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JEE Advanced · Mathematics · Statistics and Probability

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1
A bookshelf has 6 distinct Mathematics books and 5 distinct Physics books. From these 11 books, 6 are chosen at random. Let XX be the absolute difference between the number of Mathematics and Physics books chosen. If α\alpha is the mean of XX, find 77α77\alpha.
Not attempted
Your answer: —
Correct answer: 100
2
A data set of 10 observations x1,…,x10x_1,\ldots,x_{10} has mean 5 and variance 7. The first 8 observations have mean 4 and variance 3.5, and x9<x10x_9<x_{10}. Find 3x9+2x103x_9+2x_{10}.
Not attempted
Your answer: —
Correct answer: 44
3
Suppose Box I contains 6 red and 9 green balls, and Box II contains 8 red and 12 green balls. All balls are mixed and one is chosen at random. Let E1,E2E_1,E_2 be the events that the chosen ball belonged to Box I, Box II respectively, and let F1,F2F_1,F_2 be the events that the chosen ball is red, green respectively. Then which of the following statements is(are) TRUE?
Not attempted
A
The events E1E_1 and F1F_1 are independent
B
The events E2E_2 and F2F_2 are dependent
C
P(F1∣E1)=P(F1∣E2)P(F_1|E_1) = P(F_1|E_2)
D
P(F1∣E1)>P(F2∣E2)P(F_1|E_1) > P(F_2|E_2)
4
A factory has a total of three manufacturing units, M1M_1, M2M_2, and M3M_3, which produce bulbs independent of each other. The units M1,M2,M3M_1,M_2,M_3 produce bulbs in the proportions 2:2:12:2:1 respectively. It is known that 20%20\% of the bulbs produced in the factory are defective. It is also known that, of all the bulbs produced by M1M_1, 15%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 M2M_2 is 25\dfrac25. If a bulb is chosen randomly from the bulbs produced by M3M_3, then the probability that it is defective is ___________.
Not attempted
Your answer: —
Correct answer: 0.3
5
Three students S1,S2S_1, S_2, and S3S_3 are given a problem to solve. Consider the following events: UU: At least one of S1,S2,S3S_1, S_2, S_3 can solve the problem, VV: S1S_1 can solve the problem, given that neither S2S_2 nor S3S_3 can solve the problem, WW: S2S_2 can solve the problem and S3S_3 cannot solve the problem, TT: S3S_3 can solve the problem. For any event EE, let P(E)P(E) denote the probability of EE. If P(U)=12P(U) = \dfrac{1}{2}, P(V)=110P(V) = \dfrac{1}{10}, and P(W)=112P(W) = \dfrac{1}{12}, then P(T)P(T) is equal to
Not attempted
A
1336\dfrac{13}{36}
B
13\dfrac{1}{3}
C
1960\dfrac{19}{60}
D
14\dfrac{1}{4}
6
Consider the following frequency distribution: Value458961211Frequency5f1f22113\begin{array}{c|c|c|c|c|c|c|c}\text{Value} & 4 & 5 & 8 & 9 & 6 & 12 & 11 \\ \hline \text{Frequency} & 5 & f_1 & f_2 & 2 & 1 & 1 & 3\end{array} Suppose that the sum of the frequencies is 1919 and the median of this frequency distribution is 66. For the given frequency distribution, let α\alpha denote the mean deviation about the mean, β\beta denote the mean deviation about the median, and σ2\sigma^2 denote the variance. Match each entry in List-I to the correct entry in List-II and choose the correct option. List-I (P) 7f1+9f27f_1+9f_2 is equal to (Q) 19α19\alpha is equal to (R) 19β19\beta is equal to (S) 19σ219\sigma^2 is equal to List-II (1) 146146 (2) 4747 (3) 4848 (4) 145145 (5) 5555
Not attempted
A
(P) →\to (5) (Q) →\to (3) (R) →\to (2) (S) →\to (4)
B
(P) →\to (5) (Q) →\to (2) (R) →\to (3) (S) →\to (1)
C
(P) →\to (5) (Q) →\to (3) (R) →\to (2) (S) →\to (1)
D
(P) →\to (3) (Q) →\to (2) (R) →\to (5) (S) →\to (4)
7
A bag contains NN balls out of which 33 balls are white, 66 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=1,2,3i=1,2,3, let Wi,Gi,W_i,G_i, and BiB_i denote the events that the ball drawn in the ithi^{\text{th}} draw is a white ball, green ball, and blue ball, respectively. If the probability P(W1∩G2∩B3)=25NP(W_1\cap G_2\cap B_3)=\dfrac{2}{5N} and the conditional probability P(B3∣W1∩G2)=29P(B_3\mid W_1\cap G_2)=\dfrac{2}{9}, then NN equals ______.
