A bookshelf contains 6 distinct books of Mathematics and 5 distinct books of Physics. From these 11 books, 6 books are chosen at random. Let $X$ be the absolute value of the difference between the number of Mathematics books chosen and the number of Physics books chosen. If $\alpha$ is the mean of the random variable $X$, then the value of $77 \alpha$ is $\_\_\_\_$ .
Correct Answer: 100
Explanation:
Let the number of Mathematics books chosen be $a$ and the number of Physics books chosen be $b$.
Since a total of 6 books are selected, we have:
$a + b = 6$
The random variable is defined as the absolute difference between these two numbers:
$X = |a - b|$
Case 1: $X = 6$
This happens if all 6 books are Mathematics books and none are Physics books, or vice versa. But since only 6 Maths and 5 Physics books exist, only $a=6$, $b=0$ is possible.
Suppose that Box I contains 6 red balls and 9 green balls, and Box II contains 8 red balls and 12 green balls. All the balls of Box I and Box II are mixed together and a ball is chosen at random from them. Let $E_1$ be the event that the ball chosen belonged to Box I and let $E_2$ be the event that the ball chosen belonged to Box II. Let $F_1$ be the event that the ball chosen is red and let $F_2$ be the event that the ball chosen is green.
Then which of the following statements is (are) TRUE?
A.
The events $E_1$ and $F_1$ are independent
B.
The events $E_2$ and $F_2$ are dependent
C.
The conditional probability $P(F_1|E_1)$ is equal to the conditional probability $P(F_1|E_2)$
D.
The conditional probability $P(F_1|E_1)$ is greater than the conditional probability $P(F_2|E_2)$
Correct Answer: A,C
Explanation:
The first step is to define the given events clearly:
The event $ U $ indicates that at least one of the students can solve the problem. Thus, its complement $ U' $ (none of the students can solve the problem) is expressed as:
A factory has a total of three manufacturing units, $M_1, M_2$, and $M_3$, which produce bulbs independent of each other. The units $M_1, M_2$, and $M_3$ produce bulbs in the proportions of $2: 2: 1$, respectively. It is known that $20 \%$ of the bulbs produced in the factory are defective. It is also known that, of all the bulbs produced by $M_1, 15 \%$ are defective. Suppose that, if a randomly chosen bulb produced in the factory is found to be defective, the probability that it was produced by $M_2$ is $\frac{2}{5}$.
If a bulb is chosen randomly from the bulbs produced by $M_3$, then the probability that it is defective is __________.
Correct Answer: 0.27TO0.33
Explanation:
$H_1$: The bulb is produced by unit $M_1$.
$H_2$: The bulb is produced by unit $M_2$.
$H_3$: The bulb is produced by unit $M_3$.
$E$: The bulb is defective.
The unit production proportions and known probabilities are:
A student appears for a quiz consisting of only true-false type questions and answers all the questions. The student knows the answers of some questions and guesses the answers for the remaining questions. Whenever the student knows the answer of a question, he gives the correct answer. Assume that the probability of the student giving the correct answer for a question, given that he has guessed it, is $\frac{1}{2}$. Also assume that the probability of the answer for a question being guessed, given that the student's answer is correct, is $\frac{1}{6}$. Then the probability that the student knows the answer of a randomly chosen question is :
A.
$\frac{1}{12}$
B.
$\frac{1}{7}$
C.
$\frac{5}{7}$
D.
$\frac{5}{12}$
Correct Answer: C
Explanation:
We can solve this problem using Bayes' theorem. Let's define the following events :
$A$: The student knows the answer of a randomly chosen question.
$B$: The student guesses the answer of a randomly chosen question.
$C$: The student gives the correct answer to a question.
We are asked to find the probability that the student knows the answer of a randomly chosen question, i.e., $P(A | C)$. According to Bayes' theorem,
$ P(A | C) = \frac{P(C | A) \cdot P(A)}{P(C)} $
We need to find the values of $P(C | A)$, $P(A)$, and $P(C)$. Let's proceed step by step :
1. $P(C | A)$ is the probability that the student gives the correct answer given that he knows the answer. Since he knows the answer, he will definitely give the correct answer. Therefore,
$ P(C | A) = 1 $
2. $P(A)$ is the probability that the student knows the answer of a randomly chosen question. We don't have this value directly, and we will denote it as $P(A)$. Likewise, the probability that the student guesses the answer is $P(B) = 1 - P(A)$.
3. $P(C)$ is the total probability that the student gives the correct answer. This can be calculated using the Law of Total Probability :
$ P(C) = P(C | A) \cdot P(A) + P(C | B) \cdot P(B) $
Where:
$ P(C | B) = \frac{1}{2} $
(The probability that the student gives the correct answer given that he guessed it), and
$ P(B | C) = \frac{1}{6} $
(The probability that the answer was guessed given the student's answer is correct).
