Question 1011 Mark
Choose the correct answer from the given four options.
A die is thrown and a card is selected at random from a deck of $52$ playing cards. The probability of getting an even number on the die and a spade card is:
AnswerLet $E_{1 }=$ Event for getting an even number on the die
And $E_{2 }=$ Event that a spade card is selected.
$\therefore\text{P}(\text{E}_1)=\frac{3}{6}=\frac{1}{2}$ and $\text{P}(\text{E}_2)=\frac{13}{52}=\frac{1}{4}$
Then, $\text{P}(\text{E}_1\cap\text{E}_2)=\text{P}(\text{E}_1)\cdot\text{P}(\text{E}_2)$
$=\frac{1}{2}\cdot\frac{1}{4}=\frac{1}{8}$
View full question & answer→Question 1021 Mark
Choose the correct answer from the given four options.
If A and B are two events such that $\text{P}(\text{A})=\frac{1}{2},\text{P}(\text{B})=\frac{1}{3},$ $\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{1}{4},$ then $\text{P}(\text{A}'\cap\text{B}')$ equals:
Answer
- $\frac{1}{4}$
Solution:
We have, $\text{P}(\text{A})=\frac{1}{2},\text{P}(\text{B})=\frac{1}{3}$ and $\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{1}{4}$
$\Rightarrow\text{P}(\text{A}\cap\text{B})=\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)\cdot\text{P}(\text{B})$
$=\frac{1}{4}\cdot\frac{1}{3}=\frac{1}{12}$
Now, $\text{P} ({\text{A}'}\cap{\text{B}'})=1-\text{P}(\text{A}\cup{\text{B}})$
$=1-\big[\text{P}(\text{A})+\text{P}(\text{B})-\text{P}(\text{A}\cap\text{B})\big]$
$=1-\Big[\frac{1}{2}+\frac{1}{3}-\frac{1}{12}\Big]=1-\frac{9}{12}$
$=\frac{3}{12}=\frac{1}{4}$ View full question & answer→Question 1031 Mark
Choose the correct answer from the given four options.
You are given that A and B are two events such that $\text{P}(\text{B})=\frac{3}{5},\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{1}{2}$ and $\text{P}(\text{A}\cup\text{B})=\frac{4}{5},$ then P(A) equals:
Answer
- $\frac{1}{2}$
Solution:
We have, $\text{P}(\text{B})=\frac{3}{5},\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{1}{2}$ and $\text{P}(\text{A}\cup\text{B})=\frac{4}{5}$
$\therefore\text{P}(\text{A}\cap\text{B})=\Big(\frac{\text{A}}{\text{B}}\Big)\cdot\text{P}(\text{B})$
$=\frac{1}{2}\cdot\frac{3}{5}=\frac{3}{10}$
Now $\text{P}(\text{A}\cup\text{B})=\text{P}({\text{A}})+\text{P}({\text{B}})\cdot\text{P}(\text{A}\cap\text{B})$
$\Rightarrow\frac{4}{5}=\text{P}(\text{A})+\frac{3}{5}-\frac{3}{10}$
$\therefore\text{P}(\text{A})=\frac{4}{5}-\frac{3}{5}+\frac{3}{10}=\frac{1}{2}$ View full question & answer→Question 1041 Mark
India play two matches each with West indies and Australia. In any match the probability of india getting 0,1 and 2 points are 0.45, 0.05 and 0.50 respectively. Assuming that the outcomes are indepecdent, the probability of india getting at least 7 points is.
Answer
- 0.0875
Solution:
Here, there are total 5 ways by which India can get at least 7 points.
- 2 points + 2 points + 2 points + 2 points = (0.5 × 0.5 × 0.5 × 0.5)
- 1 points + 2 points + 2 points + 2 points = (0.05 × 0.5 × 0.5 × 0.5)
- 2 points + 1 points + 2 points + 2 points = (0.5 × 0.05 × 0.5 × 0.5)
- 2 points + 2 points + 1 points + 2 points = (0.5 × 0.5 × 0.05 × 0.5)
- 2 points + 2 points + 2 points + 1 points = (0.5 × 0.5 × 0.5 × 0.05)
View full question & answer→Question 1051 Mark
The probability that in a year of $22^{nd}$ century chosen at random, there will be $53$ Sunday, is
AnswerWe know a leap year is fallen within $4$ years, So its probability is $\frac{25}{100}=\frac{1}{4}$
$53^{rd}$ Sunday leap year $=\frac{1}{4}\times\frac{2}{7}=\frac{2}{28}$
Similarly probability of $53^{rd}$ Sunday in a non leap year $=\frac{75}{100}\times\frac{1}{7}=\frac{3}{4}\times\frac{1}{7}=\frac{3}{28}$
Required probability $=\frac{2}{28}+\frac{3}{28}=\frac{5}{28}$.
View full question & answer→Question 1061 Mark
Choose the correct answer from the given four options. Two dice are thrown. If it is known that the sum of numbers on the dice was less than $6,$ the probability of getting a sum $3,$ is:
AnswerLet $E_1 =$ Event that the sum of numbers on the dice was less than $6$
And $E_2=$ Event that the sum of numbers on the dice is $3.$
$\therefore E_1 = \{(1, 4), (4, 1), (2, 3), (3, 2), (2, 2), (1, 3), (3, 1), (1, 2), (2, 1), (1, 1)\}$
$\Rightarrow n(E_1) = 10$
And $E_2 = \{(1, 2), (2, 1)\}$
$\Rightarrow n(E_2) = 2$
$\therefore$ Required Probability $=\frac{2}{10}=\frac{1}{5}$
View full question & answer→Question 1071 Mark
A box contains 3 orange balls, 3 green balls and 2 blue balls. Three balls are drawn at random from the box without replacement. The probability of drawing 22 green balls and one blue ball is
Answer
- $\frac{3}{28}$
Solution:
Total balls in a box - 3orange + 3green + 2blue = 8
Three balls are drawn at random from the box then samplw space $\text{n(S)}= {^{8}}\text{C}_3=\frac{8\times7\times6}{3\times2\times1}=56$
Let A be the event that drawing 2 green and one blue ball.
