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.
Answer$\text{P}(\text{A}\cup\text{B})=\text{P(A)}+\text{P(B)}-\text{P}(\text{A}\cap\text{B})$
$=0.25+0.5-0.14$
$0.61$
P(Both A and B not happening) $=\text{P}(\text{A}\cup\text{B})'$
$=1-\text{P}(\text{A}\cup\text{B})$
$=1-0.61$
$=0.39$
View full question & answer→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
$= P\text{(WWB) + P(WBW) + P(BWW)}$
$=\frac{3}{4}\times\frac{2}{4}\times\frac{3}{4}+\frac{3}{4}\times\frac{2}{4}\times\frac{1}{4}+\frac{1}{4}\times\frac{2}{4}\times\frac{1}{4}$
$=\frac{18+6+2}{64}=\frac{13}{32}$
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, 4white 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?
AnswerThe two balls selected can be represented as 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
X(BR) = 1
X(RB) = 1
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-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 × 100 + 2 × 150 + 200 = 800
Number of Year-III's names = 3 × 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
AnswerS = [(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→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.
$\text{p}=\frac{1}{2}\Rightarrow\text{q}=\frac{1}{2}\dots(1)$
$\text{P(X}=7)=\text{P(X}=9)$
$\text{ }^{\text{n}}\text{C}_7\text{p}^7\text{q}^{\text{n}-7}=\text{ }^{\text{n}}\text{C}_9\text{p}^9\text{q}^{\text{n}-9}$
$\frac{\text{ }^{\text{n}}\text{C}_7}{\text{ }^{\text{n}}\text{C}_9}=\frac{\text{q}^{\text{n}-9}}{\text{q}^{\text{n}-7}}\times\frac{\text{p}^9}{\text{p}^7}$
$\frac{\frac{\text{n}!}{7!(\text{n}-7)!}}{\frac{\text{n}!}{9!(\text{n}-9)!}}=\text{q}^{-2}\text{p}^2$
$\frac{9!(\text{n}-9)!}{7!(\text{n}-7)!}=\frac{\text{p}^2}{\text{q}^2}$
$\frac{9\times8\times7!(\text{n}-9)!}{7!(\text{n}-7)(\text{n}-8)(\text{n}-9)!}=1\dots\big[\because\text{from (1)}\big]$
$9\times8=(\text{n}-7)(\text{n}-8)$
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→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
$\frac{135}{392}$
- ✓
$\frac{15}{56}$
- D
$\frac{15}{29}$
AnswerCorrect option: C. $\frac{15}{56}$
Probability 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
$∴ \text{A ∩ B} =0.$
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
$\frac{15}{12}$
- ✓
$\frac{7}{2}$
AnswerCorrect option: D. $\frac{7}{2}$
We 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
$\frac{1}{2}$
- ✓
$\frac{1}{2}$
AnswerCorrect option: D. $\frac{1}{2}$
$\text{P(B)}=\frac{3}{5},\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{1}{2},\text{P}(\text{A}\cup\text{B})=\frac{4}{5}$
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) =
- A
$\frac{1}{4}$
- B
$\frac{1}{2}$
- ✓
$\frac{3}{4}$
- D
$\frac{3}{8}$
AnswerCorrect option: C. $\frac{3}{4}$
$\text{P(A)}=\frac{1}{3}\text{P(A)}$
$\Rightarrow\ \text{P(B)}=3\text{P(A)}\ .....(\text{i})$
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 $1.$
$\therefore P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) +P(X = 6) + P(X = 7) +P(X = 8) = 1$
$\Rightarrow 0 + 2p + 2p + 3p + p^2 + 2p^2 + 7p^2 + 2p = 1$
$\Rightarrow 10p^2+ 9p - 1 = 0$
$\Rightarrow (10p - 1)(p + 1) = 0$
$\Rightarrow\text{p}=\frac{1}{10}\text{ or }-1 ($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 even and 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. $\frac{4}{7}$
$S = \{\text{GBB, GGB, GBG, GGG, BGG, BGB, BBG, BBB}\}$
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.
$E_1 =\{\text{GBB, GGB, GBG, GGG}\}$
$E_2 = \{\text{GBB, GGB, GBG, GGG, BGG, BGB, BBG}\}$
$\text{n}(\text{E}_1)=4,\text{n}(\text{E}_2)=7,\text{n}(\text{A}\cap\text{B})=4$
$\Rightarrow\ \text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{\text{n}(\text{A}\cap\text{B})}{\text{n(B)}}=\frac{4}{7}$
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 $=\frac 18$
$∴ $ P(not white) $= \frac{18}{10} = 95$
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$\text{p}=\frac{1}{4},\sqrt{\text{npq}}=3$
$\Rightarrow\text{q}=\frac{3}{4},\text{npq}=9$
$\Rightarrow\text{Mean = np}=\frac{9}{\text{q}}$
$\Rightarrow\text{Mean}=9\times\frac{4}{3}=12$
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 $\text{P}(\text{A}\cup\text{B})=$
AnswerWe 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 × 4 × 3 × 2 = 120
The number of favourable cases = 2(4 × 3 × 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:
AnswerWhen 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 $\text{r}=\frac{\text{n}+1}{2}$
$\text{n}=99\Rightarrow\text{r}=49 \text{ or }50$
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 × 0.2 + 0.8 × 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. $\frac{1}{3}$
$\frac{1}{3}$
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: AnswerP(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→