Not attempted
Your answer: —
Correct answer: 11
8
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 12\dfrac12. Also assume that the probability of the answer for a question being guessed, given that the student's answer is correct, is 16\dfrac16. Then the probability that the student knows the answer of a randomly chosen question is
Not attempted
A
112\dfrac{1}{12}
B
17\dfrac{1}{7}
C
57\dfrac{5}{7}
D
512\dfrac{5}{12}
9
Let XX be a random variable, and let P(X=x)P(X=x) denote the probability that XX takes the value xx. Suppose that the points (x,P(X=x))(x,P(X=x)), x=0,1,2,3,4x=0,1,2,3,4, lie on a fixed straight line in the xyxy-plane, and P(X=x)=0P(X=x)=0 for all x∈R−{0,1,2,3,4}x\in\mathbb R-\{0,1,2,3,4\}. If the mean of XX is 52\dfrac52, and the variance of XX is α\alpha, then the value of 24α24\alpha is ______.
Not attempted
Your answer: —
Correct answer: 42
10
Consider the 6×66\times6 square shown in the figure. Let A1,A2,…,A49A_1,A_2,\ldots,A_{49} be the points of intersection (the dots in the picture), taken in some order. We say AiA_i and AjA_j are friends if they are adjacent along a row or along a column. Assume each point AiA_i has an equal chance of being chosen. Let pip_i be the probability that a randomly chosen point has exactly ii friends, for i=0,1,2,3,4i=0,1,2,3,4. Let XX be a random variable such that P(X=i)=piP(X=i)=p_i for i=0,1,2,3,4i=0,1,2,3,4. Then the value of 7E(X)7E(X) is
Not attempted
Your answer: —
Correct answer: 24
11
Consider the 6×66\times6 square shown in the figure. Let A1,A2,…,A49A_1,A_2,\ldots,A_{49} be the points of intersection (the dots in the picture), taken in some order. We say AiA_i and AjA_j are friends if they are adjacent along a row or along a column. Assume each point AiA_i has an equal chance of being chosen. Two distinct points are chosen randomly out of the points A1,A2,…,A49A_1,A_2,\ldots,A_{49}. Let pp be the probability that they are friends. Then the value of 7p7p is
Not attempted
Your answer: —
Correct answer: 0.5
12
Consider an experiment of tossing a coin repeatedly until the outcomes of two consecutive tosses are the same. If the probability of a random toss resulting in a head is 13\dfrac{1}{3}, then the probability that the experiment stops with a head (i.e. the two matching consecutive tosses are both heads) is
Not attempted
A
13\dfrac{1}{3}
B
521\dfrac{5}{21}
C
421\dfrac{4}{21}
D
27\dfrac{2}{7}
13
Consider the given data with frequency distribution: xix_i: 3, 8, 11, 10, 5, 43,\ 8,\ 11,\ 10,\ 5,\ 4 fif_i: 5, 2, 3, 2, 4, 45,\ 2,\ 3,\ 2,\ 4,\ 4 Match each entry in List-I to the correct entries in List-II. List-I (P) The mean of the above data is (Q) The median of the above data is (R) The mean deviation about the mean of the above data is (S) The mean deviation about the median of the above data is List-II (1) 2.52.5 (2) 55 (3) 66 (4) 2.72.7 (5) 2.42.4 The correct option is:
