar{\text{R}}}$
$=\frac{5}{8}\cdot\frac{3}{7}\cdot\frac{4}{6}+\frac{5}{8}\cdot\frac{4}{7}\cdot\frac{3}{6}=\frac{120}{336}$
$\therefore\text{P}\Big(\frac{\text{E}_2}{\text{E}_1}\Big)=\frac{\text{P}(\text{E}_1\cap\text{E}_2)}{\text{P}(\text{E}_1)}$
$=\frac{\frac{120}{336}}{\frac{210}{336}}=\frac{4}{7}$
View full question & answer→MCQ 1221 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→MCQ 1231 Mark
Choose the correct answer from the given four options.$\text{P}\Big(\frac{\text{B}}{\text{A}}\Big)$ is equal to:
- A
$\frac{1}{5}$
- B
$\frac{3}{10}$
- C
$\frac{1}{2}$
- ✓
$\frac{3}{5}$
AnswerCorrect option: D. $\frac{3}{5}$
$\text{P}\Big(\frac{\text{B}}{\text{A}'}\Big)=\frac{\text{P}(\text{B}\cap\text{A}')}{\text{P}(\text{A}')}$
$=\frac{\text{P}(\text{B})-\text{P}(\text{B}\cap\text{A})}{1-\text{P}(\text{A})}$
$=\frac{\frac{3}{5}-\frac{3}{10}}{1-\frac{1}{2}}=\frac{\frac{6-3}{10}}{\frac{1}{2}}$
$=\frac{6}{10}=\frac{3}{5}$
View full question & answer→MCQ 1241 Mark
If $P(A) + P(B) = 1;$ then which of the following option explains the event $A$ and $B$ correctly?
- ✓
Event $A$ and $B$ are mutually exclusive, exhaustive and complementary events.
- B
Event $A$ and $B$ are mutually exclusive and exhaustive events.
- C
Event $A$ and $B$ are mutually exclusive and complementary events.
- D
Event $A$ and $B$ are exhaustive and complementary events.
AnswerCorrect option: A. Event $A$ and $B$ are mutually exclusive, exhaustive and complementary events.
Since $P(A) + P(B) = 1$
$\therefore A \cup B =0.$
Thus, event $A$ and $B$ are mutually exclusive, exhaustive and complementary events.
View full question & answer→MCQ 1251 Mark
Choose the correct answer from the given four options.If $\text{P}(\text{A})=\frac{3}{10},\text{P}(\text{B})=\frac{2}{5}$ and $\text{P}(\text{A}\cup\text{B})=\frac{3}{5},$ then $\text{P}\Big(\frac{\text{B}}{\text{A}}\Big)+\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)$ equas:
- A
$\frac{1}{4}$
- B
$\frac{1}{3}$
- C
$\frac{15}{12}$
- ✓
AnswerWe have, $\text{P}(\text{A})=\frac{3}{10},\text{P}(\text{B})=\frac{2}{5}$ and $\text{P}(\text{A}\cup\text{B})=\frac{3}{5}$
$\text{P}(\text{A}\cup\text{B})=\text{P}(\text{A})+\text{P}(\text{B})-\text{P}(\text{A}\cap\text{B})$
$\therefore\frac{3}{5}=\frac{3}{10}+\frac{2}{5}-\text{P}(\text{A}\cap\text{B})$
$\therefore\text{P}(\text{A}\cap\text{B})=\frac{1}{10}$
$\text{P}\Big(\frac{\text{B}}{\text{A}}\Big)+\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{\text{P}(\text{B}\cap\text{A})}{\text{P}(\text{A})}+\frac{\text{P}(\text{A}\cap\text{B})}{\text{P}(\text{B})}$
$=\frac{\frac{1}{10}}{\frac{3}{10}}+\frac{\frac{1}{10}}{\frac{2}{5}}$
$=\frac{1}{3}+\frac{1}{4}=\frac{7}{12}$
View full question & answer→MCQ 1261 Mark
Choose the correct answer in each of the following:
If A and B are any two events such that P(A) + P(B) – P(A and B) = P(A), then
Answer$\text{P}(\text{A|B})=1$
$\text{P}(\text{A})+\text{P}(\text{B})-\text{P}(\text{A and B})=\text{P}(\text{A})$
$\Rightarrow\ \text{P}(\text{B})=\text{P}(\text{A}\cap\text{B})\ \Rightarrow\ 1=\frac{\text{P}(\text{A}\cap\text{B})}{\text{P}(\text{B})}=1=\text{P}(\text{A|B})$
$\therefore$ (B) is correct answer.
