MCQ 512 Marks
One ticket is selected at random from 50 tickets numbered $00,01,02, \ldots, 49$. Then the probability that the sum of the digits on the selected ticket is 8 , given that the product of these digits is zero, equals
- ✓
$\frac{1}{14}$
- B
$\frac{1}{7}$
- C
$\frac{5}{14}$
- D
$\frac{1}{50}$
AnswerCorrect option: A. $\frac{1}{14}$
(A)
Consider the following events:
A = Sum of the digits on the selected tickets is 8 .
B = Product of the digits on the selected ticket is zero.
There are 14 tickets having product of digits appearing on them as zero. The numbers on such tickets are $00,01,02,03,04,05,06,07$, 08, 09, 10, 20, 30, 40.
$\therefore P ( B )=\frac{14}{50}$ and $P ( A \cap B )=\frac{1}{50}$
$\therefore $ Required probability $= P ( A / B )=\frac{ P ( A \cap B )}{ P ( B )}=\frac{1}{14}$
View full question & answer→MCQ 522 Marks
Ram is visiting a friend. Ram knows that his friend has 2 children and 1 of them is a boy. Assuming that a child is equally likely to be a boy or a girl, then the probability that the other child is a girl, is
- ✓
$\frac{1}{2}$
- B
$\frac{1}{3}$
- C
$\frac{2}{3}$
- D
$\frac{7}{10}$
AnswerCorrect option: A. $\frac{1}{2}$
(A)
Event that at least one of them is a boy → A,
Event that other is girl → B ,
So, required probability
$P(B / A)=\frac{P(B \cap A)}{P(A)}$
Now, total cases are 3 (BG, BB, GG)
$\begin{aligned}\therefore \quad & \frac{P(B \cap A)}{P(A)}=\frac{\frac{1}{3}}{\frac{2}{3}}=\frac{1}{2} \\& \ldots[\because B \cap A=\{B G\} \text { and } A=\{B G, B B\}]\end{aligned}$
View full question & answer→MCQ 532 Marks
A box contains 10 mangoes out of which 4 are spoiled 2 mangoes are taken together at random. If one of them is found to be good, then the probability that the other is also good, is
- A
$\frac{1}{3}$
- B
$\frac{2}{3}$
- C
$\frac{8}{15}$
- ✓
$\frac{5}{13}$
AnswerCorrect option: D. $\frac{5}{13}$
(D)
Probability that both mangoes are good
$=P(A)=\frac{6_{C_2}}{10_{C_2}}=\frac{1}{3}$
Probability that of only one mango is good
$=P(B)=\frac{6_{C_1} \times 4_{C_1}}{10_{C_2}}=\frac{8}{15}$
Probability that at least one magno is good
$=P(C)=P(A)+P(B)=\frac{13}{15}$
Hence, required proability $= P \left(\frac{ A }{ C }\right)$
$=\frac{ P ( A \cap C )}{ P ( C )}$
$P ( A \cap C )= P ( A )+ P ( C )- P ( A \cup C )$
Note that $P ( A \cup C )= P ( C )$
$\therefore \quad P ( A \cap C )= P ( A )$
$\therefore \quad P \left(\frac{ A }{ C }\right)=\frac{ P ( A )}{ P ( C )}=\frac{\frac{1}{3}}{\frac{13}{15}}=\frac{5}{13}$
View full question & answer→MCQ 542 Marks
It is given that the events A and B are such that $P ( A )=\frac{1}{4}, P ( A / B )=\frac{1}{2}$ and $P ( B / A )=\frac{2}{3}$. Then, $P(B)$ is
- A
$\frac{2}{3}$
- B
$\frac{1}{2}$
- C
$\frac{1}{6}$
- ✓
$\frac{1}{3}$
AnswerCorrect option: D. $\frac{1}{3}$
(D)
$P ( A \cap B )= P ( A ) P ( B / A )$
$\therefore \quad P ( A \cap B )=\frac{1}{4} \times \frac{2}{3}=\frac{1}{6}$
Now, $P ( A / B )=\frac{ P ( A \cap B )}{ P ( B )}$
$\Rightarrow \frac{1}{2}=\frac{1}{6} \times \frac{1}{ P ( B )}$
$\Rightarrow P ( B )=\frac{1}{3}$
View full question & answer→MCQ 552 Marks
A coin is tossed three times in succession. it E is the event that there are at least two heads and F is the event in which first toss is a head, then$P\left(\frac{E}{F}\right)=$
- ✓
$\frac{3}{4}$
- B
$\frac{3}{8}$
- C
$\frac{1}{2}$
- D
$\frac{1}{8}$
AnswerCorrect option: A. $\frac{3}{4}$
(A)
$S =\{HHH , HHT , HTH , THH , HTT, THT, TTH, TTT \}$
$\therefore n ( S )=8, n ( E )=4, n ( F )=4$ and $n ( E \cap F )=3$
$\therefore P\left(\frac{E}{F}\right)=\frac{P(E \cap F)}{P(F)}=\frac{\frac{3}{8}}{\frac{4}{8}}=\frac{3}{4}$
View full question & answer→MCQ 562 Marks
