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JEE • MATHEMATICS • PROBABILITY

Probability practice · Mixed difficulty

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Practice format and chapter quiz

These sets use single-correct-answer MCQs. They do not cover every JEE Main or JEE Advanced question format and are not full-length mock papers.

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Showing 721-750 of 1130 questions

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721 Medium
721 of 1130
Application +4 | -1
A bag contains 10 white and 3 black balls. Balls are drawn one-by-one without replacement till all the black balls are drawn. The probability that the procedure of drawing balls will come to an end at the seventh draw is
View answer and explanation
Correct option: B
15286\frac{{15}}{{286}}
Option B (15286\frac{{15}}{{286}}) is correct because it represents the probability that the last black ball is drawn on the seventh draw, calculated using conditional probability.
722 Easy
722 of 1130
Application +4 | -1
A die is loaded so that the probability of face ii is proportional to i,i=1,2,…,6i, i=1,2, \ldots, 6. The probability of an even number occuring when the die is rolled, is
View answer and explanation
Correct option: B
47\frac{4}{7}
Option B (4/7) is correct because it calculates the probability of an even number occurring by summing the probabilities of the even faces, given that the probability of each face is proportional to its number.
723 Easy
723 of 1130
Analyzing +4 | -1
AA is the set of all four digit natural numbers which has exactly one digit ' 7 '. A number is chosen at random from the set AA. The probability that the chosen number when divided by 5 leaves the remainder 2 , is
View answer and explanation
Correct option: A
97297\frac{97}{297}
Option A (97297\frac{{97}}{{297}}) is correct because it represents the probability calculated by dividing the number of favorable outcomes (numbers divisible by 5 with remainder 2) by the total number of four-digit numbers with exactly one digit '7'.
724 Medium
724 of 1130
Application +4 | -1
A bag contains 5 white, 3 black balls. 2 balls are drawn one by one, their colour noted and
replaced in bag then probability that one ball is white and one is black, is
View answer and explanation
Correct option: A
1532\frac{15}{32}
Option A (1532\frac{{15}}{{32}}) is correct because it calculates the probability of drawing one white and one black ball in either order, with replacement.
725 Easy
725 of 1130
Application +4 | -1
The letters of the word 'ARTICLE' are arranged at random. The Probability that the vowels may occupy the even places, is
View answer and explanation
Correct option: A
135\frac{1}{35}
Option A (135\frac{1}{35}) is correct because it represents the probability of arranging the letters in 'ARTICLE' such that the vowels occupy the even places.
726 Easy
726 of 1130
Application +4 | -1
Two numbers xx and yy are chosen at random from the set {1,2,3,…,30}\{1,2,3, \ldots, 30\}. The probability that x2−y2x^2-y^2 is divisible by 3 is
View answer and explanation
Correct option: D
none of these
Option D (none of these) is correct because the probability that x2−y2x^2-y^2 is divisible by 3 is 4787\frac{47}{87}, which is not among the given options.
727 Easy
727 of 1130
Application +4 | -1
Suppose an unbiased die has nn sides numbered i=1,2i=1,2, .... nn. The die is tossed nn times (assume independence) and a "match" is defined to be the occurrence of side ii on the its roll. The probability of at least one match is
View answer and explanation
Correct option: C
1−(1−1n)n1-\left(1-\frac{1}{n}\right)^n
Option C (1-(1-1/n)^n) is correct because it calculates the probability of at least one match by finding the complement of the probability of no matches.
728 Easy
728 of 1130
Understanding +4 | -1
What is the probability that when one die is thrown, the number appearing on top is even?
View answer and explanation
Correct option: C
1/21/2
Option C (12\frac{1}{2}) is correct because there are 3 even numbers (2, 4, 6) on a standard die, and 3/6 simplifies to 1/2.
729 Easy
729 of 1130
Application +4 | -1
A fair die is rolled repeatedly. The probability that 1 occurs for the first time at the even throw is
View answer and explanation
Correct option: B
511\frac{5}{11}
Option B (5/11) is correct because it represents the sum of probabilities of the first 1 occurring at the 2nd, 4th, 6th, ... throws, forming a geometric series.
730 Medium
730 of 1130
Application +4 | -1
A multiple choice examination has 5 questions. Each question has three alternative answers of which exactly one is correct. The probability that a student will get 4 or more correct answers just by guessing is
View answer and explanation
Correct option: C
