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JEE • MATHEMATICS • PROBABILITY

Probability practice · Mixed difficulty

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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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391 Easy
391 of 1130
Application +4 | -1
Let A={1,  3,  5,  7,  9},  B={2,  4,  6,  8}.A = \left\{ {1,\;3,\;5,\;7,\;9} \right\},\;B = \left\{ {2,\;4,\;6,\;8} \right\}. If a cartesian product A×BA \times B, if chosen at random, the probability of a+b=9a + b = 9 is
View answer and explanation
Correct option: B
15\frac{1}{5}
Option B (15\frac{1}{5}) is correct because it calculates the probability of a + b = 9 when a Cartesian product A x B is chosen at random.
392 Easy
392 of 1130
Application +4 | -1
Let A=[2,3,4,…,20]{A}=[2,3,4, \ldots, 20]. A number is chosen at random from the set A{A} and it is found to be a prime number. The probability that it is more than 10 is
View answer and explanation
Correct option: C
15\frac{1}{5}
Option C (15\frac{1}{5}) is correct because it represents the conditional probability of selecting a prime number greater than 10, given that a prime number has already been selected from the set A.
393 Easy
393 of 1130
Application +4 | -1
A doctor is called to see a sick child. The doctor knows (prior to the visit) that 90%90 \% of the sick children in that neighbourhood are sick with the flue, denoted by FF, while 10%10 \% are sick with the measles, denoted by M. A well-known symptom of measles is a rash, denoted by RR.
The probability of having a rash for a child sick with the measles is 0.950.95. However, occasionally children with the flue also develop a rash with conditional probability 0.080.08. Upon examination the child, the doctor finds a rash, then the probability that the child has the measles, is
View answer and explanation
Correct option: D
95167\frac{95}{167}
Option D (95167\frac{95}{167}) is correct because it is calculated using Bayes' theorem to find the probability of measles given a rash.
394 Hard
394 of 1130
Application +4 | -1
Two dice are thrown nn times in succession. The probability of obtaining a double six at least once is
View answer and explanation
Correct option: B
1−(3536)n1 - {\left( {\frac{{35}}{{36}}} \right)^n}
Option B (1−(3536)n1 - {(\frac{{35}}{{36}})^n}) is correct because it represents the probability of obtaining at least one double six in n throws, calculated as 1 minus the probability of not obtaining any double sixes.
395 Easy
395 of 1130
Understanding +4 | -1
Write the set A={x:xA=\{x: x is a positive prime <10}< 10\} in the tabular form
View answer and explanation
Correct option: A
{2,3,5,7}\{2,3,5,7\}
Option A ({2,3,5,7}) is correct because these are the only positive prime numbers less than 10.
396 Easy
396 of 1130
Application +4 | -1
Let AA and BB be the sets {1,2,…10}\{1,2, \ldots 10\} and {1,2,...20}\{1,2, ...20\} respectively. A function is selected randomly from AA to BB the probability that the function is non-decreasing is
View answer and explanation
Correct option: C
29C19(20)10\frac{{ }^{29} C_{19}}{(20)^{10}}
Option C (29C19(20)10\frac{{ }^{29} C_{19}}{(20)^{10}}) is correct because it represents the ratio of non-decreasing functions to the total number of functions from A to B.
397 Medium
397 of 1130
Application +4 | -1
Throwing a biased die, a person will get 5 rupees if he throws the number 5 and will get 8 rupees for any
other number, then the expected income (in Rupees) per throw is (it is given that the number 5 will appear
5 times as frequently as any other number)
View answer and explanation
Correct option: A
6.5
Option A (6.5) is correct because the expected income is calculated by considering the probabilities of each outcome and their corresponding values.
398 Easy
398 of 1130
Analyzing +4 | -1
Six teams play a match with each other. The probability that neither a team win all the matches nor a team loose all the matches is
View answer and explanation
Correct option: C
175216\frac{175}{216}
Option C (175216\frac{175}{216}) is correct because the probability is calculated by subtracting the probabilities of a team winning all matches or losing all matches from 1.
399 Easy
399 of 1130
Application +4 | -1
Three coins are tossed, then what is the probability that at least two heads appears on upper face?
View answer and explanation
Correct option: D
None of these
The correct probability is 12\frac{1}{2}, which is not listed in the options. The explanation shows the calculation of this probability by considering all possible outcomes when tossing three coins.
400 Easy
400 of 1130
Application +4 | -1
An urn contains nine balls of which three are red, four are blue and two are green. Three balls are drawn at random without replacement from the urn. The probability that the three balls have different colours, is
View answer and explanation
Correct option: B
27\frac{2}{7}
Option B (27\frac{2}{7}) is correct because it calculates the probability of drawing three balls of different colors by finding the ratio of favorable outcomes to total possible outcomes.
401 Easy
401 of 1130
Application +4 | -1
n(≥3)n(\geq 3) persons are sitting in a row. Two of them are selected at random. The probability that they are not together is
View answer and explanation
Correct option: B
1−2n1-\frac{2}{n}
Option B (1−2n1-\frac{2}{n}) is correct because it represents the probability that two randomly selected persons are not sitting together in a row of n persons.
402 Medium
402 of 1130
Application +4 | -1
Two numbers aa and bb are chosen at random from the set {1,2,3,…,3n}\{1,2,3, \ldots, 3 n\}. The probability that a2−b2a^2-b^2 is divisible by 3 , is
View answer and explanation
Correct option: B
5n−33(3n−1)\frac{5 n-3}{3(3 n-1)}
Option B (5n−33(3n−1)\frac{5 n-3}{3(3 n-1)}) is correct because it calculates the probability that a2−b2a^2 - b^2 is divisible by 3 by considering the cases where a and b belong to the same residue class modulo 3 or to different classes that satisfy the divisibility condition.