$\text{n(A)}={^{3}}\text{C}_2\times{^{2}}\text{C}_2=6$
$\text{P(A)}=\frac{6}{56}=\frac{3}{28}$ View full question & answer→Question 1081 Mark
A bouquet from $11$ different flowers is to be made so that it contains not less then three flowers. Then the number of the different ways of selecting flowers to from the bouquet.
AnswerNo. of ways $= ^{11}C_3 + ^{11}C_4 + ^{11}C_5 + ^{11}C_6 + ^{11}C_7 + ^{11}C_8 + ^{11}C_9 + ^{11}C_{10} + ^{11}C_{11}$
$\Rightarrow 165 + 330 + 462 + 462 + 330 + 165 + 55 + 11 + 1$
$\Rightarrow 1981$
View full question & answer→Question 1091 Mark
A box contains 6 nails and 10 nuts. Half of the nails and half of the nuts are rusted. If one iten is chosen ar random, the probability that it is rusted or is nail is
Answer
- $\frac{11}{16}$
Solution:
Rusted items = 3 + 5 = 8
Rusted nails = 3
Total nails = 6
P(getting a rusted item or a nail) = P(getting a rusted item) + P(getting a nail) - P(getting a rusted item and a nail)
$=\frac{8}{16}+\frac{6}{16}-\frac{3}{16}$
$=\frac{8+6-3}{16}$
$=\frac{11}{16}$ View full question & answer→Question 1101 Mark
Let X denote the number of times heads occur in n tosses of a fair coin. If P(X = 4), P(X = 5) and P(X = 6) are in AP, the value of n is:
Answer
- 7, 14
Solution:
X denotes the number of times heads occurs.
P(X = 4),P(X = 5),P(X = 6) are in AP
$\Rightarrow2\text{P(X = 4),P(X = 5),P(X = 6)}$
$\Rightarrow2\text{ }^{\text{n}}\text{C}_5\big(\frac{1}{2}\big)^5\big(\frac{1}{2}\big)^{\text{n}-5}=\text{ }^{\text{n}}\text{C}_4\big(\frac{1}{2}\big)^4\big(\frac{1}{2}\big)^{\text{n}-4}\times\text{ }^{\text{n}}\text{C}_6\big(\frac{1}{2}\big)^6\big(\frac{1}{2}\big)^{\text{n}-6}$
$\Rightarrow2\text{ }^{\text{n}}\text{C}_5\big(\frac{1}{2}\big)^{\text{n}}=\text{ }^{\text{n}}\text{C}_4\big(\frac{1}{2}\big)^{\text{n}}+\text{ }^{\text{n}}\text{C}_6\big(\frac{1}{2}\big)^{\text{n}}$
$\Rightarrow2\text{ }^{\text{n}}\text{C}_5=\text{ }^{\text{n}}\text{C}_4+\text{ }^{\text{n}}\text{C}_6$
$\Rightarrow\frac{2\text{n}!}{5!(\text{n}-5)!}=\frac{\text{n}!}{4!(\text{n}-4)!}+\frac{\text{n}!}{6!(\text{n}-6)!}$
$\Rightarrow\frac{2}{5\times4!(\text{n}-5)(\text{n}-6)!}=\frac{1}{4!(\text{n}-4)(\text{n}-5)(\text{n}-6)!}+\frac{1}{6\times5\times4!(\text{n}-6)!}$
$\Rightarrow\frac{2}{5(\text{n}-5)}=\frac{1}{(\text{n}-4)(\text{n}-5)}+\frac{1}{6\times5}$
$\Rightarrow\frac{2}{5(\text{n}-5)}=\frac{30+(\text{n}-4)(\text{n}-5)}{30(\text{n}-4)(\text{n}-5)}$
$\Rightarrow12(\text{n}-4)=30+(\text{n}-4)(\text{n}-5)$
$\Rightarrow12(\text{n}-4)-(\text{n}-4)(\text{n}-5)=30$
$\Rightarrow(\text{n}-4)(12-\text{n}+5)=30$
$\Rightarrow(\text{n}-4)(17-\text{n})=30$
Check with options by putting value of n.
$\Rightarrow\text{n}=7,14$ View full question & answer→Question 1111 Mark
If A and B are two events associated to a random experiment such that $\text{P}(\text{A}\cap\text{B})=\frac{7}{10}$ and $\text{P(B)}=\frac{17}{20}$, then P(A|B) =
Answer
- $\frac{14}{17}$
Solution:
$\text{P}(\text{A}\cap\text{B})=-\frac{7}{10},\text{P(B)}=\frac{17}{20}$
$\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{\text{P}(\text{A}\cap\text{B})}{\text{P(B)}}$
$\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{\frac{7}{10}}{\frac{17}{20}}=\frac{14}{17}$ View full question & answer→Question 1121 Mark
A flash light has 8 batteries out of which 3 are dead. IF two batteries are selected without replacement and tested, then the probability that both are dead is,
Answer
- $\frac{3}{28}$
Solution:
We have,
The total number of batteries = 8
The number of dead batteries = 3
Let A be the event of selecting the first dead battery and B be the event of selecting the second dead battery.