View full question & answer→MCQ 1271 Mark
If $\text{P(B)}=\frac{3}{5},\text{P}(\text{A}|\text{B})=\frac{1}{2}$ and $\text{P}(\text{A}\cup\text{B})=\frac{4}{5},$ then $\text{P}(\text{B}|\overline{\text{A}})=$
- A
$\frac{1}{5}$
- B
$\frac{3}{10}$
- C
$\frac{1}{2}$
- ✓
$\frac{3}{5}$
AnswerCorrect option: D. $\frac{3}{5}$
$\text{P(B)}=\frac{3}{5},\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{1}{2},\text{P}(\text{A}\cup\text{B})=\frac{4}{5}$
Consider,
$\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{1}{2}$
$\frac{\text{P}(\text{A}\cap\text{B})}{\text{P(B)}}=\frac{1}{2}$
$\frac{\text{P}(\text{A}\cap\text{B})}{\frac{3}{5}}=\frac{1}{2}$
$\text{P}(\text{A}\cap\text{B})=\frac{3}{10}$
$\text{P(A)}+\text{P(B)}-\text{P}(\text{A}\cup\text{B})=\frac{3}{10}$
$\text{P(A)}+\frac{3}{5}-\frac{4}{5}=\frac{3}{10}$
$\text{P(A)}=\frac{1}{2}$
$\text{P}\Big(\frac{\text{B}}{\overline{\text{A}}}\Big)=\frac{\text{P}(\text{B}\cap\overline{\text{A}})}{\text{P}(\overline{\text{A}})}$
$\text{P}\Big(\frac{\text{B}}{\overline{\text{A}}}\Big)=\frac{\frac{3}{5}-\frac{3}{10}}{1-\frac{1}{2}}$
$\text{P}\Big(\frac{\text{B}}{\overline{\text{A}}}\Big)=\frac{3}{5}$
View full question & answer→MCQ 1281 Mark
If S is the samle space and $\text{P(A)}=\frac{1}{3}, \text{P(B)}$ and $\text{S}=\text{A}\cup\text{B,}$ where A and B are tow mutually exclusive events, then P(A) =
- ✓
$\frac{1}{4}$
- B
$\frac{1}{2}$
- C
$\frac{3}{4}$
- D
$\frac{3}{8}$
AnswerCorrect option: A. $\frac{1}{4}$
Solution: a.
$\text{P(A)}=\frac{1}{3}\text{P(A)}$
$\Rightarrow\ \text{P(B)}=3\text{P(A)}\ .....(\text{i})$
A and B are mutually exclusive events.
$\Rightarrow\ \text{P}(\text{A}\cap\text{B}) = 0$
Now,
$\text{P}(\text{A}\cup\text{B})=\text{P(A)}+\text{P(B)}=\text{P(S)}$
$\Rightarrow\ \text{P(A)}+\text{P(B)}=1$
$\Rightarrow\ \text{P(A)}+3\text{P(A)}=1$ [From (i)]
$\Rightarrow\ 4\text{P(A)}=1$
$\Rightarrow\ \text{P(A)}=\frac{1}{4}$
View full question & answer→MCQ 1291 Mark
A random variable has the following probability distribution:
| $X = x_i$ |
$0$ |
$1$ |
$2$ |
$3$ |
$4$ |
$5$ |
$6$ |
$7$ |
| $P(X = X_i)$ |
$0$ |
$2p$ |
$2p$ |
$3p$ |
$p^2$ |
$2p^2$ |
$7p^2$ |
$2p$ |
- ✓
$\frac{1}{10}$
- B
$-1$
- C
$-\frac{1}{10}$
- D
$\frac{1}{5}$
AnswerCorrect option: A. $\frac{1}{10}$
We know that the sum of probabilities in a probability distribution is always $1.$
$\therefore P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) +P(X = 6) + P(X = 7) +P(X = 8) = 1$
$\Rightarrow 0 + 2p + 2p + 3p + p^2 + 2p^2+ 7p^2 + 2p = 1$
$\Rightarrow 10p^2+ 9p - 1 = 0$
$\Rightarrow (10p - 1)(p + 1) = 0$
$\Rightarrow\text{p}=\frac{1}{10}$ or $-1 ($Negleting $-1$ as the value of the probability cannot be negative$)$
View full question & answer→MCQ 1301 Mark
Out of $30$ consecutive integers$, 2$ are chosen at random. The probability that their sum is odd, is
- A
$\frac{14}{29}$
- B
$\frac{16}{29}$
- ✓
$\frac{15}{29}$
- D
$\frac{10}{29}$
AnswerCorrect option: C. $\frac{15}{29}$
For sum of two integers to be odd, one integer should be eve annd the other should be odd. In $30$ consecutive integers$, 15$ are even and $15$ are odd.