A bag contains $N$ balls out of which 3 balls are white, 6 balls are green, and the remaining balls are blue. Assume that the balls are identical otherwise. Three balls are drawn randomly one after the other without replacement. For $i=1,2,3$, let $W_i, G_i$, and $B_i$ denote the events that the ball drawn in the $i^{\text {th }}$ draw is a white ball, green ball, and blue ball, respectively. If the probability $P\left(W_1 \cap G_2 \cap B_3\right)=\frac{2}{5 N}$ and the conditional probability $P\left(B_3 \mid W_1 \cap G_2\right)=\frac{2}{9}$, then $N$ equals ________.
Let $X$ be a random variable, and let $P(X=x)$ denote the probability that $X$ takes the value $x$. Suppose that the points $(x, P(X=x)), x=0,1,2,3,4$, lie on a fixed straight line in the $x y$-plane, and $P(X=x)=0$ for all $x \in \mathbb{R}-\{0,1,2,3,4\}$. If the mean of $X$ is $\frac{5}{2}$, and the variance of $X$ is $\alpha$, then the value of $24 \alpha$ is _____________.
Consider an experiment of tossing a coin repeatedly until the outcomes of two consecutive tosses are same. If the probability of a random toss resulting in head is $\frac{1}{3}$, then the probability that the experiment stops with head is :
A.
$\frac{1}{3}$
B.
$\frac{5}{21}$
C.
$\frac{4}{21}$
D.
$\frac{2}{7}$
Correct Answer: B
Explanation:
We are given a biased coin with the probability of getting heads (H) as $\frac{1}{3}$ and thus, the probability of getting tails (T) is $\frac{2}{3}$ (since the total probability of getting either heads or tails should sum to 1).
The experiment involves tossing this coin repeatedly until we obtain two identical outcomes in a row. We are interested in the case where the experiment ends with two consecutive heads.
Let's consider two different scenarios for the sequences of coin tosses :
Sequences that begin with a head and end with HH (like HH, HTHH, HTHTHH, and so on).
Sequences that begin with a tail and end with HH (like THH, THTHH, THTHTHH, and so on).
Both types of sequences will result in the experiment ending, so we need to consider both in our final probability calculation.
Scenario 1 : Sequences beginning with H
The simplest sequence is getting HH right away. The probability of this event is $(\frac{1}{3})^2 = \frac{1}{9}$.
The next sequence is getting a head, then a tail, and then HH (i.e., HTHH). The probability of this sequence is $\frac{1}{3} \cdot \frac{2}{3} \cdot (\frac{1}{3})^2 = \frac{2}{81}$.
We can see a pattern in these sequences. Each sequence in this scenario can be described as getting one head, followed by some number of "tail, head" pairs, and ending with HH.
Hence, the sequences in this scenario form a geometric series, where the ratio between consecutive terms is $r = P(HT) = \frac{1}{3} \cdot \frac{2}{3} = \frac{2}{9}$.
The sum of an infinite geometric series is given by the formula $\frac{a}{1 - r}$, where $a$ is the first term of the series and $r$ is the common ratio.
In this scenario, the first term is the probability of getting HH immediately, which is $\frac{1}{9}$. So, the sum of this series is :
In this scenario, we first get a tail, and then we follow the same pattern as in the first scenario.
The simplest sequence is THH, with a probability of $\frac{2}{3} \cdot (\frac{1}{3})^2 = \frac{2}{27}$.
The next sequence is getting a tail, a head, another tail, and then HH (i.e., THTHH). The probability of this sequence is $\frac{2}{3} \cdot \frac{1}{3} \cdot \frac{2}{3} \cdot (\frac{1}{3})^2 = \frac{4}{243}$.
As before, these sequences form a geometric series, with the ratio between consecutive terms is $r = P(TH) = \frac{2}{9}$.
In this scenario, the first term is the probability of getting THH, which is $\frac{2}{27}$. So, the sum of this series is :
Finally, to calculate the total probability of the experiment ending with two consecutive heads, we need to sum up the probabilities calculated in both scenarios:
Let $X=\left\{(x, y) \in \mathbb{Z} \times \mathbb{Z}: \frac{x^2}{8}+\frac{y^2}{20}<1\right.$ and $\left.y^2<5 x\right\}$. Three distinct points $P, Q$ and $R$ are randomly chosen from $X$. Then the probability that $P, Q$ and $R$ form a triangle whose area is a positive integer, is :
A.
$\frac{71}{220}$
B.
$\frac{73}{220}$
C.
$\frac{79}{220}$
D.