Now,
P(both dead batteries are selected) $=\text{P}(\text{A}\cap\text{B})$
$=\text{P(A)}\times\text{P}(\text{B}|\text{A})$
$=\frac{3}{8}\times\frac{2}{7}$
$=\frac{3}{28}$
Hence, the correct alternative is option (a). View full question & answer→Question 1131 Mark
Choose the correct answer from the given four options.
A box contains 3 orange balls, 3 green balls and 2 blue balls. Three balls are drawn at random from the box without replacement. The probability of drawing 2 green balls and one blue ball is:
Answer
- $\frac{3}{28}$
Solution:
Probability of drawing 2 green balls and one blue ball
$=\text{P}_\text{G}\cdot\text{P}_\text{G}\cdot\text{P}_\text{B}+\text{P}_\text{B}\cdot\text{P}_\text{G}\cdot\text{P}_\text{G}+\text{P}_\text{G}\cdot\text{P}_\text{B}\cdot\text{P}_\text{G}$
$=\frac{3}{8}\cdot\frac{2}{7}\cdot\frac{2}{6}+\frac{2}{8}\cdot\frac{3}{7}\cdot\frac{2}{6}+\frac{3}{8}\cdot\frac{2}{7}\cdot\frac{2}{6}$
$=\frac{1}{28}+\frac{1}{28}+\frac{1}{28}=\frac{3}{28}$ View full question & answer→Question 1141 Mark
If A and B are two events such that $\text{P(A)}=\frac{4}{5},$ and $\text{P}(\text{A}\cap\text{B})=\frac{7}{10},$ then P(B|A) =
Answer
- $\frac{7}{8}$
Solution:
We have,
$\text{P(A)}=\frac{4}{5}$ and $\text{P}(\text{A}\cap\text{B})=\frac{7}{10}$
Now,
$\text{P}(\text{B}|\text{A})=\frac{\text{P}(\text{A}\cap\text{B})}{\text{P(A)}}$
$=\frac{\Big(\frac{7}{10}\Big)}{\Big(\frac{4}{5}\Big)}$
$=\frac{7\times5}{10\times4}$
$=\frac{7}{8}$
Hence, the correct alternative is option (c). View full question & answer→Question 1151 Mark
If the events A and B are independent, then $\text{P}(\text{A}\cap\text{B})$ is equal to,
Answer
- P(A) P(B)
Solution:
$\text{P}(\text{A}\cap\text{B})=\text{P(A)} \text{ P(B)}$ for independent events. View full question & answer→Question 1161 Mark
A box contains 10 good articles and 6 with defects. One item is drawn at random. The probability that it is either good or has a defect is,
Answer
- $\frac{64}{64}$
Solution:
P(good item) $=\frac{10}{16}$
P(defected item) $=\frac{6}{16}$
P(eitherr good or defected item) = P(good item) + P(defected item)
$=\frac{10}{16}+\frac{6}{16}$
$=\frac{16}{16}$
$=1$
$=\frac{64}{64}$ View full question & answer→Question 1171 Mark
If A and B are two independent events such that P(A) = 0.3 and $\text{P}(\text{A}\cup\text{B})=0.5,$ then P(A|B) - P(B|A) =
Answer
- $\frac{1}{70}$
Solution:
We have,
$\text{P(A)}=0.3$ and $\text{P}(\text{A}\cup\text{B})=0.5$
As, A and B are independent events
So, $\text{P}(\text{A}\cap\text{B})=\text{P(A)}\times\text{P(B)}$
$=0.3\times\text{P(B)}$
$=0.3\text{ P(B)}\ .....\text{(i)}$
Also, $\text{P}(\text{A}\cup\text{B})=\text{P(A)}+\text{P(B)}-\text{P}(\text{A}\cap\text{B})$
$\Rightarrow 0.5 = 0.3+\text{P(B)}-0.3\text{ P(B)}$ [Using (i)]
$\Rightarrow 0.5-0.3 = 0.7\text{ P(B)}$
$\Rightarrow0.7\text{ P(B)}=0.2$
$\Rightarrow\text{ P(B)}=\frac{0.2}{0.7}$
$\Rightarrow\text{ P(B)}=\frac{2}{7}$
Using (i), we get
$\text{P}(\text{A}\cap\text{B})=0.3\times\frac{2}{7}=\frac{6}{70}$
Now,
$\text{P}(\text{A}|\text{B})-\text{P}(\text{B}|\text{A})=\frac{\text{P}(\text{A}\cap\text{B})}{\text{P(B)}}-\frac{\text{P}(\text{A}\cap\text{B})}{\text{P(A)}}$
$=\frac{\Big(\frac{6}{70}\Big)}{\Big(\frac{2}{7}\Big)}-\frac{\Big(\frac{6}{70}\Big)}{0.3}$
$=\frac{6\times7}{70\times2}-\frac{6}{70\times0.3}$
$=\frac{3}{10}-\frac{2}{7}$
$=\frac{21-20}{70}$
$=\frac{1}{70}$
Hence, the correct alternative is option (c). View full question & answer→Question 1181 Mark
A bag contain 4 white and 2 black balls. Two balls are drawn at random. The probability that they are of the same colour is ________.
Answer
- $\frac{7}{15}$
Solution:
We assume that there are 4 white balls and 2 black balls.
There are $\big(\frac{6}{2}\big)=15$ total possible ways of drawing two balls from these given 6 balls.
We are interested in the event where the two drawn balls are of the same colour.