$P($Sum is odd$) = P($first integer is odd and second is even$) + P($first integer is even and second integer is odd$)$
$=\frac{15}{30}\times\frac{15}{29}+\frac{15}{30}\times\frac{15}{29}$
$=\frac{450}{30\times29}$
$=\frac{15}{29}$
View full question & answer→MCQ 1311 Mark
Assume that in a family, each chold is equally likely to be a boy or a girl. A family with tree cgildren is chosen at random. Tere probability that the eldest child is a girl given that the family has at least oe girl.
- A
$\frac{1}{2}$
- B
$\frac{1}{3}$
- C
$\frac{2}{3}$
- ✓
$\frac{4}{7}$
AnswerCorrect option: D. $\frac{4}{7}$
Solution: d. $\frac{4}{7}$
S = {GBB, GGB, GBG, GGG, BGG, BGB, BBG, BBB}
Let E$_1$ be the event that choosing a family with a girl as eldest child. E$_2$ be the event that choosing a family with at least one girl.
E$_1$ = {GBB, GGB, GBG, GGG}
E$_2$ = {GBB, GGB, GBG, GGG, BGG, BGB, BBG}
$\text{n}(\text{E}_1)=4,\text{n}(\text{E}_2)=7,\text{n}(\text{A}\cap\text{B})=4$
$\Rightarrow\ \text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{\text{n}(\text{A}\cap\text{B})}{\text{n(B)}}=\frac{4}{7}$
View full question & answer→MCQ 1321 Mark
Choose the correct answer in each of the following:
If P(A|B) > P(A), then which of the following is correct :
Answer$\text{P}(\text{B|A})>\text{P}(\text{B})$
$\text{P}(\text{A|B})>\text{P}(\text{A})$
$\Rightarrow\ \frac{\text{P}(\text{A}\cap\text{B})}{\text{P}(\text{B})}>\text{P}(\text{A})\ \ \Rightarrow\ \ \frac{\text{P}(\text{A}\cap\text{B})}{\text{P}(\text{A})}=\text{P}(\text{B})$
$\Rightarrow\ {\text{P}(\text{B}|\text{A})}>{\text{P}(\text{B})}.$
(C) is correct answer.
View full question & answer→MCQ 1331 Mark
Let $A$ and $B$ be two events. If $\text{P(A)}=0.2,\text{P(B)}=0.4,\text{P}(\text{A}\cup\text{B})=0.6$ then $P(A|B)$ is equal to
Answer$\text{P(A)}=0.2,\text{P(B)}=0.4,\text{P}(\text{A}\cup\text{B})=0.6$
$\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{\text{A}(\text{A}\cap\text{B})}{\text{P(B)}}$
$\Rightarrow \text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{\text{P(A)}+\text{P(B)}-\text{P}(\text{A}\cup\text{B})}{\text{P(B)}}$
$\Rightarrow\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{0.2+0.4-0.6}{\text{P(B)}}$
$\Rightarrow\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=0$
View full question & answer→MCQ 1341 Mark
If $A$ and $B$ are two events such that $\text{A}\neq\phi,\text{B}=\phi,$ then,
- ✓
$\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{\text{P}(\text{A}\cap\text{B})}{\text{P(B)}}$
- B
$\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\text{P(A)}\text{ P(B)}$
- C
$\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\text{P}\Big(\frac{\text{B}}{\text{A}}\Big)=1$
- D
$\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{\text{P(A)}}{\text{P(B)}}$
AnswerCorrect option: A. $\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{\text{P}(\text{A}\cap\text{B})}{\text{P(B)}}$
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→MCQ 1351 Mark
A bag contains six red four green and eight white balls If a ball is picked at random the probability that it is not white is:
- A
$\frac13$
- B
$\frac49$
- ✓
$\frac59$
- D
$\frac23$
AnswerCorrect option: C. $\frac59$
Number of balls that are not white $= 10$
Total $=\frac 18$
$\therefore P($not white$) = \frac{18}{10} = \frac{9}{5}$