$\frac{83}{220}$
Correct Answer: B
Explanation:
$
\frac{x^2}{8}+\frac{y^2}{20}<1 \text { and } y^2<5 x
$
Let $X$ be the set of all five digit numbers formed using 1,2,2,2,4,4,0. For example, 22240 is in $X$ while 02244 and 44422 are not in $X$. Suppose that each element of $X$ has an equal chance of being chosen. Let $p$ be the conditional probability that an element chosen at random is a multiple of 20 given that it is a multiple of 5 . Then the value of $38 p$ is equal to :
Correct Answer: 31
Explanation:
Number of elements in $\mathrm{X}$ which are multiple of 5
Among these 38 elements, let us calculate when element is not divisible by 20
Let $p_i$ be the probability that a randomly chosen point has $i$ many friends, $i=0,1,2,3,4$. Let $X$ be a random variable such that for $i=0,1,2,3,4$, the probability $P(X=i)=p_i$. Then the value of $7 E(X)$ is :
Correct Answer: 24
Explanation:
Number of points having 0 friend $=0$
Number of points having 1 friend $=0$
Number of points having 2 friends $=4$
Number of points having 3 friends $=5 \times 4=20$
Number of points having 4 friends $=49-24=25$
$\mathrm{P}_{\mathrm{i}}=$ Probability that randomly selected points has friends
Two distinct points are chosen randomly out of the points $A_1, A_2, \ldots, A_{49}$. Let $p$ be the probability that they are friends. Then the value of $7 p$ is :
Correct Answer: 0.50
Explanation:
Total number of ways of selecting 2 persons $={ }^{49} \mathrm{C}_2$
Number of ways in which 2 friends are selected $=6 \times 7 \times 2=84$
Box-II contains 24 red, 9 blue and 15 green balls,
Box-III contains 1 blue, 12 green and 3 yellow balls,
Box-IV contains 10 green, 16 orange and 6 white balls.
A ball is chosen randomly from Box-I; call this ball $b$. If $b$ is red then a ball is chosen randomly from Box-II, if $b$ is blue then a ball is chosen randomly from Box-III, and if $b$ is green then a ball is chosen randomly from Box-IV. The conditional probability of the event 'one of the chosen balls is white' given that the event 'at least one of the chosen balls is green' has happened, is equal to
Two players, $P_{1}$ and $P_{2}$, play a game against each other. In every round of the game, each player rolls a fair die once, where the six faces of the die have six distinct numbers. Let $x$ and $y$ denote the readings on the die rolled by $P_{1}$ and $P_{2}$, respectively. If $x>y$, then $P_{1}$ scores 5 points and $P_{2}$ scores 0 point. If $x=y$, then each player scores 2 points. If $x < y$, then $P_{1}$ scores 0 point and $P_{2}$ scores 5 points. Let $X_{i}$ and $Y_{i}$ be the total scores of $P_{1}$ and $P_{2}$, respectively, after playing the $i^{\text {th }}$ round.
List-I
List-II
(I) Probability of $\left(X_{2} \geq Y_{2}\right)$ is
(P) $\frac{3}{8}$
(II) Probability of $\left(X_{2}>Y_{2}\right)$ is
(Q) $\frac{11}{16}$
(III) Probability of $\left(X_{3}=Y_{3}\right)$ is
Consider three sets E1 = {1, 2, 3}, F1 = {1, 3, 4} and G1 = {2, 3, 4, 5}. Two elements are chosen at random, without replacement, from the set E1, and let S1 denote the set of these chosen elements. Let E2 = E1 $-$ S1 and F2 = F1 $\cup$ S1. Now two elements are chosen at random, without replacement, from the set F2 and let S2 denote the set of these chosen elements.
Let G2 = G1 $\cup$ S2. Finally, two elements are chosen at random, without replacement, from the set G2 and let S3 denote the set of these chosen elements.
Let E3 = E2 $\cup$ S3. Given that E1 = E3, let p be the conditional probability of the event S1 = {1, 2}. Then the value of p is
A.
${1 \over 5}$
B.
${3 \over 5}$
C.
${1 \over 2}$
D.
${2 \over 5}$
Correct Answer: A
Explanation:
To find : Probability P = ${{P({S_1} \cap ({E_1} = {E_3}))} \over {P({E_1} = {E_3})}} = {{P({A_{1,2}})} \over {P(A)}}$
where $P(A) = P({A_{1,2}}) + P({A_{1,3}}) + P({A_{2,3}})$
Also, A1, 2 represents 1, 2 chosen at start and similarly others.
A number of chosen at random from the set {1, 2, 3, ....., 2000}. Let p be the probability that the chosen number is a multiple of 3 or a multiple of 7. Then the value of 500p is __________.
Correct Answer: 214
Explanation:
Given, set = {1, 2, 3, ...., 2000}
Let E1 = Event that it is a multiple of 3 = {3, 6, 9, ...., 1998}
$\therefore$ n(E1) = 666
and E2 = Event that it is a multiple of 7 = {7, 14, ..., 1995}
Three numbers are chosen at random, one after another with replacement, from the set S = {1, 2, 3, ......, 100}. Let p1 be the probability that the maximum of chosen numbers is at least 81 and p2 be the probability that the minimum of chosen numbers is at most 40.
The value of ${{625} \over 4}{p_1}$ is ___________.
Three numbers are chosen at random, one after another with replacement, from the set S = {1, 2, 3, ......, 100}. Let p1 be the probability that the maximum of chosen numbers is at least 81 and p2 be the probability that the minimum of chosen numbers is at most 40.