Not attempted
A
(P)→(3)(Q)→(2)(R)→(4)(S)→(5)(P)\to(3)\quad(Q)\to(2)\quad(R)\to(4)\quad(S)\to(5)
B
(P)→(3)(Q)→(2)(R)→(1)(S)→(5)(P)\to(3)\quad(Q)\to(2)\quad(R)\to(1)\quad(S)\to(5)
C
(P)→(2)(Q)→(3)(R)→(4)(S)→(1)(P)\to(2)\quad(Q)\to(3)\quad(R)\to(4)\quad(S)\to(1)
D
(P)→(3)(Q)→(3)(R)→(5)(S)→(5)(P)\to(3)\quad(Q)\to(3)\quad(R)\to(5)\quad(S)\to(5)
14
Let X={(x,y)∈Z×Z:x28+y220<1 and y2<5x}X = \left\{(x,y)\in\mathbb{Z}\times\mathbb{Z} : \dfrac{x^2}{8}+\dfrac{y^2}{20}<1 \text{ and } y^2<5x\right\}. Three distinct points PP, QQ and RR are randomly chosen from XX. Then the probability that PP, QQ and RR form a triangle whose area is a positive integer, is
Not attempted
A
71220\dfrac{71}{220}
B
73220\dfrac{73}{220}
C
79220\dfrac{79}{220}
D
83220\dfrac{83}{220}
15
Suppose that Box-I contains 8 red, 3 blue and 5 green balls, 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 bb. If bb is red then a ball is chosen randomly from Box-II, if bb is blue then a ball is chosen randomly from Box-III, and if bb 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
Not attempted
A
15256\dfrac{15}{256}
B
316\dfrac{3}{16}
C
552\dfrac{5}{52}
D
18\dfrac{1}{8}
16
Two players, P1P_1 and P2P_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 xx and yy denote the readings on the die rolled by P1P_1 and P2P_2, respectively. If x>yx>y, then P1P_1 scores 55 points and P2P_2 scores 00 point. If x=yx=y, then each player scores 22 points. If x<yx<y, then P1P_1 scores 00 point and P2P_2 scores 55 points. Let XiX_i and YiY_i be the total scores of P1P_1 and P2P_2, respectively, after playing the ithi^{th} round. List-I (I) Probability of (X2≥Y2)(X_2\ge Y_2) is (II) Probability of (X2>Y2)(X_2>Y_2) is (III) Probability of (X3=Y3)(X_3=Y_3) is (IV) Probability of (X3>Y3)(X_3>Y_3) is List-II (P) 38\dfrac38 (Q) 1116\dfrac{11}{16} (R) 516\dfrac5{16} (S) 355864\dfrac{355}{864} (T) 77432\dfrac{77}{432} The correct option is:
Not attempted
A
(I)→(Q); (II)→(R); (III)→(T); (IV)→(S)(I)\to(Q);\ (II)\to(R);\ (III)\to(T);\ (IV)\to(S)
B
(I)→(Q); (II)→(R); (III)→(T); (IV)→(T)(I)\to(Q);\ (II)\to(R);\ (III)\to(T);\ (IV)\to(T)
C
(I)→(P); (II)→(R); (III)→(Q); (IV)→(S)(I)\to(P);\ (II)\to(R);\ (III)\to(Q);\ (IV)\to(S)
D
(I)→(P); (II)→(R); (III)→(Q); (IV)→(T)(I)\to(P);\ (II)\to(R);\ (III)\to(Q);\ (IV)\to(T)
17
A number is chosen at random from the set {1,2,3,…,2000}\{1,2,3,\ldots,2000\}. Let pp be the probability that the chosen number is a multiple of 33 or a multiple of 77. Then the value of 500p500p is ___.
Not attempted
Your answer: —
Correct answer: 214
18
Let E,FE, F and GG be three events having probabilities P(E)=18P(E)=\frac{1}{8}, P(F)=16P(F)=\frac{1}{6} and P(G)=14P(G)=\frac{1}{4}, and let P(E∩F∩G)=110P(E\cap F\cap G)=\frac{1}{10}. For any event HH, if HcH^c denotes its complement, then which of the following statements is (are) TRUE?