1135\frac{11}{3^5}
Option C (1135\frac{11}{3^5}) is correct because it calculates the probability of getting 4 or 5 correct answers by guessing using the binomial distribution formula.
731 Easy
731 of 1130
Application +4 | -1
The probability of choosing randomly a number cc from the set {1, 2, 3, ….., 9} such that the quadratic equation x2+4x+c=0{x^2} + 4x + c = 0 has real roots is
View answer and explanation
Correct option: D
49\frac{4}{9}
Option D (49\frac{4}{9}) is correct because it represents the probability of choosing a number c such that the quadratic equation has real roots, calculated by finding the values of c that satisfy the discriminant condition.
732 Medium
732 of 1130
Application +4 | -1
A fair coin is tossed a fixed number of times. If the probability of getting exactly 3 heads equals the probability of getting exactly 5 heads, then probability of getting exactly one head is
View answer and explanation
Correct option: B
132\frac{1}{32}
Option B (132\frac{1}{32}) is correct because it represents the probability of getting exactly one head, given that the probability of getting exactly 3 heads equals the probability of getting exactly 5 heads.
733 Easy
733 of 1130
Application +4 | -1
A die is thrown 5 times. Getting an odd number is consider a success. The variance of the distribution is
View answer and explanation
Correct option: D
54\frac{5}{4}
Option D (5/4) is correct because the variance of a binomial distribution is calculated as npq, where n=5 and p=1/2.
734 Medium
734 of 1130
Understanding +4 | -1
A fair coin is tossed 2019 times. Probability of getting at least 1010 heads is
View answer and explanation
Correct option: A
12\frac{1}{2}
Option A (12\frac{1}{2}) is correct because the probability of getting at least 1010 heads is the same as the probability of getting less than or equal to 1009 heads, and these two probabilities must sum to 1.
735 Medium
735 of 1130
Application +4 | -1
The odds in favour of standing first of three student appearing in an enamination are 1:2,2:51: 2,2: 5 and 1:71: 7 respectively. The probability that either of them will stand first, is
View answer and explanation
Correct option: A
125168\frac{125}{168}
Option A (125/168) is correct because it calculates the probability that either of the three students will stand first by summing their individual probabilities of standing first, as the events are mutually exclusive.
736 Easy
736 of 1130
Application +4 | -1
If 4P(A)=6P(B)=10P(A∩B)=1,4P\left( A \right) = 6P\left( B \right) = 10P\left( {A \cap B} \right) = 1, then P(BA)P\left( {\frac{B}{A}} \right) is equal to
View answer and explanation
Correct option: A
25\frac{2}{5}
Option A (25\frac{2}{5}) is correct because it is calculated by using the conditional probability formula P(B/A)=P(A∩B)P(A)P(B/A) = \frac{P(A \cap B)}{P(A)}.
737 Medium
737 of 1130
Understanding +4 | -1
A locker can be opened by dialing a fixed three-digit code (between 000 and 999). A stranger, who does not know the code, tries to open the locker by dialing three digits at random. If pp is the probability that the stranger succeeds at the kk th trial, then 1000p1000 p is equal to. (Assume that the stranger does not repeat unsuccessful combinations.)
View answer and explanation
Correct option: A
1
Option A (1) is correct because the probability of success on the kth trial is 1/1000, leading to 1000p = 1.
738 Easy
738 of 1130
Application +4 | -1
Three six-faced dice are thrown together. The probability that the sum of the numbers appearing on the dice is k(9≤k≤14)k(9 \leq k \leq 14) is
View answer and explanation
Correct option: A
21k−k2−83216\frac{21 k-k^2-83}{216}
Option A (21k−k2−83216\frac{{21 k-k^2-83}}{{216}}) is correct because it calculates the probability by considering the coefficient of xk−3x^{k-3} in the expansion of (1−x6)3(1−x)−3(1-x^6)^3(1-x)^{-3} for the given range of k.
739 Easy
739 of 1130
Application +4 | -1
A dice is thrown two times. If getting the odd number is considered as success, then the probability of two successes is
View answer and explanation
Correct option: D
14\frac{1}{4}
Option D (14\frac{1}{4}) is correct because the probability of getting an odd number on a dice is 1/2, and since the dice are thrown twice, the probability of two successes is (1/2)*(1/2).
740 Easy
740 of 1130
Application +4 | -1
A man takes a step forward with probability 0.4 and back-ward with probability 0.6. The probability that at the end of eleven steps he is one step away from the starting point is
View answer and explanation
Correct option: A
  11C6(0.24)5{\;^{11}}{C_6}{\left( {0.24} \right)^5}
Option A (  11C6(0.24)5{\;^{11}}{C_6}{\left( {0.24} \right)^5}) is correct because it accounts for the probability of ending one step away by moving either one step forward or one step backward.