The value of ${{125} \over 4}{p_2}$ is ___________.
Let E, F and G be three events having probabilities $P(E) = {1 \over 8}$, $P(F) = {1 \over 6}$ and $P(G) = {1 \over 4}$, and let P (E $\cap$ F $\cap$ G) = ${1 \over {10}}$. For any event H, if Hc denotes the complement, then which of the following statements is (are) TRUE?
Let C1 and C2 be two biased coins such that the probabilities of getting head in a single toss are ${{2 \over 3}}$ and ${{1 \over 3}}$, respectively. Suppose $\alpha $ is the number of heads that appear when C1 is tossed twice, independently, and suppose $\beta $ is the number of heads that appear when C2 is tossed twice, independently. Then the probability that the roots of the quadratic polynomial x2 $-$ ax + $\beta $ are real and equal, is
A.
${{40} \over {81}}$
B.
${{20} \over {81}}$
C.
${{1} \over {2}}$
D.
${{1} \over {4}}$
Correct Answer: B
Explanation:
It is given that $\alpha $ is the number of heads that appear when C1 is tossed twice, the probability distribution of random variable $\alpha $ is
Similarly, it is given that $\beta $ is the number of heads that appear when C2 is tossed twice, so probability distribution of random variable $\beta $ is
Now, as the roots of quadratic polynomial x2 $-$ $\alpha $x + $\beta $ are real and equal, so D = $\alpha $2 $-$ 4$\beta $ = 0 and it is possible if ($\alpha $, $\beta $) = (0, 0) or (2, 1)
The probability that a missile hits a target successfully is 0.75. In order to destroy the target completely, at least three successful hits are required. Then the minimum number of missiles that have to be fired so that the probability of completely destroying the target is NOT less than 0.95, is ............
Correct Answer: 6
Explanation:
It is given that the probability, a missile hits a target successfully $p = {3 \over 4}$, so the probability to not hits the target is ${1 \over 4}$. And it is also given that to destroy the target completely, at least three successful hits are required.
Now, according to the question, let the minimum number of missiles required to fired is n, so
Two fair dice, each with faces numbered 1, 2, 3, 4, 5 and 6, are rolled together and the sum of the numbers on the faces is observed. This process is repeated till the sum is either a prime number or a perfect square. Suppose the sum turns out to be a perfect square before it turns out to be a prime number. If p is the probability that this perfect square is an odd number, then the value of 14p is ..........
Correct Answer: 8
Explanation:
Let an event E of sum of outputs are perfect square (i.e., 4 or 9), so
There are three bags B1, B2 and B3. The bag B1 contains 5 red and 5 green balls, B2 contains 3 red and 5 green balls, and B3 contains 5 red and 3 green balls. Bags B1, B2 and B3 have probabilities ${3 \over {10}}$, ${3 \over {10}}$ and ${4 \over {10}}$ respectively of being chosen. A bag is selected at random and a ball is chosen at random from the bag. Then which of the following options is/are correct?
A.
Probability that the chosen ball is green, given that the selected bag is B3, equals ${3 \over 8}$.
B.
Probability that the selected bag is B3, given that the chosen ball is green, equals ${5 \over 13}$.
C.
Probability that the chosen ball is green equals ${39 \over 80}$.
D.
Probability that the selected bag is B3 and the chosen ball is green equals ${3 \over 10}$.
Correct Answer: A,C
Explanation:
It is given that there are three bags B1, B2 and B3 and probabilities of being chosen B1, B2 and B3 are respectively.
There are five students S1, S2, S3, S4 and S5 in a music class and for them there are five seats R1, R2, R3, R4 and R5 arranged in a row, where initially the seat Ri is allotted to the student Si, i = 1, 2, 3, 4, 5. But, on the examination day, the five students are randomly allotted the five seats.
(There are two questions based on Paragraph "A", the question given below is one of them)
The probability that, on the examination day, the student S1 gets the previously allotted seat R1, and NONE of the remaining students gets the seat previously allotted to him/her is
A.
${3 \over {40}}$
B.
${1 \over 8}$
C.
${7 \over 40}$
D.
${1 \over 5}$
Correct Answer: A
Explanation:
Here, five students S1, S1, S3, S4 and S5 and five seats R1, R2, R3, R4 and R5
$ \therefore $ Total number of arrangement of sitting five students is 5! = 120
Here, S1 gets previously allotted seat R1
$ \therefore $ S2, S3, S4 and S5 not get previously seats.
Total number of way S2, S3, S4 and S5 not get previously seats is
There are five students S1, S2, S3, S4 and S5 in a music class and for them there are five seats R1, R2, R3, R4 and R5 arranged in a row, where initially the seat Ri is allotted to the student Si, i = 1, 2, 3, 4, 5. But, on the examination day, the five students are randomly allotted the five seats.