Not attempted
A
P(E∩F∩Gc)≤140P(E\cap F\cap G^c)\le\frac{1}{40}
B
P(Ec∩F∩G)≤115P(E^c\cap F\cap G)\le\frac{1}{15}
C
P(E∪F∪G)≤1324P(E\cup F\cup G)\le\frac{13}{24}
D
P(Ec∩Fc∩Gc)≤512P(E^c\cap F^c\cap G^c)\le\frac{5}{12}
19
Three numbers are chosen at random, one after another with replacement, from the set S={1,2,3,…,100}S=\{1,2,3,\ldots,100\}. Let p1p_1 be the probability that the maximum of the chosen numbers is at least 8181 and p2p_2 be the probability that the minimum of the chosen numbers is at most 4040. The value of 6254p1\frac{625}{4}p_1 is ____.
Not attempted
Your answer: —
Correct answer: 76.25
20
Three numbers are chosen at random, one after another with replacement, from the set S={1,2,3,…,100}S=\{1,2,3,\ldots,100\}. Let p1p_1 be the probability that the maximum of the chosen numbers is at least 8181 and p2p_2 be the probability that the minimum of the chosen numbers is at most 4040. The value of 1254p2\frac{125}{4}p_2 is ____.
Not attempted
Your answer: —
Correct answer: 24.5
21
Consider three sets E1={1,2,3}E_1=\{1,2,3\}, F1={1,3,4}F_1=\{1,3,4\} and G1={2,3,4,5}G_1=\{2,3,4,5\}. Two elements are chosen at random, without replacement, from the set E1E_1, and let S1S_1 denote the set of these chosen elements. Let E2=E1−S1E_2=E_1-S_1 and F2=F1∪S1F_2=F_1\cup S_1. Now two elements are chosen at random, without replacement, from the set F2F_2 and let S2S_2 denote the set of these chosen elements. Let G2=G1∪S2G_2=G_1\cup S_2. Finally, two elements are chosen at random, without replacement, from the set G2G_2 and let S3S_3 denote the set of these chosen elements. Let E3=E2∪S3E_3=E_2\cup S_3. Given that E1=E3E_1=E_3, let pp be the conditional probability of the event S1={1,2}S_1=\{1,2\}. Then the value of pp is
Not attempted
A
15\frac{1}{5}
B
35\frac{3}{5}
C
12\frac{1}{2}
D
25\frac{2}{5}
22
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 pp is the probability that this perfect square is an odd number, then the value of 14p14p is ____
Not attempted
Your answer: —
Correct answer: 8
23
The probability that a missile hits a target successfully is 0.750.75. 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 0.950.95, is ____
Not attempted
Your answer: —
Correct answer: 6
24
Let C1C_1 and C2C_2 be two biased coins such that the probabilities of getting head in a single toss are 23\frac{2}{3} and 13\frac{1}{3}, respectively. Suppose α\alpha is the number of heads that appear when C1C_1 is tossed twice, independently, and suppose β\beta is the number of heads that appear when C2C_2 is tossed twice, independently. Then the probability that the roots of the quadratic polynomial x2−αx+βx^2 - \alpha x + \beta are real and equal, is
Not attempted
A
4081\frac{40}{81}
B
2081\frac{20}{81}
C
12\frac{1}{2}
D
14\frac{1}{4}
25
Let ∣X∣|X| denote the number of elements in a set XX. Let S={1,2,3,4,5,6}S=\{1,2,3,4,5,6\} be a sample space, where each element is equally likely to occur. If AA and BB are independent events associated with SS, then the number of ordered pairs (A,B)(A,B) such that 1≤∣B∣<∣A∣1\le|B|<|A|, equals ____
Not attempted
Your answer: —
Correct answer: 422
26
There are three bags B1B_1, B2B_2 and B3B_3. The bag B1B_1 contains 55 red and 55 green balls, B2B_2 contains 33 red and 55 green balls, and B3B_3 contains 55 red and 33 green balls. Bags B1B_1, B2B_2 and B3B_3 have probabilities 310,310\dfrac{3}{10},\dfrac{3}{10} and 410\dfrac{4}{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?