If two events A and B are such that $P \left( A ^{ c }\right)=0.3$, $P ( B )=0.4$ and $P \left( AB ^{ c }\right)=0.5$, then $P \left[ B /\left( A \cup B ^c\right)\right]$ is equal to
- A
$\frac{1}{2}$
- B
$\frac{1}{3}$
- ✓
$\frac{1}{4}$
- D
AnswerCorrect option: C. $\frac{1}{4}$
(C)
$P \left( B /\left( A \cup B ^{ c }\right)\right]=\frac{ P \left( B \cap\left( A \cup B ^{ c }\right)\right)}{ P \left( A \cup B ^{ c }\right)}$
$=\frac{ P ( A \cap B )}{ P ( A )+ P \left( B ^{ c }\right)- P \left( A \cap B ^{ c }\right)}$
$=\frac{ P ( A )- P \left( A \cap B ^{ c }\right)}{ P ( A )+ P \left( B ^{ c }\right)- P \left( A \cap B ^{ c }\right)}$
$=\frac{0.7-0.5}{0.7+0.6-0.5}=\frac{1}{4}$
View full question & answer→MCQ 572 Marks
If A and B are two events with $P ( A )=\frac{1}{4}, P ( A / B )=\frac{1}{4}$ and $P ( B / A )=\frac{1}{2}$ then
- A
A and B are mutually exclusive
- ✓
- C
- D
Answer(B)
$P ( A \cap B )= P ( A ) \cdot P ( B / A )=\frac{1}{4} \times \frac{1}{2}=\frac{1}{8}$
Since, $P ( A \cap B )= P ( B ) P ( A / B )$
$\therefore \frac{1}{8}= P ( B ) \times \frac{1}{4}$
$\Rightarrow P ( B )=\frac{1}{2}$
$\therefore P ( A ) \cdot P ( B )=\frac{1}{4} \times \frac{1}{2}=\frac{1}{8}= P ( A \cap B )$
$\therefore A$ and B are independent
View full question & answer→MCQ 582 Marks
Let A and B be two non-null events such that $A \subset B$. Then, which of the following statements is always correct?
- A
$P ( A \mid B )= P ( B )- P ( A )$
- ✓
$P ( A \mid B ) \geq P ( A )$
- C
$P ( A \mid B ) \leq P ( A )$
- D
$P ( A \mid B )=1$
AnswerCorrect option: B. $P ( A \mid B ) \geq P ( A )$
(B)
$A \subset B \quad\ldots[Given]$
$\Rightarrow A \cap B=A \quad\ldots(i)$
$\therefore P ( A \mid B )=\frac{ P ( A \cap B )}{ P ( B )}$
$=\frac{ P ( A )}{ P ( B )} \quad\ldots[From (i)]$
$\therefore P ( A \mid B ) \geq P ( A ) \quad \ldots[\because 0< P ( B ) \leq 1]$
View full question & answer→MCQ 592 Marks
In a class, $40 \%$ of students study maths and science and $60 \%$ of student study maths. What is the probability of a student studying science given the student is already studying maths?
- A
$\frac{1}{3}$
- B
$\frac{1}{6}$
- ✓
$\frac{2}{3}$
- D
$\frac{1}{5}$
AnswerCorrect option: C. $\frac{2}{3}$
(C)
M: student studying maths
S: student studying science
$\begin{aligned} \therefore \quad & P(M \cap S)=40 \%=0.4
\\ & P(M)=60 \%=0.6\end{aligned}$
Probability of student studying science given the student is already studying maths
$\begin{aligned}= P ( S / M ) & = P ( M \cap S ) / P ( M ) \\ & =\frac{0.4}{0.6}=\frac{2}{3}\end{aligned}$
View full question & answer→MCQ 602 Marks
If $A$ and $B$ are two events such that $P(A)=\frac{1}{3}$, $P(B)=\frac{1}{4}$ and $P(A \cap B)=\frac{1}{5}$, then $P\left(\frac{\bar{B}}{\bar{A}}\right)=$
- ✓
$\frac{37}{40}$
- B
$\frac{37}{45}$
- C
$\frac{23}{40}$
- D
$\frac{1}{3}$
AnswerCorrect option: A. $\frac{37}{40}$
(A)
$\begin{aligned} P(A \cup B) & =P(A)+P(B)-P(A \cap B) \\ & =\frac{1}{3}+\frac{1}{4}-\frac{1}{5}=\frac{23}{60}\end{aligned}$
$P\left(\frac{\bar{B}}{\bar{A}}\right)=\frac{1-P(A \cup B)}{P(\bar{A})}=\frac{1-\frac{23}{60}}{1-\frac{1}{3}}=\frac{37}{60} \times \frac{3}{2}=\frac{37}{40}$
View full question & answer→MCQ 612 Marks
If A and B are two events such that $A \subseteq B$, then $P\left(\frac{B}{A}\right)=$
Answer(B)
Since, $A \subseteq B \Rightarrow A \cap B = B \cap A = A$
Hence, $P \left(\frac{ B }{ A }\right)=\frac{ P ( B \cap A )}{ P ( A )}=\frac{ P ( A )}{ P ( A )}=1$
View full question & answer→MCQ 622 Marks
If A and B are two independent events, then $P\left(\frac{A}{B}\right)=$
Answer(C)
$P \left(\frac{ A }{ B }\right)=\frac{ P ( A \cap B )}{ P ( B )}=\frac{ P ( A ) \cdot P ( B )}{ P ( B )}= P ( A )$.