(There are two questions based on Paragraph "A", the question given below is one of them)
For i = 1, 2, 3, 4, let Ti denote the event that the students Si and Si+1 do NOT sit adjacent to each other on the day of the examination. Then, the probability of the event ${T_1} \cap {T_2} \cap {T_3} \cap {T_4}$ is
Three randomly chosen nonnegative integers x, y and z are found to satisfy the equation x + y + z = 10. Then the probability that z is even, is
A.
${1 \over {2}}$
B.
${36 \over {55}}$
C.
${6 \over {11}}$
D.
${5 \over {11}}$
Correct Answer: C
Explanation:
To solve this problem, we need to find the total number of possible solutions to the equation $ x + y + z = 10 $ where $ x, y, $ and $ z $ are nonnegative integers.
First, we employ the "stars and bars" method to determine the number of nonnegative integer solutions to this equation. This method states that the number of ways to partition $ n $ identical items (our sum) into $ k $ distinct groups (our variables) is given by:
$ \binom{n + k - 1}{k - 1} $
For $ x + y + z = 10 $, we have $ n = 10 $ and $ k = 3 $. Plugging into the formula, we get:
So, there are 66 possible solutions to the equation $ x + y + z = 10 $.
Next, we need to find the number of solutions for which $ z $ is even. Let $ z = 2k $ where $ k $ is a nonnegative integer. Then the equation becomes:
$ x + y + 2k = 10 $
Rearranging it, we get:
$ x + y = 10 - 2k $
Here, $ 10 - 2k $ must be nonnegative, so $ k $ can take values $ 0, 1, 2, 3, 4, 5 $ making total 6 possible values of k.
For each value of $ k $, $ x + y $ must equal the corresponding $ 10 - 2k $. The number of nonnegative integer solutions to $ x + y = m $ for any nonnegative integer $ m $ is given by:
$ \binom{m + 1}{1} = m + 1 $
We will sum the solutions for each valid $ k $:
When $ k = 0 $: $ x + y = 10 $, the number of solutions is $ 11 $.
When $ k = 1 $: $ x + y = 8 $, the number of solutions is $ 9 $.
When $ k = 2 $: $ x + y = 6 $, the number of solutions is $ 7 $.
When $ k = 3 $: $ x + y = 4 $, the number of solutions is $ 5 $.
When $ k = 4 $: $ x + y = 2 $, the number of solutions is $ 3 $.
When $ k = 5 $: $ x + y = 0 $, the number of solutions is $ 1 $.
Summing these, we get:
$ 11 + 9 + 7 + 5 + 3 + 1 = 36 $
Thus, there are 36 solutions where $ z $ is even out of a total of 66 solutions. Hence, the probability that $ z $ is even is given by:
Football teams ${T_1}$ and ${T_2}$ have to play two games against each other. It is assumed that the outcomes of the two games are independent. The probabilities of ${T_1}$ winning, drawing and losing a game against ${T_2}$ are ${1 \over 2},{1 \over 6}$ and ${1 \over 3}$ respectively. Each team gets $3$ points for a win, $1$ point for a draw and $0$ point for a loss in a game. Let $X$ and $Y$ denote the total points scored by teams ${T_1}$ and ${T_2}$ respectively after two games.
$\,\,\,\,P\,\left( {X > Y} \right)$ is
A.
${1 \over 4}$
B.
${5 \over 12}$
C.
${1 \over 2}$
D.
${7 \over 12}$
Correct Answer: B
Explanation:
$\bullet$ Probability of wining of T1 against T2 is = 1/2.
$\bullet$ Probability of drawing of T1 against T2 is = 1/6.
$\bullet$ Probability of losing of T1 against T2 is = 1/3.
Football teams ${T_1}$ and ${T_2}$ have to play two games against each other. It is assumed that the outcomes of the two games are independent. The probabilities of ${T_1}$ winning, drawing and losing a game against ${T_2}$ are ${1 \over 2},{1 \over 6}$ and ${1 \over 3}$ respectively. Each team gets $3$ points for a win, $1$ point for a draw and $0$ point for a loss in a game. Let $X$ and $Y$ denote the total points scored by teams ${T_1}$ and ${T_2}$ respectively after two games.
$P\,\left( {X = Y} \right)$ is
A.
${{11} \over {36}}$
B.
${{1} \over {3}}$
C.
${{13} \over {36}}$
D.
${{1} \over {2}}$
Correct Answer: C
Explanation:
$\bullet$ Probability of wining of T1 against T2 = 1/2.
$\bullet$ Probability of drawing of T1 against T2 is = 1/6.
$\bullet$ Probability of losing of T1 against T2 is = 1/3.
A computer producing factory has only two plants ${T_1}$ and ${T_2}.$ Plant ${T_1}$ produces $20$% and plant ${T_2}$ produces $80$% of the total computers produced. $7$% of computers produced in the factory turn out to be defective. It is known that $P$ (computer turns out to be defective given that it is produced in plant ${T_1}$)
$ = 10P$ (computer turns out to be defective given that it is produced in plant ${T_2}$),
where $P(E)$ denotes the probability of an event $E$. A computer produced in the factory is randomly selected and it does not turn out to be defective. Then the probability that it is produced in plant ${T_2}$ is
A.