Not attempted
A
Probability that the selected bag is B3B_3 and the chosen ball is green equals 310\dfrac{3}{10}
B
Probability that the chosen ball is green equals 3980\dfrac{39}{80}
C
Probability that the chosen ball is green, given that the selected bag is B3B_3, equals 38\dfrac38
D
Probability that the selected bag is B3B_3, given that the chosen ball is green, equals 513\dfrac{5}{13}
27
Let SS be the sample space of all 3×33\times3 matrices with entries from the set {0,1}\{0,1\}. Let the events E1E_1 and E2E_2 be given by E1={A∈S:det⁡A=0}E_1=\{A\in S:\det A=0\} and E2={A∈S:sum of entries of A is 7}E_2=\{A\in S:\text{sum of entries of }A\text{ is }7\}. If a matrix is chosen at random from SS, then the conditional probability P(E1∣E2)P(E_1\mid E_2) equals _____
Not attempted
Your answer: —
Correct answer: 0.5
28
There are five students S1,S2,S3,S4S_1, S_2, S_3, S_4 and S5S_5 in a music class and for them there are five seats R1,R2,R3,R4R_1, R_2, R_3, R_4 and R5R_5 arranged in a row, where initially the seat RiR_i is allotted to the student SiS_i, i=1,2,3,4,5i=1,2,3,4,5. But, on the examination day, the five students are randomly allotted the five seats. The probability that, on the examination day, the student S1S_1 gets the previously allotted seat R1R_1, and NONE of the remaining students gets the seat previously allotted to him/her is
Not attempted
A
340\dfrac{3}{40}
B
18\dfrac{1}{8}
C
740\dfrac{7}{40}
D
15\dfrac{1}{5}
29
There are five students S1,S2,S3,S4S_1, S_2, S_3, S_4 and S5S_5 in a music class and for them there are five seats R1,R2,R3,R4R_1, R_2, R_3, R_4 and R5R_5 arranged in a row, where initially the seat RiR_i is allotted to the student SiS_i, i=1,2,3,4,5i=1,2,3,4,5. But, on the examination day, the five students are randomly allotted the five seats. For i=1,2,3,4i=1,2,3,4, let TiT_i denote the event that the students SiS_i and Si+1S_{i+1} do NOT sit adjacent to each other on the day of the examination. Then, the probability of the event T1∩T2∩T3∩T4T_1\cap T_2\cap T_3\cap T_4 is
Not attempted
A
115\dfrac{1}{15}
B
110\dfrac{1}{10}
C
760\dfrac{7}{60}
D
15\dfrac{1}{5}
30
Three randomly chosen nonnegative integers x,yx, y and zz are found to satisfy the equation x+y+z=10x + y + z = 10. Then the probability that zz is even, is
Not attempted
A
3655\frac{36}{55}
B
611\frac{6}{11}
C
12\frac{1}{2}
D
511\frac{5}{11}
31
Let XX and YY be two events such that P(X)=13P(X)=\dfrac{1}{3}, P(X∣Y)=12P(X\mid Y)=\dfrac{1}{2} and P(Y∣X)=25P(Y\mid X)=\dfrac{2}{5}. Then
Not attempted
A
P(Y)=415P(Y)=\dfrac{4}{15}
B
P(X′∣Y)=12P(X'\mid Y)=\dfrac{1}{2}
C
P(X∩Y)=15P(X\cap Y)=\dfrac{1}{5}
D
P(X∪Y)=25P(X\cup Y)=\dfrac{2}{5}
32
Comprehension\textbf{Comprehension}: Football teams T1T_1 and T2T_2 play two games against each other; the outcomes of the two games are independent. The probabilities of T1T_1 winning, drawing and losing a game against T2T_2 are 12,16,13\dfrac12,\dfrac16,\dfrac13 respectively. Each team gets 33 points for a win, 11 point for a draw and 00 points for a loss in a game. Let XX and YY denote the total points scored by T1T_1 and T2T_2 respectively after two games. P(X>Y)P(X>Y) is
Not attempted
A
14\dfrac14
B
512\dfrac{5}{12}
C
12\dfrac12
D
712\dfrac{7}{12}
33