For this, we note that the number of ways of drawing 2 white balls is $\big(\frac{4}{2}\big)=6$ whereas the number of ways of drawing 2 black balls is$\big(\frac{2}{2}\big)=1.$
So, the probability that the two drawn balls are of the same colour is $\frac{6+1}{15}=\frac{7}{15}.$ View full question & answer→Question 1191 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
AnswerThe 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→Question 1201 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→Question 1211 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→Question 1221 Mark
If A and B are two events such that $\text{A}\subset\text{B}$ and $\text{P}(\text{B})\neq0,$ then which of the following is correct?
Answer$\text{A}\subset\text{B}\ \Rightarrow\ \ \text{A}\cap\text{B}=\text{A P}\ \ \text{and}\ \ \text{P}(\text{B})\neq0$
$\Rightarrow\ \text{P}(\text{A}|\text{B})=\frac{\text{P}(\text{A}\cap\text{B})}{\text{P}(\text{B})}=\frac{\text{P}(\text{A})}{\text{P}(\text{B})}$
Since $\text{P}(\text{B})\neq\ 0$
$\therefore\ \frac{\text{P}(\text{A})}{\text{P}(\text{B})}<1\ \ \ \ \ \Rightarrow\ \text{P}(\text{A})<\text{P}(\text{B})\ \Rightarrow\ \text{P}(\text{A}|\text{B})\geq\text{P}(\text{A})$
Hence, option (C) is correct.
View full question & answer→Question 1231 Mark
In each of the following choose the correct answer:If A and B are events such that $\text{P}(\text{A}|\text{B})=\text{P}(\text{B}|\text{A}),\ \text{then}:$
Answer$\text{P}(\text{A}|\text{B})=\text{P}(\text{B}|\text{A})$ $\Rightarrow\ \frac{\text{P}(\text{A}\cap\text{B})}{\text{P}(\text{B})}=\frac{\text{P}(\text{B}\cap\text{A})}{\text{P}(\text{A})}$$\Rightarrow\ \text{P}(\text{A})=\text{P}(\text{B})$
Therefore, option (D) is correct.
View full question & answer→Question 1241 Mark
A bag contains 5 brown and 4 white socks. A man pulls out two socks. The probability that these are of the sane colour is.
Answer
- $\frac{48}{108}$
Solution:
Total number of balls = 5brown + 4white = 9
Required probability $=\frac{5}{9}\times\frac{4}{8}+\frac{4}{9}\times\frac{3}{8}=\frac{4}{9}$
$\Rightarrow\ \frac{4\times12}{9\times12}=\frac{48}{108}$ View full question & answer→Question 1251 Mark
Choose the correct answer in each of the following:
The mean of the numbers obtained on throwing a die having written 1 on three faces, 2 on two faces and 5 on one face is
Answer
| $\text{x}_i$ |
$\text{p}_i$ |
$\text{p}_i\text{x}_i$ |
$1$
$2$
$5$ |
$\frac{3}{6}$
$\frac{2}{6}$
$\frac{1}{6}$ |
$\frac{3}{6}$
$\frac{4}{6}$
$\frac{5}{6}$ |
| |
|
$\sum\text{p}_i\text{x}_i=\frac{12}{6}=2$ |
Therefore, option (B) is correct. View full question & answer→Question 1261 Mark
Choose the correct answer from the given four options.
Two events E and F are independent. If $\text{P}(\text{E})=0.3,\text{P}(\text{E}\cup\text{F})=0.5,$ then $\text{P}\Big(\frac{\text{E}}{\text{F}}\Big)-\text{P}\Big(\frac{\text{F}}{\text{E}}\Big)$ equal:
Answer
- $\frac{1}{70}$
Solution:
We have, $\text{P}(\text{E})=0.3,\text{P}(\text{E}\cup\text{F})=0.5$
Also E and F are independent.
Now $\text{P}(\text{E}\cup\text{F})=\text{P}(\text{E})+\text{P}(\text{F})-\text{P}(\text{E}\cap\text{F})$
$\Rightarrow0.5=0.3+\text{P}(\text{F})-0.3\text{P}(\text{F})$
$\Rightarrow\text{P}(\text{F})=\frac{0.5-0.3}{0.7}=\frac{2}{7}$
$\therefore\text{P}\Big(\frac{\text{E}}{\text{F}}\Big)-\text{P}\Big(\frac{\text{F}}{\text{E}}\Big)$
$=\text{P}(\text{E})-\text{P}(\text{F})$ (as E and F are indepandent)
$=\frac{3}{10}-\frac{2}{7}=\frac{1}{70}$ View full question & answer→Question 1271 Mark
A rifleman is firing at a distant target and has only 10% chance of hiting it. the least number of round he must fire in order to have more than 50% chance of hitting it at least once is:
Answer
- 7
Solution:
Given $\text{p}=\frac{1}{10}\Rightarrow\text{q}=\frac{9}{10}$
Let n be the number of rounds.
$\text{P(X}\geq1)=1-\text{P(X}=0)$
$\Rightarrow\text{P(X}\geq1)\geq0.5$
$\Rightarrow1-\text{P(X}=0)\geq0.5$
$\Rightarrow\text{P(X}=0)\leq0.5$
$\Rightarrow0.9^{\text{n}}\leq0.5$
Using log table,
$\text{n}\leq6.572\approx7$
He must fire in order to have more than
50% chance of hitting the target at least once. View full question & answer→Question 1281 Mark
A and B draw two cards each, one after another, from a pack of well-shuffled pack of 52 cards. The probability that all the four cards drawn are of the same suit is
Answer
- $\frac{44}{85\times49}$
Solution:
Total cards = 52 There are four suits of cards in a pack, i.e. diamond, heart, spade and club.