${{36} \over {73}}$
B.
${{47} \over {79}}$
C.
${{78} \over {93}}$
D.
${{75} \over {83}}$
Correct Answer: C
Explanation:
Let $P_1$ be the defective computers that are produced from plant $T_1$ and $P_2$ be that from plant $T_2$.
The total percentage of the defective computers produced is $7 \%$.
The minimum number of times a fair coin needs to be tossed, so that the probability of getting at least two heads is at least $0.96,$ is
Correct Answer: 8
Explanation:
Let the coin is tossed $n$ times.
$\because p$ (at least 2 heads) $=1-[p$ (one heads) $+p$ (No heads)$\}$
As we know, by binominal probability theorem the probability of getting $r$ success in $n$ trials with $p$ being the probability of success and $q$ be the probability of failure, is given by ${ }^n \mathrm{C}_r(p)^r(q)^{n-r}$.
Let head be considered as the success and tail be the failure probability of getting head in a toss $=p=\frac{1}{2}$ and probability of getting tail in a toss $=q=\frac{1}{2}$
Let ${n_1}$ and ${n_2}$ be the number of red and black balls, respectively, in box ${\rm I}$. Let ${n_3}$ and ${n_4}$ be the number of red and black balls, respectively, in box ${\rm I}{\rm I}.$
A ball is drawn at random from box ${\rm I}$ and transferred to box ${\rm I}$${\rm I}.$ If the probability of drawing a red ball from box ${\rm I},$ after this transfer, is ${1 \over 3},$ then the correct option(s) with the possible values of ${n_1}$ and ${n_2}$ is(are)
A.
${n_1} = 4$ and ${n_2} = 6$
B.
${n_1} = 2$ and ${n_2} = 3$
C.
${n_1} = 10$ and ${n_2} = 20$
D.
${n_1} = 3$ and ${n_2} = 6$
Correct Answer: C,D
Explanation:
$\therefore$ P (drawing red ball from B1) = ${1 \over 3}$
Let ${n_1}$ and ${n_2}$ be the number of red and black balls, respectively, in box ${\rm I}$. Let ${n_3}$ and ${n_4}$ be the number of red and black balls, respectively, in box ${\rm I}{\rm I}.$
One of the two boxes, box ${\rm I}$ and box ${\rm I}{\rm I},$ was selected at random and a ball was drawn randomly out of this box. The ball was found to be red. If the probability that this red ball was drawn from box ${\rm I}{\rm I}$ is ${1 \over 3},$ then the correct option(s) with the possible values of ${n_1}$ ${n_2},$ ${n_3}$ and ${n_4}$ is (are)
Box $1$ contains three cards bearing numbers $1,2,3;$ box $2$ contains five cards bearing numbers $1,2,3,4,5;$ and box $3$ contains seven cards bearing numbers $1,2,3,4,5,6,7.$ A card is drawn from each of the boxes. Let ${x_i}$ be number on the card drawn from the ${i^{th}}$ box, $i=1,2,3.$
The probability that ${x_1} + {x_2} + {x_3}$ is odd, is
A.
${{29} \over {105}}$
B.
${{53} \over {105}}$
C.
${{57} \over {105}}$
D.
${{1} \over {2}}$
Correct Answer: B
Explanation:
Given three Boxes:
Given, that a card is drow from each Box and $x_i$ denote the number on the card drawn from $i^{\text {tht }}$ box, $i=1,2,3$.
Box $1$ contains three cards bearing numbers $1,2,3;$ box $2$ contains five cards bearing numbers $1,2,3,4,5;$ and box $3$ contains seven cards bearing numbers $1,2,3,4,5,6,7.$ A card is drawn from each of the boxes. Let ${x_i}$ be number on the card drawn from the ${i^{th}}$ box, $i=1,2,3.$
The probability that ${x_1},$, ${x_2},$ ${x_3}$ are in an arithmetic progression, is
A.
${{9} \over {105}}$
B.
${{10} \over {105}}$
C.
${{11} \over {105}}$
D.
${{7} \over {105}}$
Correct Answer: C
Explanation:
Given three Boxes :
Given, that a card is drow from each box and $x_i$ denote the number on the card drawn from $i^{\text {th }}$ box, $i=1,2,3$.
$2 x_2=x_1+x_3$ indicate, $x_1+x_3$ is even and for every even value of $x_1+x_3$ there is only one possible value of $x_2$. Now, $x_1+x_3$ is even in two cases:
Three boys and two girls stand in a queue. The probability, that the number of boys ahead of every girl is at least one more than the number of girls ahead of her, is
A.
${1 \over 2}$
B.
${1 \over 3}$
C.
${2 \over 3}$
D.