Comprehension\textbf{Comprehension}: Football teams T1T_1 and T2T_2 play two games against each other; the outcomes of the two games are independent. The probabilities of T1T_1 winning, drawing and losing a game against T2T_2 are 12,16,13\dfrac12,\dfrac16,\dfrac13 respectively. Each team gets 33 points for a win, 11 point for a draw and 00 points for a loss in a game. Let XX and YY denote the total points scored by T1T_1 and T2T_2 respectively after two games. P(X=Y)P(X=Y) is
Not attempted
A
1136\dfrac{11}{36}
B
13\dfrac13
C
1336\dfrac{13}{36}
D
12\dfrac12
34
A computer producing factory has only two plants T1T_1 and T2T_2. Plant T1T_1 produces 20\% and plant T2T_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 T1)=10 P(computer turns out to be defective given that it is produced in plant T2)P(\text{computer turns out to be defective given that it is produced in plant } T_1)=10\,P(\text{computer turns out to be defective given that it is produced in plant } T_2), where P(E)P(E) denotes the probability of an event EE. 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 T2T_2 is
Not attempted
A
3673\dfrac{36}{73}
B
4779\dfrac{47}{79}
C
7893\dfrac{78}{93}
D
7583\dfrac{75}{83}
35
Setup (shared with Q.58)\textbf{Setup (shared with Q.58)}: Let n1n_1 and n2n_2 be the number of red and black balls, respectively, in box I. Let n3n_3 and n4n_4 be the number of red and black balls, respectively, in box II. One of the two boxes, box I and box II, 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 II is 13\dfrac13, then the correct option(s) with the possible values of n1,n2,n3n_1,n_2,n_3 and n4n_4 is(are)
Not attempted
A
n1=3, n2=3, n3=5, n4=15n_1=3,\ n_2=3,\ n_3=5,\ n_4=15
B
n1=3, n2=6, n3=10, n4=50n_1=3,\ n_2=6,\ n_3=10,\ n_4=50
C
n1=8, n2=6, n3=5, n4=20n_1=8,\ n_2=6,\ n_3=5,\ n_4=20
D
n1=6, n2=12, n3=5, n4=20n_1=6,\ n_2=12,\ n_3=5,\ n_4=20
36
Setup (shared with Q.58)\textbf{Setup (shared with Q.58)}: Let n1n_1 and n2n_2 be the number of red and black balls, respectively, in box I. Let n3n_3 and n4n_4 be the number of red and black balls, respectively, in box II. A ball is drawn at random from box I and transferred to box II. If the probability of drawing a red ball from box I, after this transfer, is 13\dfrac13, then the correct option(s) with the possible values of n1n_1 and n2n_2 is(are)
Not attempted
A
n1=4n_1=4 and n2=6n_2=6
B
n1=2n_1=2 and n2=3n_2=3
C
n1=10n_1=10 and n2=20n_2=20
D
n1=3n_1=3 and n2=6n_2=6
37
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 0.960.96, is
Not attempted
Your answer: —
Correct answer: 8
38
Comprehension\textbf{Comprehension}: Box 1 contains three cards bearing numbers 1,2,31,2,3; box 2 contains five cards bearing numbers 1,2,3,4,51,2,3,4,5; and box 3 contains seven cards bearing numbers 1,2,3,4,5,6,71,2,3,4,5,6,7. A card is drawn from each of the boxes. Let xix_i be the number on the card drawn from the ithi^{\text{th}} box, i=1,2,3i=1,2,3. The probability that x1+x2+x3x_1+x_2+x_3 is odd, is
Not attempted
A
29105\dfrac{29}{105}
B
53105\dfrac{53}{105}
C
57105\dfrac{57}{105}
D
12\dfrac12
39
Comprehension\textbf{Comprehension}: Box 1 contains three cards bearing numbers 1,2,31,2,3; box 2 contains five cards bearing numbers 1,2,3,4,51,2,3,4,5; and box 3 contains seven cards bearing numbers 1,2,3,4,5,6,71,2,3,4,5,6,7. A card is drawn from each of the boxes. Let xix_i be the number on the card drawn from the ithi^{\text{th}} box, i=1,2,3i=1,2,3. The probability that x1,x2,x3x_1,x_2,x_3 are in an arithmetic progression, is