Pall 4 cards are of same suit = Pall 4 cards are of diamond + Pall 4 cards are of heart + Pall 4 cards are of spade + Pall 4 cards are of club.
$=4\times\frac{13}{52}\times\frac{12}{51}\times\frac{11}{50}\times\frac{10}{59}$
$=4\times\frac{11}{85\times49}$
$=\frac{44}{85\times49}$ View full question & answer→Question 1291 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?
Answer
- 0, 1, 2
Solution:
The 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→Question 1301 Mark
Choose the correct answer from the given four options.
In a college, 30% students fail in physics, 25% fail in mathematics and 10% fail in both. One student is chosen at random. The probability that she fails in physics if she has failed in mathematics is:
Answer
- $\frac{2}{5}$
Solution:
Here, $\text{P}_{(\text{Ph})}=\frac{30}{100}=\frac{3}{10}$
$\text{P}_{(\text{M})}=\frac{25}{100}=\frac{1}{4}$
And $\text{P}_{(\text{M}\cap\text{Ph})}=\frac{10}{100}=\frac{1}{10}$
$\therefore\text{P}\Big(\frac{\text{Ph}}{\text{M}}\Big)=\frac{\text{P}(\text{Ph}\cap\text{M})}{\text{P}(\text{M})}$
$=\frac{\frac{1}{10}}{\frac{1}{4}}=\frac{2}{5}$ View full question & answer→Question 1311 Mark
Choose the correct answer from the given four options.
| $\text{X}$ |
$1$ |
$2$ |
$3$ |
$4$ |
| $\text{P}(\text{X})$ |
$\frac{1}{10}$ |
$\frac{1}{5}$ |
$\frac{3}{10}$ |
$\frac{2}{5}$ |
For the following probability distribution $E(X^2)$ is equal to$:$ Answer$\text{E}(\text{X}^2)=\sum\text{X}^2\text{P}(\text{X})$
$=1\cdot\frac{1}{10}+4\cdot\frac{1}{5}+9\cdot\frac{3}{10}+16\cdot\frac{2}{5}$
$=\frac{1}{10}+\frac{4}{5}+\frac{27}{10}+\frac{32}{5}$
$=\frac{1+8+27+64}{10}$
$=10$
View full question & answer→Question 1321 Mark
Choose the correct answer from the given four options.
If A and B are two independent events with $\text{P}(\text{A})=\frac{3}{5}$ and $\text{P}(\text{A})=\frac{4}{9},$ then $\text{P}(\text{A'}\cap\text{B'})$ equals:
Answer
- $\frac{2}{9}$
Solution:
Since A and B are independent events, A' And B' are aslo independent.
$\therefore\text{P}(\text{A}'\cap\text{B}')=\text{P}(\text{A})\cdot\text{P}(\text{B})$
$=\big[1-\text{P}(\text{A})\big]\big[1-\text{P}(\text{B})\big]$
$=\Big(1-\frac{3}{5}\Big)\Big(1-\frac{4}{9}\Big)$
$=\frac{2}{5}\cdot\frac{5}{9}=\frac{2}{9}$ View full question & answer→Question 1331 Mark
Choose the correct answer from the given four options.
|
X
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-4
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-3
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-2
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-1
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0
|
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P(X)
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0.1
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0.2
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0.3
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0.2
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0.2
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For the following probability distribution E(X) is equal to:
Answer
- -1.8
Solution:
$\text{E}(\text{X})=\sum\text{P}(\text{X})$
$= -4\times(0.1)+(-3 \times0.2)+(-2\times0.3)+(-1\times0.2)+(0\times0.2)$
$=-0.4-0.6-0.6-0.2=-1.8$ View full question & answer→Question 1341 Mark
An urn contains 9 balls two of which are red, three blue and four black. Three balls are drawn at random. The probability that they are of the same colour is,
Answer
- $\frac{5}{84}$
Solution:
Given:
Red balls = 2
Blue balls = 3
Black balls = 4
P(All three balls are of same colour) = P(all three are blue) + P(all three are black)
$=\frac{3}{9}\times\frac{2}{8}\times\frac{1}{7}+\frac{4}{9}\times\frac{3}{8}\times\frac{2}{7}$
$=\frac{1}{84}+\frac{4}{84}$
$=\frac{5}{84}$ View full question & answer→Question 1351 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?
Answer
- $\frac{2}{5}$
Solution:
Total number of outcomes = 5 × 4 × 3 × 2 = 120
he number of favourable cases = 2 (4 × 3 × 2) - = 48 (i.e., odd numbers)
herefore,Required probability $=\frac{48}{120}=\frac{2}{5}.$ View full question & answer→Question 1361 Mark
Choose the correct answer from the given four options.