${3 \over 4}$
Correct Answer: A
Explanation:
Given, 3 boys and 2 girls stand in a queue.
Sample space $n(s)=\left| \!{\underline {\,
{5} \,}} \right.=120$
According to the given condition, following cases may arise.
A box ${B_1}$ contains $1$ white ball, $3$ red balls and $2$ black balls. Another box ${B_2}$ contains $2$ white balls, $3$ red balls and $4$ black balls. A third box ${B_3}$ contains $3$ white balls, $4$ red balls and $5$ black balls.
If $2$ balls are drawn (without replacement) from a randomly selected box and one of the balls is white and the other ball is red, the probability that these $2$ balls are drawn from box ${B_2}$ is
A.
${{116} \over {181}}$
B.
${{126} \over {181}}$
C.
${{65} \over {181}}$
D.
${{55} \over {181}}$
Correct Answer: D
Explanation:
Let $E$ be the event of one white and one red balls $A_1, A_2, A_3$ be the events of both balls are from box $\mathrm{B}_1, \mathrm{~B}_2, \mathrm{~B}_3$ respectively
Now, the probability of one white and one red both balls are drawn from box $B_2$ is $P\left(\frac{A_2}{E}\right)$.
A box ${B_1}$ contains $1$ white ball, $3$ red balls and $2$ black balls. Another box ${B_2}$ contains $2$ white balls, $3$ red balls and $4$ black balls. A third box ${B_3}$ contains $3$ white balls, $4$ red balls and $5$ black balls.
If $1$ ball is drawn from each of the boxex ${B_1},$ ${B_2}$ and ${B_3},$ the probability that all $3$ drawn balls are of the same colour is
A.
${{82} \over {648}}$
B.
${{90} \over {648}}$
C.
${{558} \over {648}}$
D.
${{566} \over {648}}$
Correct Answer: A
Explanation:
$\begin{aligned}
& \text { Let } \mathrm{P} \text { be the probability of same colour balls } \\
& \text { drawn from each boxes } \mathrm{B}_1, \mathrm{~B}_2, \mathrm{~B}_3 \\
& \therefore \quad \mathrm{P}=\mathrm{P} \text { (all are white) }+\mathrm{P} \text { (all are red) }+\mathrm{P} \text { (all } \\
& \quad \text { are Black) } \\
& \Rightarrow P=\frac{1}{6} \times \frac{2}{9} \times \frac{3}{12}+\frac{3}{6} \times \frac{3}{9} \times \frac{4}{12}+\frac{2}{6} \times \frac{4}{9} \times \frac{5}{12} \\
& \Rightarrow P=\frac{6}{648}+\frac{36}{648}+\frac{40}{648} \\
& \Rightarrow P=\frac{82}{648}
\end{aligned}$
Hints:
The probability of one ball is drawn from each boxes $B_1, B_2, B_3$ is equal to the sum of probabilities of all are white balls, all are red balls and all are Black balls.
Four persons independently solve a certain problem correctly with probabilities ${1 \over 2},{3 \over 4},{1 \over 4},{1 \over 8}.$ Then the probability that the problem is solved correctly by at least one of them is
A.
${{235} \over {256}}$
B.
${{21} \over {256}}$
C.
${{3} \over {256}}$
D.
${{253} \over {256}}$
Correct Answer: A
Explanation:
Let A, B, C, D are the events of solving a problem.
Of the three independent events ${E_1},{E_2}$ and ${E_3},$ the probability that only ${E_1}$ occurs is $\alpha ,$ only ${E_2}$ occurs is $\beta $ and only ${E_3}$ occurs is $\gamma .$ Let the probability $p$ that none of events ${E_1},{E_2}$ or ${E_3}$ occurs satisfy the equations $\left( {\alpha -2\beta } \right)p = \alpha \beta $ and $\left( {\beta - 3\gamma } \right)p = 2\beta \gamma .$ All the given probabilities are assumed to lie in the interval $(0, 1)$.
Then ${{\Pr obability\,\,of\,\,occurrence\,\,of\,\,{E_1}} \over {\Pr obability\,\,of\,\,occurrence\,\,of\,\,{E_3}}}$
Correct Answer: 6
Explanation:
Given, three independent events $\mathrm{E}_1, \mathrm{E}_2$ and $\mathrm{E}_3$.
Four fair dice ${D_1,}$ ${D_2,}$ ${D_3}$ and ${D_4}$ ; each having six faces numbered $1, 2, 3, 4, 5$ and $6$ are rolled simultaneously. The probability that ${D_4}$ shows a number appearing on one of ${D_1},$ ${D_2}$ and ${D_3}$ is
A.
${{91} \over {216}}$
B.
${{108} \over {216}}$
C.
${{125} \over {216}}$
D.
${{127} \over {216}}$
Correct Answer: A
Explanation:
For the given condition, the sample space $=6^4$
For favourable condition
Case I : For $D_4$ there are ${ }^6 C_1$ way. Now, it appears on any one of $D_1, D_2, D_3$ i.e. ${ }^3 C_1 \cdot 1$, for other two there are $5 \times 5$ ways.