View full question & answer→MCQ 632 Marks
The probability of India winning a test match against West Indies is $\frac{1}{2}$ Assuming independence from match to match, the probability that in a 5 match series India's second win occurs at the third test, is
- A
$\frac{2}{3}$
- B
$\frac{1}{2}$
- ✓
$\frac{1}{4}$
- D
$\frac{1}{8}$
AnswerCorrect option: C. $\frac{1}{4}$
(C)
The sample space is [LWW, WLW]
$\therefore P (L W W)+ P (W L W)$
$=$ Probability that in 5 match series, it is India's second win
$= P ( L ) P ( W ) P ( W )+ P ( W ) P ( L ) P ( W )$
$=\frac{1}{8}+\frac{1}{8}=\frac{2}{8}=\frac{1}{4}$
View full question & answer→MCQ 642 Marks
A student appears for tests I, II and III. The student is successful if he passes in tests I, II or III. The probabilities of the student passing in tests, I, II and III are respectively $p, q$ and $\frac{1}{2}$. If the probability of the student to be successful is $\frac{1}{2}$. Then
AnswerCorrect option: A. $p(1+q)=1$
(A)
Let $A _1$ - student passes in Test - I
$A _2$ - student passes in Test - II
$A _3$ - student passes in Test - III
A - student is successful
$A-\left(A_1 \cap A_2 \cap A_3^{\prime}\right) \cup\left(A_1 \cap A_2^{\prime} \cap A_3\right)$$\cup\left( A _1 \cap A_2 \cap A_3\right)$
$\begin{aligned} \therefore \quad P ( A ) & = P \left( A _1\right) \cdot P \left( A _2\right) \cdot P \left( A _3^{\prime}\right) \\ & + P \left( A _1\right) \cdot P \left( A _2^{\prime}\right) \cdot P \left( A _3\right) \\ & + P \left( A _1\right) \cdot P \left( A _2\right) \cdot P \left( A _3\right)\end{aligned}$
$\therefore \quad \frac{1}{2}=p \cdot q \cdot \frac{1}{2}+p \cdot(1-q) \cdot \frac{1}{2}+p \cdot q \cdot \frac{1}{2}$
$\therefore p+p q=1 \quad \Rightarrow p(1+q)=1$
View full question & answer→MCQ 652 Marks
$\begin{array}{l}\text { If } P ( B )=\frac{3}{4}, P (\overline{ A } \cap B \cap \overline{ C })=\frac{1}{3}, \\ P ( A \cap B \cap \overline{ C })=\frac{1}{3}, \text { then } P ( B \cap C )=\end{array}$
- ✓
$\frac{1}{12}$
- B
$\frac{3}{4}$
- C
$\frac{5}{12}$
- D
$\frac{23}{60}$
AnswerCorrect option: A. $\frac{1}{12}$
(A)
$P ( B \cap C )= P ( B )-[ P ( A \cap B \cap \overline{ C })+ P (\overline{ A } \cap B \cap \overline{ C })]$
$=\frac{3}{4}-\frac{2}{3}=\frac{1}{12}$
View full question & answer→MCQ 662 Marks
Two aeroplanes I and II bomb a target in succession. The probabilities of I and II scoring a hit correctly are 0.3 and 0.2 respectively. The second plane will bomb only if the first misses the target. The probability that the target is hit by second plane is
Answer(D)
Let $A _{ i }( i =1,2)$ denote the event that $i ^{\text {th }}$ plane hits the target.
Clearly, $A _1$ and $A _2$ are independent events.
Required probability $= P \left(\overline{ A }_1 \cap A_2\right)$
$= P \left(\overline{ A }_1\right) P \left( A _2\right)$
$=(1-0.3)(0.2)=0.14$
View full question & answer→MCQ 672 Marks
The probability of solving a question by three students are $\frac{1}{2}, \frac{1}{4}, \frac{1}{6}$ respectively. Probability of question being solved will be
- ✓
$\frac{33}{48}$
- B
$\frac{35}{48}$
- C
$\frac{31}{48}$
- D
$\frac{37}{48}$
AnswerCorrect option: A. $\frac{33}{48}$
(A)
i. This question can also be solved by one student
ii. This question can be solved by two students simultaneously
ii. This question can be solved by three students all together.
We have, $P ( A )=\frac{1}{2}, P ( B )=\frac{1}{4}, P ( C )=\frac{1}{6}$
$\begin{aligned} \therefore \quad & P ( A \cup B \cup C )= P ( A )+ P ( B )+ P ( C ) \\ & -[ P ( A ) \cdot P ( B )+ P ( B ) \cdot P ( C )+ P ( C ) \cdot P ( A )] \\ & +[ P ( A ) \cdot P ( B ) \cdot P ( C )]\end{aligned}$
$=\frac{1}{2}+\frac{1}{4}+\frac{1}{6}-\left[\frac{1}{2} \times \frac{1}{4}+\frac{1}{4} \times \frac{1}{6}+\frac{1}{6} \times \frac{1}{2}\right]+\left[\frac{1}{2} \times \frac{1}{4} \times \frac{1}{6}\right]$
$=\frac{33}{48}$
View full question & answer→MCQ 682 Marks
A problem in mathematics is given to 4 students whose chances of solving individually are $\frac{1}{2}, \frac{1}{3}, \frac{1}{4}$ and $\frac{1}{5}$. Then probability that the problem will be solved at least by one student is
- A
$\frac{2}{3}$
- B
$\frac{3}{5}$
- ✓
$\frac{4}{5}$
- D
$\frac{3}{4}$
AnswerCorrect option: C. $\frac{4}{5}$
(C)
Let $A , B , C$ and D be the events that the problem will be solved by $1^{\text {st }}, 2^{\text {nd }}, 3^{\text {rd }}$ and $4^{\text {th }}$ students respectively.