Suppose a random variable X follows the binomial distribution with parameters n and p, where 0 < p < 1. If $\frac{\text{P}(\text{x}=\text{r})}{\text{P}(\text{x}=\text{n}–\text{r})}$ is independent of n and r, then p equals:
Answer
- $\frac{1}{2}$
Solution:
$\text{P}(\text{x}=\text{r})={^\text{n}}\text{C}_\text{r}(\text{p})^\text{r}(\text{q})^{\text{n}-\text{r}}$
$=\frac{\text{n}!}{(\text{n}-\text{r})!\text{r}!}(\text{p})^\text{r}(1-\text{p})^{\text{n}-\text{r}}[\therefore\text{q}=1-\text{p}]$
Now, $\frac{\text{P}(\text{x}=\text{r})}{\text{P}(\text{x}=\text{n}-\text{r})}=\frac{{^\text{n}\text{C}_\text{r}\text{p}^\text{r}(1-\text{p})^{\text{n}-\text{r}}}}{{{^\text{n}}\text{C}}_{\text{n}-\text{r}}\text{p}^{\text{n}-\text{r}}(1-\text{p})^{\text{r}}}$
$=\frac{\text{P}^\text{r}(1-\text{P})^{\text{n}-\text{r}}}{\text{p}^{\text{n}-\text{r}}(1-\text{p})^\text{r}}$ $\big[\text{as}{^\text{n}}\text{C}_\text{r}={^\text{n}}\text{C}_{\text{n}-\text{r}}\big]$
$=\Big(\frac{1-\text{p}}{\text{p}}\Big)^{\text{n}-2\text{r}}$
Above expression is independent of n and r, if
$\frac{1-\text{p}}{\text{p}}=1\Rightarrow\text{p}=\frac{1}{2}$ View full question & answer→Question 1371 Mark
If A and B are two events such that $\text{P(A)}=\frac{3}{8},\text{P(B)}=\frac{5}{4}.$ and $\text{P}(\text{A}|\text{B})\times\text{P}(\overline{\text{A}}\cap\text{B})$ is equals to.
Answer
- $\frac{6}{25}$
Solution:
$\text{P(A)}=\frac{3}{8},\text{P(B)}=\frac{5}{8},\text{P}\Big({\text{A}}\cup{\text{B}}\Big)=\frac{3}{4}$
$\text{P}(\text{A}\cap\text{B})=\text{P(A)}+\text{P(B)}-\text{P}(\text{A}\cap\text{B})$
$\Rightarrow\ \frac{3}{4}=\frac{3}{8}+\frac{5}{8}-\text{P}(\text{A}\cap\text{B})$
$\Rightarrow\ \text{P}(\text{A}\cap\text{B})=\frac{1}{4}$
$\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)\text{ P}\Big(\frac{\overline{\text{A}}}{\text{B}}\Big)=\frac{\text{P}(\text{A}\cap\text{B})}{\text{P(B)}}\times\frac{\text{P}(\overline{\text{A}}\cap\text{B})}{\text{P(B)}}$
$\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)\text{ P}\Big(\frac{\overline{\text{A}}}{\text{B}}\Big)=\frac{\text{P}(\text{A}\cap\text{B})}{\text{P(B)}}\times\frac{\text{P(B)}-\text{P}(\text{A}\cap\text{B})}{\text{P(B)}}$
$\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)\text{ P}\Big(\frac{\overline{\text{A}}}{\text{B}}\Big)=\frac{\frac{1}{4}}{\frac{5}{8}}\times\frac{\big(\frac{5}{8}-\frac{1}{4}\big)}{\frac{5}{8}}$
$\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)\text{ P}\Big(\frac{\overline{\text{A}}}{\text{B}}\Big)=\frac{6}{25}$ View full question & answer→Question 1381 Mark
If the mean and variance of a binomial variate X are 2 and 1 respectively, then the probability that X takes a value greater than 1 is:
Answer
- $\frac{15}{16}$
Solution:
$\text{E(X)}=2, \text{V(X})=1$
$\text{np}=2,\text{npq}=1$
$\Rightarrow\text{q}=\frac{1}2{}=\text{p}$
$\Rightarrow\text{n}=4$
$\text{P(X}\geq1)=1-\text{P(X}=0)$
$\text{P(X}\geq1)=1-\big(\frac{1}{2}\big)^4$
$\text{P(X}\geq1)=1-\frac{1}{16}=\frac{15}{16}$ View full question & answer→Question 1391 Mark
A die is thrown and a card is selected ar random from a deck pf $52$ playing cards. The probability of getting an even number of the die and a spade card is
AnswerA Sample space when a die is thrown,
$S_1 = \{1, 2, 3, 4, 5, 6\} $
$\Rightarrow n(S_1) = 6$
Let $A$ be the event that getting even number.
$A = \{2, 4, 6\} $
$\Rightarrow n(A) = 3$
$\Rightarrow\ \text{P(A)}=\frac{3}{6}=\frac{1}{2}$
$A$ card is selected from a deck of $52$ cards.
$\text{n}(\text{S}_2)= {^{52}}\text{C}_2=52$
Let $B$ be the event that getting spade card.
$\text{n(B)}= {^{13}}\text{C}_2=13$
$\Rightarrow\ \text{P(B)}=\frac{13}{52}=\frac{1}{4}$
Required probability $= P(A) \times P(B)$
$=\frac{1}{2}\times\frac{1}{4}=\frac{1}{8}$
View full question & answer→Question 1401 Mark
Five persone entered the lift cabin on the ground floor of an $8$ floor house. Suppose that each of them independently and with equal probability can leave the cabin at any flor beginning with the first, then the probability of all $5$ persons leaving at different floors is,
AnswerFive persons can leave different floors
By $^7P_5$ ways.
Possible ways of leavinf the lift $= 7^5$
Required probability $=\frac{^{7}\text{P}_5}{7^5}$
View full question & answer→Question 1411 Mark
$A$ and $B$ are two students. Their chances of solving a problem correctly are $\frac{1}{3}$ and $\frac{1}{4}$ respectively. If the probability of their making common error is $\frac{1}{20}$ and they obtain the same answer, then the probability of their answer to be correct is.
AnswerLet $E_1$ be the event that Both $A$ and $B$ solve the problem.