Case II : For $\mathrm{D}_4$ there are ${ }^6 \mathrm{C}_1$ ways. Now, it appears on any two of $D_1, D_2, D_3$ i.e. ${ }^3 \mathrm{C}_2 .1 .1$, for other one there are 5 ways.
Let $X$ and $Y$ be two events such that $P\left( {X|Y} \right) = {1 \over 2},$ $P\left( {Y|X} \right) = {1 \over 3}$ and $P\left( {X \cap Y} \right) = {1 \over 6}.$ Which of the following is (are) correct ?
A.
$P\left( {X \cup Y} \right) = {2 \over 3}$
B.
$X$ and $Y$ are independent
C.
$X$ and $Y$ are not independent
D.
$P\left( {{X^c} \cap Y} \right) = {1 \over 3}$
Correct Answer: A,B
Explanation:
Let's analyze the given conditions and evaluate which options are correct.
Given:
$ P(X|Y) = \frac{1}{2} $
$ P(Y|X) = \frac{1}{3} $
$ P(X \cap Y) = \frac{1}{6} $
Now, let's proceed step by step through each option.
Option A: $ P(X \cup Y) = \frac{2}{3} $
We can use the formula for the union of two events:
$ P(X \cup Y) = P(X) + P(Y) - P(X \cap Y) $
To find $ P(X) $ and $ P(Y) $, we use the definitions of conditional probability:
$ P(X|Y) = \frac{P(X \cap Y)}{P(Y)} $
$ \frac{1}{2} = \frac{\frac{1}{6}}{P(Y)} $
$ P(Y) = \frac{1}{6} \times 2 = \frac{1}{3} $
Similarly,
$ P(Y|X) = \frac{P(X \cap Y)}{P(X)} $
$ \frac{1}{3} = \frac{\frac{1}{6}}{P(X)} $
$ P(X) = \frac{1}{6} \times 3 = \frac{1}{2} $
Now, substituting these values into the union formula:
A ship is fitted with three engines ${E_1},{E_2}$ and ${E_3}$. The engines function independently of each other with respective probabilities ${1 \over 2},{1 \over 4}$ and ${1 \over 4}$. For the ship to be operational at least two of its engines must function. Let $X$ denote the event that the ship is operational and Let ${X_1},{X_2}$ and ${X_3}$ denote respectively the events that the engines ${E_1},{E_2}$ and ${E_3}$ are functioning. Which of the following is (are) true?
A.
$P\left[ {X_1^c|X} \right] = {3 \over {16}}$
B.
$P$ [exactly two engines of the ship are functioning $\left. {|X} \right] = {7 \over 8}$
Let $E$ and $F$ be two independent events. The probability that exactly one of them occurs is $\,{{11} \over {25}}$ and the probability of none of them occurring is $\,{{2} \over {25}}$. If $P(T)$ denotes the probability of occurrence of the event $T,$ then
A.
$P\left( E \right) = {4 \over 5},P\left( F \right) = {3 \over 5}$
B.
$P\left( E \right) = {1 \over 5},P\left( F \right) = {2 \over 5}$
C.
$P\left( E \right) = {2 \over 5},P\left( F \right) = {1 \over 5}$
D.
$P\left( E \right) = {3 \over 5},P\left( F \right) = {4 \over 5}$
Correct Answer: A,D
Explanation:
Let $P(E) = e$ and $P(F) = f$.
$P(E \cup F) - P(E \cap F) = {{11} \over {25}}$
$ \Rightarrow e + f - 2ef = {{11} \over {25}}$ ...... (1)
$P(\overline E \cap \overline F ) = {2 \over {25}}$
$ \Rightarrow (1 - e)(1 - f) = {2 \over {25}}$
$ \Rightarrow 1 - e - f + ef = {2 \over {25}}$ ...... (2)
From Eqs. (1) and (2), we get
$ef = {{12} \over {25}}$ and $e + f = {7 \over 5}$
Let $\omega $ be a complex cube root of unity with $\omega \ne 1.$ A fair die is thrown three times. If ${r_1},$ ${r_2}$ and ${r_3}$ are the numbers obtained on the die, then the probability that ${\omega ^{{r_1}}} + {\omega ^{{r_2}}} + {\omega ^{{r_3}}} = 0$ is
A.
${1 \over 18}$
B.
${1 \over 9}$
C.
${2 \over 9}$
D.
${1 \over 36}$
Correct Answer: C
Explanation:
Sample space A dice is thrown thrice, $n(s) = 6 \times 6 \times 6$.
A signal which can be green or red with probability ${4 \over 5}$ and ${1 \over 5}$ respectively, is received by station A and then transmitted to station $B$. The probability of each station receving the signal correctly is ${3 \over 4}$. If the signal received at atation $B$ is green, then the probability that the original signal was green is