View full question & answer→MCQ 1361 Mark
In a binomial distribution, the probability of getting success is $\frac{1}{4}$ and standard deviation is $3.$ Then, its mean is:
Answer$\text{p}=\frac{1}{4},\sqrt{\text{npq}}=3$
$\Rightarrow\text{q}=\frac{3}{4},\text{npq}=9$
$\Rightarrow$Mean $= np=\frac{9}{\text{q}}$
$\Rightarrow$Mean$=9\times\frac{4}{3}=12$
View full question & answer→MCQ 1371 Mark
A bag contains $5$ red and $3$ blue balls. If $3$ balls are drawn at random without replacement, then the probability that exactly two of the three balls were red, the first ball being red, is
- A
$\frac{1}{3}$
- ✓
$\frac{4}{7}$
- C
$\frac{15}{28}$
- D
$\frac{5}{28}$
AnswerCorrect option: B. $\frac{4}{7}$
Total number of balls $= 5$ red $+ 3$ Blue $= 8$
Probability of getting exacctly two red balls given that first ball should be red
Required probability $=\text{P}\Big(\frac{\text{R}_2\text{B}_2}{\text{R}_1}\Big)+\text{P}\Big(\frac{\text{R}_1\text{B}_2}{\text{R}_1}\Big)$
Required probability $=\frac{4}{7}\times\frac{3}{6}+\frac{3}{7}\times\frac{4}{6}=\frac{4}{7}$
View full question & answer→MCQ 1381 Mark
If $P(A) = 0.4, P(B) = 0.8$ and $P(B|A) = 0.6$ then $\text{P}(\text{A}\cup\text{B})=$
- A
$0.24$
- B
$0.3$
- C
$0.48$
- ✓
$0.96$
AnswerCorrect option: D. $0.96$
We have,
$P(A) = 0.4, P(B) = 0.8$ and $P(B|A) = 0.6$
As$, P(B|A) = 0.6$
$\Rightarrow\ \frac{\text{P}(\text{A}\cap\text{B})}{\text{P(A)}}=0.6$
$\Rightarrow\ \text{P}(\text{A}\cap\text{B})=0.6\times\text{P(A)}$
$\Rightarrow\ \text{P}(\text{A}\cap\text{B})=0.6\times0.4$
$\Rightarrow\ \text{P}(\text{A}\cap\text{B})=0.24$
Now, $\text{P}(\text{A}\cup\text{B})=\text{P(A)}+\text{P(B)}-\text{P}(\text{A}\cap\text{B})$
$=0.4+0.8-0.24$
$=1.2-0.24$
$=0.96$
Hence, the correct alternative is option $(d).$
View full question & answer→MCQ 1391 Mark
Three faces of aj ordinary dice are yellow, two faces are red and one face is blue. The dice is rolled $3$ times. The probability that yellow red and blue face appear in the first second and third throws respectively, is
- ✓
$\frac{1}{36}$
- B
$\frac{1}{6}$
- C
$\frac{1}{30}$
- D
AnswerCorrect option: A. $\frac{1}{36}$
$P($yellow face$) =\frac{3}{6}=\frac{1}{2}$
$P($red face$) =\frac{2}{6}=\frac{1}{3}$
$P($one face$) =\frac{1}{6}$
$P($yellow face, red face and blue face appear in the required order$) =\frac{1}{2}\times\frac{1}{3}\times\frac{1}{6}=\frac{1}{36}$
View full question & answer→MCQ 1401 Mark
Choose the correct answer from the given four options.Eight coins are tossed together. The probability of getting exactly $3$ heads is:
- A
$\frac{1}{256}$
- ✓
$\frac{7}{32}$
- C
$\frac{5}{32}$
- D
$\frac{3}{32}$
AnswerCorrect option: B. $\frac{7}{32}$
We know that, probaility distribution $\text{P}(\text{X}=\text{r})={^\text{n}\text{C}}_\text{r}(\text{P})^{\text{r}}\text{q}^{\text{n}-\text{r}}$
Here, $\text{n}=8,\text{r}=3,\text{p}=\frac{1}{2}$ and $\text{q}=\frac{1}{2}$
$\therefore$ Reuaired probability $={^8}\text{C}_3\Big(\frac{1}{2}\Big)^3\Big(\frac{1}{2}\Big)^{8-3}=\frac{8!}{5!3!}\Big(\frac{1}{2}\Big)^8 $
$=\frac{8\cdot7\cdot6}{3\cdot2}\cdot\frac{1}{2^8}=\frac{7}{32}$
View full question & answer→MCQ 1411 Mark
The probability that a leap year will have $53$ sundays is:
- A
$\frac17$
- ✓
$\frac27$
- C
$\frac57$
- D
$\frac67$
AnswerCorrect option: B. $\frac27$
A leap year has $52$ weeks and $2$ days.