Not attempted
A
9105\dfrac{9}{105}
B
10105\dfrac{10}{105}
C
11105\dfrac{11}{105}
D
7105\dfrac{7}{105}
40
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
Not attempted
A
12\dfrac12
B
13\dfrac13
C
23\dfrac23
D
34\dfrac34
41
A box B1B_1 contains 1 white ball, 3 red balls and 2 black balls. Another box B2B_2 contains 2 white balls, 3 red balls and 4 black balls. A third box B3B_3 contains 3 white balls, 4 red balls and 5 black balls. If 1 ball is drawn from each of the boxes B1B_1, B2B_2 and B3B_3, the probability that all 3 drawn balls are of the same colour is
Not attempted
A
82648\dfrac{82}{648}
B
90648\dfrac{90}{648}
C
558648\dfrac{558}{648}
D
566648\dfrac{566}{648}
42
A box B1B_1 contains 1 white ball, 3 red balls and 2 black balls. Another box B2B_2 contains 2 white balls, 3 red balls and 4 black balls. A third box B3B_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 B2B_2 is
Not attempted
A
116181\dfrac{116}{181}
B
126181\dfrac{126}{181}
C
65181\dfrac{65}{181}
D
55181\dfrac{55}{181}
43
Of the three independent events E1,E2E_1,E_2 and E3E_3, the probability that only E1E_1 occurs is α\alpha, only E2E_2 occurs is β\beta and only E3E_3 occurs is γ\gamma. Let the probability pp that none of the events E1,E2E_1,E_2 or E3E_3 occurs satisfy the equations (α−2β)p=αβ(\alpha-2\beta)p=\alpha\beta and (β−3γ)p=2βγ(\beta-3\gamma)p=2\beta\gamma. All the given probabilities are assumed to lie in the interval (0,1)(0,1). Then Probability of occurrence of E1Probability of occurrence of E3=\dfrac{\text{Probability of occurrence of }E_1}{\text{Probability of occurrence of }E_3}=
Not attempted
Your answer: —
Correct answer: 6
44
Four persons independently solve a certain problem correctly with probabilities 12,34,14,18\dfrac{1}{2},\dfrac{3}{4},\dfrac{1}{4},\dfrac{1}{8}. Then the probability that the problem is solved correctly by at least one of them is
Not attempted
A
235256\dfrac{235}{256}
B
21256\dfrac{21}{256}
C
3256\dfrac{3}{256}
D
253256\dfrac{253}{256}
P(at least one)=1−P(none solves it)P(\text{at least one}) = 1 - P(\text{none solves it}), and by independence P(none)=∏(1−pi)P(\text{none}) = \prod (1-p_i).
Deconstruction\textbf{Deconstruction}: Four independent solvers with given individual success probabilities; find the probability that at least one succeeds. Strategy\textbf{Strategy}: Complement rule for independent events: P(at least one)=1−∏i=14(1−pi)P(\text{at least one})=1-\prod_{i=1}^4(1-p_i). Execution\textbf{Execution}: 1−p1=12, 1−p2=14, 1−p3=34, 1−p4=781-p_1=\dfrac12,\ 1-p_2=\dfrac14,\ 1-p_3=\dfrac34,\ 1-p_4=\dfrac78. - Product =12⋅14⋅34⋅78=1⋅1⋅3⋅72⋅4⋅4⋅8=21256=\dfrac{1}{2}\cdot\dfrac{1}{4}\cdot\dfrac{3}{4}\cdot\dfrac{7}{8}=\dfrac{1\cdot1\cdot3\cdot7}{2\cdot4\cdot4\cdot8}=\dfrac{21}{256}. - P(at least one)=1−21256=235256P(\text{at least one}) = 1-\dfrac{21}{256}=\dfrac{235}{256}. Verification\textbf{Verification}: Each factor (1−pi)∈(0,1)(1-p_i)\in(0,1) so the product is a valid probability less than any individual factor (it is ≤1/2\le 1/2, matching 21256<12\dfrac{21}{256}<\dfrac12); the final probability 235256\dfrac{235}{256} is close to 1, sensible given one solver already has p=3/4p=3/4. Final Answer\textbf{Final Answer}: (A) 235256\dfrac{235}{256}