$A$ and $B$ are independent,
$\Rightarrow\ \text{P}(\text{E}_1)=\text{P(A)}\times\text{P(B)}$
$\Rightarrow\ \text{P}(\text{E}_1)=\frac{1}{3}\times\frac{1}{4}=\frac{1}{12}$
Let $E_2$ both $A$ and $B$ got wrong solution.
$\text{P}(\text{E}_2)=\Big(1-\frac{1}{3}\Big)\times\Big(1-\frac{1}{4}\Big)=\frac{1}{2}$
Let $E$ be the event getting same answer.
$\text{P}\Big(\frac{\text{E}}{\text{E}_1}\Big)=1,\text{P}\Big(\frac{\text{E}}{\text{E}_2}\Big)=\frac{1}{20}$
$\Rightarrow\text{P}\Big(\frac{\text{E}_1}{\text{E}}\Big)=\frac{\frac{1}{12}\times1}{\frac{1}{12}\times1+\frac{1}{2}\times\frac{1}{20}}=\frac{10}{13}$
View full question & answer→Question 1421 Mark
Choose the correct answer from the given four options.
The probability of guessing correctly at least 8 out of 10 answers on a true-false type examination is:
Answer
- $\frac{7}{128}$
Solution:
We know that, $\text{P}(\text{x}=\text{r})={^\text{n}}\text{C}_\text{r}(\text{P}){^\text{r}.(\text{q})6^{\text{n}-\text{r}}}$
Here, $\text{n}=10,\text{p}=\frac{1}{2},\text{q}=\frac{1}{2}$
and $\text{r}\geq8\text{ i.e,}\text{ r}=8,9,10$
$\Rightarrow\text{P}(\text{X}=\text{r})=\text{P}(\text{r}=8)+\text{P}(\text{r}=9)+\text{P}(\text{r}=10) $
$={^{10}}\text{C}_8\Big(\frac{1}{2}\Big)^8\Big(\frac{1}{2}\Big)^{10-8}+{^{10}}\text{C}_9\Big(\frac{1}{2}\Big)^9\Big(\frac{1}{2}\Big)^{10-9}+{^{10}}\text{C}_{10}\Big(\frac{1}{2}\Big)^{10}\Big(\frac{1}{2}\Big)^{10-10}$
$=\Big(\frac{10!}{8!2!}+\frac{10!}{9!1!}+1\Big)\Big(\frac{1}{2}\Big)^{10}$
$=[45+10+1]\Big(\frac{1}{2}\Big)^{10}$
$=56\Big(\frac{1}{2}\Big)^{10}=\frac{7}{128}$ View full question & answer→Question 1431 Mark
Three persons, A, B and C fine a target in turn starting with A. Their probability of hitting the target are 0.4, 0.2 and 0.2, respectively. The probability of two hits is
Answer
- 0.188
Solution:
Let:
A be the event of hitting the target by the person A,
B be the event of hitting the target by the person B and
C be the event of hitting the target by the person C
We have,
P(A) = 0.4, P(B) = 0.3 and P(C) = 0.2
Also,
$\text{P}(\overline{\text{A}})=1-\text{P(A)}=1-0.4=-0.6,$
$\text{P}(\overline{\text{B}})=1-0.3=0.7$ and
$\text{P}(\overline{\text{C}})=1-0.2=0.8$
Now,
$\text{P(Two hits)}=\text{P}(\text{AB}\overline{\text{C}})+\text{P}(\text{A}\overline{\text{B}}\text{C})+\text{P}(\overline{\text{A}}\text{BC})$
$=\text{P(A)}\times\text{P(B)}\times\text{P}(\overline{\text{C}})+\text{P(A)}\times(\overline{\text{B}})\times\text{P(C)}\\+\text{P}(\overline{\text{A}})\times\text{P(B)}\times\text{P(C)}$
$=0.4\times0.3\times0.8+0.4\times0.7\times0.2+0.6\times0.3\times0.2$
$=0.096+0.056+0.036$
$=0.188$
Hence, the correct alternative is option (d). View full question & answer→Question 1441 Mark
If A and B are such that $\text{P}(\text{A}\cup\text{B})=\frac{5}{9}$ and $\text{P}(\overline{\text{A}}\cup\overline{\text{B}})=\frac{2}{3},$ then $\text{P}(\overline{\text{A}})+\text{P}(\overline{\text{B}})=$
Answer
- $\frac{10}{9}$
Solution:
$\text{P}(\text{A}\cup\text{B})=\frac{5}{9},\text{P}(\overline{\text{A}}\cup\overline{\text{B}})=\frac{2}{3}$
Consider,
$\text{P}(\overline{\text{A}}\cup\overline{\text{B}})=\text{P}(\overline{\text{A}\cup\text{B}})$
$\Rightarrow\text{P}(\overline{\text{A}}\cup\overline{\text{B}})=\frac{2}{3}$
$\Rightarrow 1-\text{P}(\text{A}\cap\text{B})=\frac{2}{3}$
$\Rightarrow\text{P}(\text{A}\cap\text{B})=\frac{1}{3}$
$\Rightarrow\ \text{P(A)}+\text{P(B)}-\text{P}(\text{A}\cup\text{B})=\frac{1}{3}$
$\Rightarrow\ \text{P(A)}+\text{P(B)}-\frac{5}{9}=\frac{1}{3}$
$\Rightarrow\ \text{P(A)}+\text{P(B)}=\frac{8}{9}$
$\text{P}(\overline{\text{A}})+\text{P}(\overline{\text{B}})=1-\text{P(A)}+1-\text{P(B)}$
$\Rightarrow\text{P}(\overline{\text{A}})+\text{P}(\overline{\text{B}})=2-\big[\text{P(A)}+\text{P(B)}\big]$
$\Rightarrow\text{P}(\overline{\text{A}})+\text{P}(\overline{\text{B}})=2-\frac{8}{9}$
$\Rightarrow\text{P}(\overline{\text{A}})+\text{P}(\overline{\text{B}})=\frac{10}{9}$ View full question & answer→Question 1451 Mark
If X follows a binomial distribution with parameter $\text{n}=100$ and $\text{p}=\frac{1}{3},$ then P(X = r) is maximum when r =
Answer
- 33
Solution:
$\text{n}=100,\text{p}=\frac{1}{3}\Rightarrow\text{q}=\frac{2}{3}$
$\text{np}=\frac{100}{3}=33+\frac{1}{3}$
⇒ Probability is maximum at 33. View full question & answer→Question 1461 Mark
If the random variable X has the following distribution:
| X: |
0 |
1 |
2 |
3 |
4 |
5 |
6 |
7 |
8 |
| P(X): |
a |
3a |
5a |
7a |
9a |
11a |
13a |
15a |
17a |
then the value of a is:
Answer
- $\frac{1}{81}$
Solution:
We know that the sum of probsabilities in a probability distribution is always 1.