The $53^{rd}$ Sunday will be from these extra two days
These $2$ days can be $($Sunday, Monday$)$ or $($Mon, Tue$)$ or $($Tue, Wed$).....($Sat, Sun$)$
There are $7$ possibilities for these $2$ days
Out of which Sunday is coming in $2$ possibilities.
$\therefore P(2$ sundays in leap year$) =\frac27$
View full question & answer→MCQ 1421 Mark
Three integers are chosen at random from the first $20$ integers. The probability that their product is even is,
- A
$\frac{2}{19}$
- B
$\frac{3}{29}$
- ✓
$\frac{17}{19}$
- D
$\frac{4}{19}$
AnswerCorrect option: C. $\frac{17}{19}$
Required probability that product of two integers should be even.
$10$ integers are odd out of first $20$ integers.
Required probability $= 1 -$ Probability of product is odd
Product of three integers is odd if two numbers are odd
Required probability $=1-\frac{10}{20}\times\frac{9}{19}\times\frac{8}{18}=\frac{17}{19}$
View full question & answer→MCQ 1431 Mark
Choose the correct answer from the given four options.If $\text{P}(\text{A})=\frac{4}{5},$ and $\text{P}(\text{A}\cap\text{B})=\frac{7}{10},$ then $\text{P}\Big(\frac{\text{B}}{\text{A}}\Big)$ is equal to:
- A
$\frac{1}{10}$
- B
$\frac{1}{8}$
- ✓
$\frac{7}{8}$
- D
$\frac{17}{20}$
AnswerCorrect option: C. $\frac{7}{8}$
$\text{P}(\text{A})=\frac{4}{5},\ \text{P}(\text{A}\cap\text{B})=\frac{7}{10}$
$\therefore\text{P}\Big(\frac{\text{B}}{\text{A}}\Big)=\frac{\text{P}(\text{A}\cap\text{B})}{\text{P}(\text{A})}$
$=\frac{\frac{7}{10}}{\frac{4}{5}}=\frac{7}{8}$
View full question & answer→MCQ 1441 Mark
The probability that an automobile will be stolen and found within one week is $0.0006.$ The probability that an automobile will be stolen is $0.0015.$ The probability that a stolen automobile will be found in one week is:
AnswerLet $P(S)$ be the probability of automobile stolen.
And $P(F)$ be the probability of automobile found.
According to the question,
$\text{P(S∩F) = 0.0006, P(S) = 0.0015}$
We know,
$\text{P}\Big(\frac{\text{F}}{\text{S}}\Big)=\frac{\text{P(F}\cap\text{S)}}{\text{P(S)}}=\frac{0.0006}{0. 0015}=0.4$
View full question & answer→MCQ 1451 Mark
A four$-$digit number is formed by using the digits $1, 2, 4, 8$ and $9$ without repitition. If one number is selected from those numbers, then what is the probability that it will be an odd number?