Probability that no student solve the problem
$= P (\overline{ A } \cap \overline{ B } \cap \overline{ C } \cap \overline{ D })$
$=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)\left(1-\frac{1}{5}\right)=\frac{1}{2} \times \frac{2}{3} \times \frac{3}{4} \times \frac{4}{5}=\frac{1}{5}$
$\ldots[$ As $\overline{ A }, \overline{ B }, \overline{ C }, \overline{ D }$ are independent events $]$
⇒ Probability that the problem will be solved by at least one student $= P ( A \cup B \cup C \cup D)$
$=1- P (\overline{ A } \cap \overline{ B } \cap \overline{ C } \cap \overline{ D })$
$=1-\frac{1}{5}=\frac{4}{5}$
View full question & answer→MCQ 692 Marks
A man and his wife appear for an interview for two posts. The probability of the husband selection is $\frac{1}{7}$ and that of the wife's selection is $\frac{1}{5}$. What is the probability that only one of them will be selected
- A
$\frac{1}{7}$
- ✓
$\frac{2}{7}$
- C
$\frac{3}{7}$
- D
AnswerCorrect option: B. $\frac{2}{7}$
(B)
The probability of husband is not selected$=1-\frac{1}{7}=\frac{6}{7}$
The probability that wife is not selected$=1-\frac{1}{5}=\frac{4}{5}$
The probability that only husband selected$=\frac{1}{7} \times \frac{4}{5}=\frac{4}{35}$
The probability that only wife selected$=\frac{1}{5} \times \frac{6}{7}=\frac{6}{35}$
$\begin{aligned} \text { Hence, required probability } & =\frac{6}{35}+\frac{4}{35}=\frac{10}{35}
\\ & =\frac{2}{7}\end{aligned}$
View full question & answer→MCQ 702 Marks
A bag contains 3 black and 4 white balls. Two balls are drawn one by one at random without replacement. The probability that the second drawn ball is white, is
- A
$\frac{4}{49}$
- B
$\frac{1}{7}$
- ✓
$\frac{4}{7}$
- D
$\frac{12}{49}$
AnswerCorrect option: C. $\frac{4}{7}$
(C)
Second white ball can draw in two ways.
i. First is white and second is white
Probability $=\frac{4}{7} \times \frac{3}{6}=\frac{2}{7}$
ii. First is black and second is white
Probability $=\frac{3}{7} \times \frac{4}{6}=\frac{2}{7}$
Hence, required probability $=\frac{2}{7}+\frac{2}{7}=\frac{4}{7}$.
View full question & answer→MCQ 712 Marks
The probability that A speaks truth is $\frac{4}{5}$, while this probability for $B$ is $\frac{3}{4}$. The probability that they contradict each other when asked to speak on a fact
- A
$\frac{4}{5}$
- B
$\frac{1}{5}$
- ✓
$\frac{7}{20}$
- D
$\frac{3}{20}$
AnswerCorrect option: C. $\frac{7}{20}$
(C)
Here, $P ( A )=\frac{3}{4}, P ( B )=\frac{4}{5}$
$\therefore P (\overline{ A })=\frac{1}{4}$ and $P (\overline{ B })=\frac{1}{5}$
$\therefore $ Required probability
$= P (A) \cdot P (\overline{ B }) +\, P (\bar{A}) \cdot P ( B )=\frac{7}{20}.$
View full question & answer→MCQ 722 Marks
If the probabilities that $A$ and $B$ will die within a year are $p$ and $q$ respectively, then probability that only one of them will be alive at the end of the year is,
Answer(B)
Here, $P ( A )= p$
$\Rightarrow P (\overline{ A })=1- p$
and $P(B)=q \Rightarrow P (\overline{ B })=1- q$
Probability that one person is alive is the sum of two cases A dies B lives and A lives B dies
$=p(1-q)+q(1-p)=p+q-2 p q$
View full question & answer→MCQ 732 Marks
The probability that a man will live 10 more years is $\frac{1}{4}$ and the probability that his wife will live 10 more years is $\frac{1}{3}$. Then the probability that neither will be alive in 10 years is,
- A
$\frac{5}{12}$
- ✓
$\frac{1}{2}$
- C
$\frac{7}{12}$
- D
$\frac{11}{12}$
AnswerCorrect option: B. $\frac{1}{2}$
(B)
$P(M)=\frac{1}{4} \Rightarrow P\left(M^{\prime}\right)=\frac{3}{4}$
and $P ( W )=\frac{1}{3} \Rightarrow P \left( W ^{\prime}\right)=\frac{2}{3}$
Both events are independent so that probability that no one will be alive is
$P \left( W ^{\prime} \cap M ^{\prime}\right)= P \left( W ^{\prime}\right) P \left( M ^{\prime}\right)=\frac{3}{4} \times \frac{2}{3}=\frac{1}{2}$
View full question & answer→MCQ 742 Marks
If A and B are independent events of a random experiment such that $P(A \cap B)=\frac{1}{6}$ and $P (\bar{A} \cap \bar{B})=\frac{1}{3}$, then $P ( A )$ is equal to
- A
$\frac{1}{4}$
- ✓
$\frac{1}{3}$
- C
$\frac{1}{6}$
- D
$\frac{2}{3}$
AnswerCorrect option: B. $\frac{1}{3}$
(B)
$P ( A \cap B )=\frac{1}{6}$ and $P (\overline{ A } \cap \overline{ B })=\frac{1}{3}$
$\Rightarrow P ( A ) P ( B )=\frac{1}{6}$ and $P (\overline{ A }) P (\overline{ B })=\frac{1}{3}$