741 Medium
741 of 1130
Application +4 | -1
A JEE aspirant estimates that he will be successful with an 80% chance if he studies 10 hr/day with 60%
chance if he studies 7 hr/day and with 40% chance if he studies 4 hr/day. Further, he believes that he will
study 10 hr, 7 hr and 4 hr/day with probability 0.1, 0.2 and 0.7 respectively. Given that he is successful,
the probability that he studies for 4 hr/day equals.
View answer and explanation
Correct option: C
712\frac{7}{12}
Option C (712\frac{7}{12}) is correct because it's calculated using Bayes' theorem or conditional probability, considering the probabilities of success given different study hours and the probabilities of studying those hours.
742 Medium
742 of 1130
Analyzing +4 | -1
Let AA and BB two events such that P(A∪B‾)=16,P(A∩B)=14P\left( {\overline {A \cup B} } \right) = \frac{1}{6},P\left( {A \cap B} \right) = \frac{1}{4} and P(Aˉ)=14,P\left( {\bar A} \right) = \frac{1}{4}, where Aˉ\bar A stands for complement of event AA. Then, events AA and BB are
View answer and explanation
Correct option: B
Independent but not equally likely
Option B (Independent but not equally likely) is correct because the events A and B are independent as P(A ∩ B) = P(A) * P(B), but P(A) and P(B) have different values.
743 Medium
743 of 1130
Application +4 | -1
In a random experiment, a fair die is rolled until two fours are obtained in succession. The probability that the experiment will end in the fifth throw of the die is equal to
View answer and explanation
Correct option: A
17565\frac{{175}}{{{6^5}}}
Option C (cos⁡−1(1935)\cos ^{ - 1}(\frac{{19}}{{35}})) is correct because it represents the angle between the faces calculated using the dot product of their normal vectors.
744 Easy
744 of 1130
Analyzing +4 | -1
In x=33n,nx = {33^n},n is a positive integral value, then what is the probability that xx will have 3 at its units place?
View answer and explanation
Correct option: B
01-Apr
Option B, 01-Apr, is correct because the probability that 33n33^n has 3 at its units place is 1/4, as the units digit cycles through 3, 9, 7, and 1.
745 Easy
745 of 1130
Application +4 | -1
A man draws a card from a pack of 52 cards and then replaces it. After shuffling the pack, he again draws a card. This he repeats a number of times. The probability that he will draw a heart for the first time in the third draw is
View answer and explanation
Correct option: A
964\frac{9}{64}
Option A (964\frac{9}{64}) is correct because it calculates the probability of not drawing a heart in the first two draws and then drawing a heart in the third draw, with replacement.
746 Medium
746 of 1130
Analyzing +4 | -1
AA is a set containing n elements. A subset PP of AA is chosen at random. The set AA is reconstructed by replacing the elements of PP. A subset QQ is again chosen at random. The probability that P=QP=Q, is
View answer and explanation
Correct option: A
12n\frac{1}{2^n}
Option A (12n\frac{1}{2^n}) is correct because it calculates the probability that two randomly chosen subsets P and Q of A are equal.
747 Easy
747 of 1130
Application +4 | -1
If XX is a binomial variate with the range {0,1,2,3,4,5,6} and P(X=2)=4P(X=4),P\left( {X = 2} \right) = 4P\left( {X = 4} \right), then the parameter pp of XX is
View answer and explanation
Correct option: A
13\frac{1}{3}
Option A (13\frac{1}{3}) is correct because it is the parameter p of the binomial variate X calculated using the given relationship between P(X=2) and P(X=4).
748 Easy
748 of 1130
Application +4 | -1
A letter is taken out at random from `ASSISTANT’ and another is taken out from `STATISTICS’. The probability that they are the same letters is
View answer and explanation
Correct option: C
1990\frac{{19}}{{90}}
Option C (1990\frac{{19}}{{90}}) is correct because it sums the probabilities of drawing the same letter (A, I, S, or T) from both words.
749 Medium
749 of 1130
Application +4 | -1
A letter is known to have come from CHENNAI, JAIPUR, NAINITAL, DUBAI and MUMBAI. On the post mark only two consecutive letters AI are legible. Then the probability that it come from MUMBAI, is
View answer and explanation
Correct option: B
84403\frac{84}{403}
Option B is correct because the probability is calculated using conditional probability, considering the likelihood of \"AI\" appearing in each city and then finding the probability it came from MUMBAI.
750 Easy
750 of 1130
Application +4 | -1
Five coins are tossed simultaneously. The probability that at least one head turning up, is
View answer and explanation
Correct option: D
31/3231 / 32
Option D (3132\frac{31}{32}) is correct because it's the probability of at least one head, calculated as 1 minus the probability of no heads.
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