- A
$\frac15$
- ✓
$\frac25$
- C
$\frac35$
- D
$\frac45$
AnswerCorrect option: B. $\frac25$
Total number of outcomes $= 5 \times 4 \times 3 \times 2 = 120$
The number of favourable cases $= 2(4 \times 3 \times 2) = 48 ($i.e., odd numbers$)$
Therefore,
Required probability $\frac{48}{120}=\frac25$
View full question & answer→MCQ 1461 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. $49, 50$
When a coin is tossed $\text{p = q}=\frac{1}{2}$
$\Rightarrow\text{P(X = r})^{\text{ }^{\text{n}}}\text{C}_{\text{r}}\times0.5^{\text{n}}$
Coin is tossed $99$ times.
For odd number of $n$ maximum terms at
$\text{r}=\frac{\text{n}-1}{2}$ and $\text{r}=\frac{\text{n}+1}{2}$
$\text{n}=99$
$\Rightarrow\text{r}=49$ or $50$
View full question & answer→MCQ 1471 Mark
A speaks truth in $75\%$ cases and $B$ seaks truth in $80\%$ cases. Probability that they contradict each other in a statement, is
- ✓
$\frac{7}{20}$
- B
$\frac{13}{20}$
- C
$\frac{3}{5}$
- D
$\frac{2}{5}$
AnswerCorrect option: A. $\frac{7}{20}$
$P(A$ speaks truth$) = 0.75$
$P(A$ lies$) = 1 - 0.75 = 0.25$
$P(B$ speaks truth$) = 0.8$
$P(B$ lies$) = 1 - 0.8 = 0.2$
$P($contradicting each other in a statement$) = P(A$ speaks truth and lies$) + P(B$ speaks truth and $A$ lies$)$
$= 0.75 \times 0.2 + 0.8 \times 0.25$
$= 0.15 + 0.2$
$= 0.35$
$=\frac{35}{100}$
$=\frac{7}{20}$
View full question & answer→MCQ 1481 Mark
Choose the correct answer from the given four options.If $\text{P}(\text{A}\cap\text{B})=\frac{7}{10},$ and $\text{P}(\text{B})=\frac{17}{20},$ then $\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)$ equas:
- ✓
$\frac{14}{17}$
- B
$\frac{17}{20}$
- C
$\frac{7}{8}$
- D
$\frac{1}{8}$
AnswerCorrect option: A. $\frac{14}{17}$
Here, $\text{P}(\text{A}\cap\text{B})=\frac{7}{10}$ and $\text{P}(\text{B})=\frac{17}{20}$
$\therefore\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{\text{P}(\text{A}\cap\text{B})}{\text{P}(\text{B})}$
$=\frac{\frac{7}{10}}{\frac{17}{20}}=\frac{14}{17}$
View full question & answer→MCQ 1491 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
$\frac{1}{5}$
- ✓
$\frac{1}{3}$
- D
$\frac{2}{3}$
AnswerCorrect option: C. $\frac{1}{3}$
View full question & answer→MCQ 1501 Mark
A random variable $X$ has the following probability distribution:
| $X:$ |
$1$ |
$2$ |
$3$ |
$4$ |
$5$ |
$6$ |
$7$ |
$8$ |
| $P(X):$ |
$0.15$ |
$0.23$ |
$0.12$ |
$0.10$ |
$0.20$ |
$0.08$ |
$0.07$ |
$0.05$ |
Find the events $E = \{X : X$ is a prime number$\}, F\{X :X < 4\},$ the probability $\text{P}(\text{E}\cup\text{F})$ is: - ✓
$0.50$
- B
$0.77$
- C
$0.35$
- D
$0.87$
AnswerCorrect option: A. $0.50$
$P(E) = P(2) + P(3) + P(5) + P(7)$
$P(E) = 0.23 + 0.12 + 0.20 + 0.07$
$P(E) = 0.62$
And
$P(F) = P(1) + P(2) + P(3)$
$P(F) = 0.15 + 0.23 + 0.12$
$P(F) = 0.5$
Also,
$\text{P}(\text{E}\cap\text{F})=\text{P}(2)+\text{P}(3)$
$\text{P}(\text{E}\cap\text{F})=0.23+0.12$
$\text{P}(\text{E}\cap\text{F})=0.35$
$\text{P}(\text{E}\cup\text{F})=\text{P}(\text{E})+\text{P(F)}-\text{P}(\text{E}\cap\text{F})$
$\text{P}(\text{E}\cup\text{F})=0.62+0.5-0.35$
$\text{P}(\text{E}\cup\text{F})=0.77$
View full question & answer→