$\Rightarrow x y=\frac{1}{6}$ and $(1-x)(1-y)=\frac{1}{3}$,
where $P ( A )=x, P ( B )=y$
$\Rightarrow x y=\frac{1}{6}$ and $1-x-y+\frac{1}{6}=\frac{1}{3}$
$\Rightarrow x y=\frac{1}{6}$ and $x+y=\frac{5}{6}$
$\Rightarrow x=\frac{1}{2}$ and $y=\frac{1}{3}$ or $x=\frac{1}{3}$ and $y=\frac{1}{2}$
View full question & answer→MCQ 752 Marks
Let A and B be two events such that $P (\overline{ A \cup B })=\frac{1}{6}, P ( A \cap B )=\frac{1}{4}$ and $P (\overline{ A })=\frac{1}{4}$, where $\bar{A}$ stands for complement of event $A$. Then events $A$ and $B$ are
- A
mutually exclusive and independent
- ✓
independent but not equally likely
- C
equally likely but not independent
- D
equally likely and mutually exclusive
AnswerCorrect option: B. independent but not equally likely
(B)
$P (\overline{ A \cup B })=\frac{1}{6}$
$\Rightarrow 1- P ( A \cup B )=\frac{1}{6}$
$\Rightarrow P(A \cup B)=\frac{5}{6}$
$\Rightarrow P ( A )+ P ( B )- P ( A \cap B )=\frac{5}{6}$
$\Rightarrow \frac{3}{4}+ P ( B )-\frac{1}{4}=\frac{5}{6} \Rightarrow P ( B )=\frac{1}{3}$
Clearly, $P ( A \cap B )=\frac{1}{4}=\frac{3}{4} \times \frac{1}{3}= P ( A ) P ( B )$
So, A and B are independent.
Also, $P ( A ) \neq P ( B )$. So, A and B are not equally likely.
View full question & answer→MCQ 762 Marks
For any two independent events $E _1$ and $E _2$, $P \left\{\left( E _1 \cup E _2\right) \cap\left(\overline{ E }_1 \cap \overline{ E }_2\right)\right\}$ is
- ✓
$<\frac{1}{4}$
- B
$>\frac{1}{4}$
- C
$\geq \frac{1}{2}$
- D
AnswerCorrect option: A. $<\frac{1}{4}$
(A)
Since $\overline{ E _1} \cap \overline{ E _2}=\overline{ E _1 \cup E _2}$
and $\left( E _1 \cup E _2\right) \cap\left(\overline{ E _1 \cup E _2}\right)=\phi$
$\therefore \quad P \left\{\left( E _1 \cup E _2\right) \cap\left( E _1 \cap E _2\right)\right\}= P (\phi)=0<\frac{1}{4}$
View full question & answer→MCQ 772 Marks
If A and B are independent events such that $P ( B )=\frac{2}{7}, P ( A \cup \overline{ B })=0.8$, then $P ( A )=$
Answer(C)
$P(A \cup \bar{B})=0.8$ and $P(B)=\frac{2}{7} \Rightarrow P(\bar{B})=\frac{5}{7}$
$\Rightarrow P ( A )+ P (\overline{ B })- P ( A \cap \overline{ B })=0.8$
$\Rightarrow P ( A )+ P (\overline{ B })- P ( A ) \cdot P (\overline{ B })=0.8$
$\ldots[$ As A and $\overline{ B }$ are independent events]
$\Rightarrow P ( A )+\frac{5}{7}-\frac{5}{7} P ( A )=0.8$
$\Rightarrow \frac{2}{7} P ( A )=\frac{3}{35} \Rightarrow P ( A )=0.3$
View full question & answer→MCQ 782 Marks
In a class, $60 \%$ of the students know lesson I $40 \%$ know lesson II and $20 \%$ know lesson I and II. A student is selected at random. The probability that the student does not know lesson I and lesson II is
- A
$0$
- B
$\frac{4}{5}$
- C
$\frac{3}{5}$
- ✓
$\frac{1}{5}$
AnswerCorrect option: D. $\frac{1}{5}$
(D)
A: Student who know lesson I
B: Student who know lesson II
$P(A)=0.6, P(B)=0.4, P(A \cap B)=0.2$
Required probability $=1- P ( A \cup B )$
$=1-[ P ( A )+ P ( B )- P ( A \cap B )]$
$=1-(0.6+0.4-0.2)$
$=0.2=\frac{1}{5}$
View full question & answer→MCQ 792 Marks
If A and B are events having probabilities, $P ( A )=0.6, P ( B )=0.4$ and $( A \cap B )=0$, then the probability that neither A nor B occurs is
- A
$\frac{1}{4}$
- B
- C
$\frac{1}{2}$
- ✓
$0$
Answer(D)
$P \left( A ^{\prime} \cap B ^{\prime}\right)=1- P ( A \cup B )$
$=1-[ P ( A )+ P ( B )- P ( A \cap B )]$
$=1-1=0$
View full question & answer→MCQ 802 Marks
If A and B are two events. The probability that exactly one of them occurs is equal to
- ✓
$P ( A )+ P ( B )-2 P ( A \cap B )$
- B
$P ( A )+ P ( B )+ P ( A \cap B )$
- C
$P ( A )+ P ( B )$
- D
$P ( A )+ P ( B )- P ( A \cap B )$
AnswerCorrect option: A. $P ( A )+ P ( B )-2 P ( A \cap B )$
(A)
Required Probability
$= P \left[\left( A \cap B ^{\prime}\right) \cup\left( A ^{\prime} \cap B \right)\right]$
$= P \left( A \cap B ^{\prime}\right)+ P \left( A ^{\prime} \cap B \right)$
$= P ( A )- P ( A \cap B )+ P ( B )- P ( A \cap B )$
$= P ( A )+ P ( B )-2 P ( A \cap B )$
View full question & answer→MCQ 812 Marks
Let A and B be events for which $P ( A )=x$, $P ( B )=y, P ( A \cap B )= z$, then $P (\overline{ A } \cap B )$ equals
- A
$(1-x) y$
- B
$1-x+y$
- ✓
$y- z$
- D
$1-x+y-z$
AnswerCorrect option: C. $y- z$
(C)
$P (\overline{ A } \cap B )= P ( B )- P ( A \cap B )=y- z$.