$\therefore$ P(X = 0) + 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
⇒ a + 3a+ 5a+ 7a+ 9a + 11a + 13a + 15a + 17a = 1
⇒ 81a = 1
$\Rightarrow\text{a}=\frac{1}{81}$ View full question & answer→Question 1471 Mark
A biased coin with probabilty p, 0 < p < 1, of heads is tossed until a head appears for the first time. If the probability that the number of tosses required is even is $\frac{2}{5},$ then p equals:
Answer
- $\frac{1}{3}$
Solution:
p is the probability of getting head.
q = 1 - p is the probability of getting tail.
The number of tosses required is even.
$\Rightarrow\text{qp+q}^3\text{p+q}^5\text{p+q}^7\text{p+q}^9\text{p}\dots$
$\Rightarrow\text{qp}\Big(\frac{1}{1-\text{q}^2}\Big)$
$\Rightarrow\frac{(1-\text{p})\text{p}}{1-(1-\text{p})^2}$
$\Rightarrow\frac{(1-\text{p})\text{p}}{1-(1-2\text{p + p}^2)}$
$\Rightarrow\frac{1-\text{p}}{2-\text{p}}$
Given $\frac{1-\text{p}}{2-\text{p}}=\frac{2}{5}$
$\Rightarrow\text{p}=\frac{1}{3}$ View full question & answer→Question 1481 Mark
If A and B are two events such that $\text{P}(\text{A}|\text{B})=\text{p},\text{P(A)}=\text{p},\text{P(B)}=\frac{1}{3}$ and $\text{P}(\text{A}\cup\text{B})=\frac{5}{9},$ then p =
Answer
- $\frac{1}{3}$
Solution:
$\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\text{p},\text{P(A)}=\text{p},\text{P(B)}=\frac{1}{3},\text{P}(\text{A}\cup\text{B})=\frac{5}{9}$
Consider,
$\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\text{p}$
$\Rightarrow \frac{\text{P}(\text{A}\cap\text{B})}{\text{P(B)}}=\text{P}$
$\Rightarrow \frac{\text{P(A)}+\text{P(B)}-\text{P}(\text{A}\cup\text{B})}{\text{P(B)}}=\text{P}$
$\Rightarrow\frac{\text{p}+\frac{1}{3}-\frac{5}{9}}{\frac{1}{3}}=\text{p}$
$\Rightarrow\text{p}+\frac{1}{3}-\frac{5}{9}=\frac{\text{p}}{3}$
$\Rightarrow\frac{-2}{9}=\frac{\text{p}}{3}-\text{p}$
$=\frac{-2}{3}\text{p}=\frac{-2}{9}$
$\Rightarrow\text{p}=\frac{1}{3}$ View full question & answer→Question 1491 Mark
A pack of playing cards was found to contain only 51 cards. If the first 13 cards which are examined are all red, then the probability thatthe missing card is black, is:
Answer
- $\frac23$
Solution:
Total number of cards = 52
Number of lost cards = 1
13 cards are surley red therfore, from the remaining 39 cards 26 are black and 13 are red.
So probabilityof lost card being black $=\frac{(261)}{(391)}=\frac{26}{39}=\frac{2}{3}$ View full question & answer→Question 1501 Mark
Choose the correct answer from the given four options:
let $\text{P}(\text{A})=\frac{7}{13},\text{P}(\text{B})=\frac{9}{13}$ and $\text{P}(\text{A}\cup\text{B})=\frac{4}{13}.$ Then $\text{P}\Big(\frac{\text{A'}}{\text{B}}\Big)$ is equal to:
Answer
- $\frac{5}{9}$
Solution:
Here, $\text{P}(\text{A})=\frac{7}{13},\text{P}(\text{B})=\frac{9}{13}$ and $\text{P}(\text{A}\cup\text{B})=\frac{4}{13}$
$\therefore\text{P}\Big(\frac{\text{A}'}{\text{B}}\Big)=\frac{\text{P}(\text{A}'\cap\text{B})}{\text{P}(\text{B})}=\frac{\text{P}(\text{B})-\text{P}(\text{A}\cap\text{B})}{\text{P}(\text{B})}$
$=\frac{\frac{9}{13}-\frac{4}{13}}{\frac{9}{13}}=\frac{\frac{5}{13}}{\frac{9}{13}}=\frac{5}{9}$ View full question & answer→