View full question & answer→MCQ 822 Marks
If $P ( A \cap B )=\frac{1}{2}$ and $P \left( A ^{\prime} \cap B ^{\prime}\right)=\frac{1}{3}$, $P ( A )= p$ and $P ( B )=2 p$, then the value of p is
- ✓
$\frac{7}{18}$
- B
$\frac{1}{3}$
- C
$\frac{4}{9}$
- D
$\frac{1}{9}$
AnswerCorrect option: A. $\frac{7}{18}$
(A)
$P \left( A ^{\prime} \cap B ^{\prime}\right)=\frac{1}{3}$
$\Rightarrow P \left[( A \cup B )^{\prime}\right]=\frac{1}{3}$
$\therefore 1-P(A \cup B)=\frac{1}{3}$
$\Rightarrow P ( A \cup B )=1-\frac{1}{3}=\frac{2}{3}$
$\therefore P ( A )+ P ( B )- P ( A \cap B )=\frac{2}{3}$
$\therefore p+2 p-\frac{1}{2}=\frac{2}{3}$
$\Rightarrow 3 p=\frac{2}{3}+\frac{1}{2}=\frac{7}{6} \Rightarrow p=\frac{7}{18}$
View full question & answer→MCQ 832 Marks
If (A) and (B) are two events and $P(A)=\frac{3}{8}$, $P ( B )=\frac{1}{2}, P ( A \cap B )=\frac{1}{4}$, then $P \left( A ^{\prime} \cup B ^{\prime}\right)=$
- A
$\frac{3}{8}$
- ✓
$\frac{3}{4}$
- C
$\frac{1}{4}$
- D
$\frac{5}{8}$
AnswerCorrect option: B. $\frac{3}{4}$
(B)
$P \left( A ^{\prime} \cup B ^{\prime}\right) = P \left[( A \cap B )^{\prime}\right]$
$=1- P ( A \cap B )=1-\frac{1}{4}=\frac{3}{4}$
View full question & answer→MCQ 842 Marks
A class consists of 80 students 25 of them are girls and 55 boys. If 10 are rich and remaining poor and also 20 of them are intelligent, then the probability of selecting intelligent rich girls is,
- A
$\frac{5}{128}$
- B
$\frac{25}{128}$
- ✓
$\frac{5}{512}$
- D
$\frac{5}{64}$
AnswerCorrect option: C. $\frac{5}{512}$
(C)
$P ( G )=\frac{25}{80}, P ( R )=\frac{10}{80}, P ( I )=\frac{20}{80}$
Since events are independent,
$\therefore \quad P$ (selecting rich and intelligent girls)
$- P ( G ) \cdot P ( R ) \cdot P ( I )-\frac{5}{512}$
View full question & answer→MCQ 852 Marks
There are 3 bags A,B and C. Bag A contains 2 white and 3 black balls, bag B contains 4 white and 2 black balls and bag C contains 3 white and 2 black balls. If a ball is drawn at random from a randomly chosen bag, then the probablitiy that the ball drawn is black is
- A
$\frac{2}{3}$
- ✓
$\frac{4}{9}$
- C
$\frac{5}{9}$
- D
$\frac{1}{9}$
AnswerCorrect option: B. $\frac{4}{9}$
(B)
Let E be the event of getting black ball.
Let $E_1$ be the event of selecting bag $A$.
Let $E _2$ be the event of selecting bag B .
Let $E _3$ be the event of selecting bag C .
Required Probability
$\begin{aligned}=P\left(E_1\right) \times P\left(\frac{E}{E_1}\right)+P\left(E_2\right) & \times P\left(\frac{E}{E_2}\right)+P\left(E_3\right) \times P\left(\frac{E}{E_3}\right)\end{aligned}$
$=\frac{1}{3} \times \frac{{ }^3 C _1}{{ }^5 C _1}+\frac{1}{3} \times \frac{{ }^2 C _1}{{ }^6 C _1}+\frac{1}{3} \times \frac{{ }^2 C _1}{{ }^5 C _1}$
$=\frac{1}{3} \times \frac{3}{5}+\frac{1}{3} \times \frac{2}{6}+\frac{1}{3} \times \frac{2}{5}$
$=\frac{4}{9}$
View full question & answer→MCQ 862 Marks
A bag contains 4 red and 6 black balls. A ball is drawn at random from the bag, its colour is observed and this ball along with two additional balls of the same colour is returned to the bag. If now a ball is drawn at random from the bag, then the probability that this drawn ball is red, is
- ✓
$\frac{2}{5}$
- B
$\frac{1}{5}$
- C
$\frac{3}{4}$
- D
$\frac{3}{10}$
AnswerCorrect option: A. $\frac{2}{5}$
(A)
Let $R _1$ be the event that the first ball drawn is red,
$B _1$ be the event that the first ball drawn is black,
$R _2$ be the event that the second ball drawn is red.
Required probability
$= P \left( R _1\right) \cdot P \left(\frac{ R _2}{ R _1}\right)+ P \left( B _1\right) \cdot P \left(\frac{ R _2}{B_1}\right)$
$=\frac{4}{10} \times \frac{6}{12}+\frac{6}{10} \times \frac{4}{12}=\frac{2}{5}$
View full question & answer→MCQ 872 Marks
A letter is taken from the word MULTIPLE and another letter is taken from the word CHOICE, the probability that both letters chosen are vowels is
- A
$5 \frac{5}{8}$
- B
$\frac{1}{2}$
- C
$\frac{1}{6}$
- ✓
$\frac{3}{16}$
AnswerCorrect option: D. $\frac{3}{16}$
(D)
In the word 'MULTIPLE' there are 3 vowels, out of total of 8, 1 vowel can be chosen in ${ }^3 C _1$ ways. In the word 'CHOICE' there are 3 vowels, out of the total of 6,1 vowel can be chosen in ${ }^3 C _1$ ways.
$\therefore \quad$ Required probability $=\frac{{ }^3 C _1}{8} \times \frac{{ }^3 C _1}{6}=\frac{3}{16}$
View full question & answer→MCQ 882 Marks
You are given a box with 20 cards in it. 10 of these cards have the letter I printed on them. The other ten have the letter T printed on them. If you pick up 3 cards at random and keep them in the same order, the probability of making the word IIT is
- A
$\frac{9}{80}$
- B
$\frac{1}{8}$
- C
$\frac{4}{27}$
- ✓
$\frac{5}{38}$
AnswerCorrect option: D. $\frac{5}{38}$
(D)
We have to consider order for IIT
$\therefore \quad$ Required probability $=\frac{10}{20} \times \frac{9}{19} \times \frac{10}{18}=\frac{5}{38}$
View full question & answer→MCQ 892 Marks
There are two childrens in a family The probability that both of them are boys is
- A
$\frac{1}{2}$
- B
$\frac{1}{3}$
- ✓
$\frac{1}{4}$
- D
$\frac{2}{3}$
AnswerCorrect option: C. $\frac{1}{4}$
(C)
For both to be boys, the probability
$=\left(\frac{1}{2}\right)\left(\frac{1}{2}\right)=\frac{1}{4}$
View full question & answer→MCQ 902 Marks
Three athletes A, B and C participate in a race competition. The probability of winning for A and B is twice of winning for C. Then the probability that the race is won by A or B, is
- A
$\frac{2}{3}$
- B
$\frac{1}{2}$
- ✓
$\frac{4}{5}$
- D
$\frac{1}{3}$
AnswerCorrect option: C. $\frac{4}{5}$
(C)
Here, $P ( A )= P ( B )=2 P ( C )$,
and $P ( A )+ P ( B )+ P ( C )=1$
$\Rightarrow P ( C )=\frac{1}{5}$ and $P ( A )= P ( B )=\frac{2}{5}$
Hence, $P ( A \cup B )= P ( A )+ P ( B )=\frac{2}{5}+\frac{2}{5}=\frac{4}{5}$
View full question & answer→MCQ 912 Marks
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. $\frac{3}{7}$
(B)
In a leap year, there are 366 days in which 52 weeks and two days. The combination of 2 days may be: Sun-Mon, Mon-Tue, Tue-Wed, Wed-Thu, Thu-Fri, Fri-Sat, Sat-Sun.
$\therefore P (53 \ fri )=\frac{2}{7} ; P (53 \ Sat )=\frac{2}{7}$
There is one combination in common i.e., (Fri-Sat)
$\therefore P(53 \text { Fri and } 53 \text { Sat}$ $)=\frac{1}{7}$
$\therefore P (53 \text { Fri or } 53 \text { Sat })= P (53 \text { Fri })+ P (53 \text { Sat })- P (53 \text { Fri and Sat })$
$=\frac{2}{7}+\frac{2}{7}-\frac{1}{7}=\frac{3}{7}$
View full question & answer→MCQ 922 Marks
If the probability of X to fail in the examination is 0. 3 and that for Y is 0.2, then the probability that either X or Y fails in the examination is
Answer(B)
Here, $P ( X )=0.3 ; P ( Y )=0.2$
Now $P ( X \cup Y )= P ( X )+ P ( Y )- P ( X \cap Y )$
Since, these are independent events
$\therefore \quad P ( X \cap Y )= P ( X ) \cdot P ( Y )$
Thus, required probability
$=0.3+0.2-0.06=0.44$
View full question & answer→MCQ 932 Marks
The chances to fail in Physics are 20% and the chances to fail in Mathematics are 10%. What are the chances to fail in at least one subject
Answer(A)
Let $P ( A )=\frac{20}{100}=\frac{1}{5}, P ( B )=\frac{10}{100}=\frac{1}{10}$
Since, events are independent and we have to find $P ( A \cup B )= P ( A )+ P ( B )- P ( A ) \cdot P ( B )$
$=\frac{1}{5}+\frac{1}{10}-\frac{1}{5} \times \frac{1}{10}$
$=\frac{3}{10}-\frac{1}{50}=\frac{14}{50} \times 100=28 \%$
View full question & answer→MCQ 942 Marks
Two dice are thrown and two coins are tossed simultaneously. The probability of getting prime numbers on both the dice along with a head and a tail on the two coin is
- ✓
$\frac{1}{8}$
- B
$\frac{1}{2}$
- C
$\frac{3}{16}$
- D
$\frac{1}{4}$
AnswerCorrect option: A. $\frac{1}{8}$
(A)
When two dice are thrown simultaneously, $n(S)=36$
A: Event that both the numbers on top are prime number
$\begin{aligned} \therefore A = & \{(2,2),(2,3),(2,5),(3,2),(3,3),(3,5),(5,2),(5,3)(5,5)\}\end{aligned}$
$\therefore n ( A )=9$
$\therefore P(A)=\frac{9}{36}=\frac{1}{4}$
When two coins are tossed simultaneously, $n(S)=4$
B : Event that we get one head and one tail
$\therefore n ( B )=2$
$\therefore P(B)=\frac{2}{4}=\frac{1}{2}$
Since both the events are independent of each other,
$\therefore $ Required probabiity $=P(A) \cdot P(B)=\frac{1}{8}$
View full question & answer→MCQ 952 Marks
A coin is tossed three times.
Event A: two head comes
Event B: last should be head
Then A and B are
Answer(B)
$P ( A )=\frac{3}{8}$ and $P ( B )=\frac{1}{2}$
$\therefore P ( A ) P ( B )=\frac{3}{8} \cdot \frac{1}{2}=\frac{3}{16}$
and $P ( A \cap B )=\frac{2}{8}=\frac{1}{4} \neq P ( A ) \cdot P ( B )$
$\therefore A$ and B are dependent.
View full question & answer→MCQ 962 Marks
The event A is independent of itself if and only if P(A) =
Answer(C)
A is independent of itself, if
$P ( A \cap A )= P ( A ) \cdot P ( A )$
$\Rightarrow P ( A )= P ( A )^2$
$\Rightarrow P ( A )=0,1$
View full question & answer→MCQ 972 Marks
If three numbers are drawn at random successively without replacement from a set $S=\{1,2, \ldots, 10\}$, then the probability that the minimum of the chosen numbers is 3 or their maximum is 7 is
- A
$\frac{1}{40}$
- B
$\frac{3}{40}$
- C
$\frac{5}{40}$
- ✓
$\frac{11}{40}$
AnswerCorrect option: D. $\frac{11}{40}$
(D)
$n ( S )={ }^{10} C _3$
A: event that minimum of chosen numbers is 3
B: event that maximum of chosen number is 7 .
$P ( A )=\frac{{ }^7 C _2}{{ }^{10} C _3}, P ( B )=\frac{{ }^6 C _2}{{ }^{10} C _3}, P ( A \cap B )=\frac{{ }^3 C _1}{{ }^{10} C _3}$
$=\frac{{ }^7 C _2}{{ }^{10} C _3}+\frac{{ }^6 C _2}{{ }^{10} C _3}-\frac{{ }^3 C _1}{{ }^{10} C _3}$
$=\frac{11}{40}$
View full question & answer→MCQ 982 Marks
A die is thrown. Let A be the event that the number obtained is greater than 3 . Let B be the event that the number obtained is less than 5 . Then, $P(A \cup B)$ is
- ✓
- B
$\frac{2}{5}$
- C
$\frac{3}{5}$
- D
$0$
Answer(A)
Here, $P ( A )=\frac{3}{6}=\frac{1}{2}, P ( B )=\frac{4}{6}=\frac{2}{3}$
and $P ( A \cap B )=$ Probability of getting a number greater than 3 and less than 5
$=$ Probability of getting $4=\frac{1}{6}$
$\therefore P ( A \cup B )= P ( A )+ P ( B )- P ( A \cap B )$
$=\frac{1}{2}+\frac{2}{3}-\frac{1}{6}=1$
View full question & answer→MCQ 992 Marks
$P ( A \cup B )= P ( A \cap B )$ if and only if the relation between $P ( A )$ and $P ( B )$ is
AnswerCorrect option: C. $P ( A )= P ( B )$
(C)
If $P ( A )= P ( B )$
As this gives,
$P ( A \cup B )= P ( A )+ P ( B )- P ( A \cap B )$
or
$P ( A )=2 P ( A )- P ( A )$
$\Rightarrow P(A \cup B)=P(A \cap B)$
View full question & answer→MCQ 1002 Marks
For two given events A and $B , P ( A \cap B )=$
- A
Not less than $P ( A )+ P ( B )-1$
- B
Not greater than $P ( A )+ P ( B )$
- C
Equal to $P ( A )+ P ( B )- P ( A \cup B )$
- ✓
View full question & answer→