403 Medium
403 of 1130
Application +4 | -1
An anti-aircraft gun can take a maximum of four shots at any plane moving away from it. The probabilities of hitting the plane at the 1st, 2nd, 3rd and 4th shots are 0.4, 0.3, 0.2 and 0.1 respectively. What is the probability that at least one shot hits the plane?
View answer and explanation
Correct option: A
0.6976
Option A (Nov-32) is correct because it is the sum of probabilities of getting exactly 4, 5, or 6 heads when tossing six coins simultaneously.
404 Medium
404 of 1130
Analyzing +4 | -1
Let R{R} be the region containing all the points within or on the sides of an equilateral triangle. A point PP is randomly selected from the region RR, the probability that PP lies near to internal angle bisector than any of the sides is equal to
View answer and explanation
Correct option: A
4−234-2 \sqrt{3}
The correct answer is {0} because the sine function will always evaluate to zero due to the greatest integer function, making the entire expression zero.
405 Easy
405 of 1130
Analyzing +4 | -1
A die is rolled thrice, the probability of getting a larger number each time than the previous number, is
View answer and explanation
Correct option: A
554\frac{5}{54}
Option A (554\frac{{5}}{{54}}) is correct because it represents the probability calculated by dividing the number of favorable outcomes (increasing numbers) by the total number of possible outcomes when rolling a die thrice.
406 Medium
406 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∪Q=AP \cup Q=A, is
View answer and explanation
Correct option: D
n3(34)n\frac{n}{3}\left(\frac{3}{4}\right)^n
Option B ((34)n\left(\frac{3}{4}\right)^n) is correct because it calculates the probability that the union of two randomly chosen subsets P and Q of A equals A.
407 Easy
407 of 1130
Application +4 | -1
Entries of a 2×22 \times 2 determinant are chosen from the set {−1,1}\{-1,1\}. The probability that determinant has zero value is
View answer and explanation
Correct option: C
12\frac{1}{2}
Option C (12\frac{1}{2}) is correct because half of the possible 2×22 \times 2 determinants with entries from {−1,1}\{-1, 1\} have a value of zero.
408 Easy
408 of 1130
Application +4 | -1
India plays a two ODI matches each with Australia and Pakistan. The probability of India getting points 0, 1, 2 are 0.45, 0.05, 0.50. The probability of India getting at least 7 points in the series is
View answer and explanation
Correct option: C
0.0875
Option C (0.0875) is correct because it calculates the probability of India getting at least 7 points in the series, considering the probabilities of getting 0, 1, or 2 points in each match.
409 Medium
409 of 1130
Application +4 | -1
Among 15 players, 8 are batsman and 7 are bowlers. The probability that a team is chosen of 6 batsman and 5 bowlers, is
View answer and explanation
Correct option: A
  8C6×  7C5  15C11\frac{{{\;^8}{C_6} \times {\;^7}{C_5}}}{{{\;^{15}}{C_{11}}}}
Option A (  8C6×  7C5  15C11\frac{{{\;^8}{C_6} \times {\;^7}{C_5}}}{{{\;^{15}}{C_{11}}}}) is correct because it represents the number of ways to choose 6 batsmen from 8 and 5 bowlers from 7, divided by the total number of ways to choose 11 players from 15.
410 Easy
410 of 1130
Application +4 | -1
The mean and variance of a random variable XX having a binomial distribution are 4 and 2 respectively, then P(X=1)P\left( {X = 1} \right) is
View answer and explanation
Correct option: A
132\frac{1}{{32}}
Option A (132\frac{1}{{32}}) is correct because it is the result of calculating the binomial probability P(X=1)P(X=1) using the given mean and variance to find nn and pp.
411 Easy
411 of 1130
Application +4 | -1
A draws two cards at random from a pack of 52 cards. After returning them to the pack and shuffling it, B draws two cards at random. The probability that their draws contain exactly one common cards is
View answer and explanation
Correct option: B
50663\frac{50}{663}
Option B (50663\frac{50}{663}) is correct because it calculates the probability that two draws of two cards each contain exactly one common card.
412 Easy
412 of 1130
Understanding +4 | -1
Five horses are in a race. Mr. A selects two of the horses at random and bets on them. The probability that Mr. A selected the winning horse is
View answer and explanation
Correct option: A
25\frac{2}{5}
Option A (2/5) is correct because it represents the probability that at least one of the two horses Mr. A selected is the winning horse.
413 Easy
413 of 1130
Application +4 | -1
In a college, 25% of the boys and 10% of the girls offer Mathematics. The girls constitute 60% of the total number of students. If a student is selected at random and is found to be studying Mathematics. The probability that the student is a girl is
View answer and explanation
Correct option: B
03-Aug
The correct answer is B (03-Aug), which represents 3/8. This is calculated using conditional probability, considering the proportion of girls and boys studying mathematics.
414 Easy
414 of 1130
Application +4 | -1
A card is drawn from a pack of 52 playing cards. The card is replaced and the pack is reshuffled. If this is done six times, the probability that 2 hearts, 2 diamonds and 2 black cards are drawn is
View answer and explanation
Correct option: C
90×(12)1090 \times\left(\frac{1}{2}\right)^{10}
The correct answer is 90×(12)1090 \times\left(\frac{1}{2}\right)^{10} because it calculates the probability of drawing 2 hearts, 2 diamonds, and 2 black cards in 6 draws with replacement.
415 Easy
415 of 1130
Analyzing +4 | -1
Let A, B, C be three mutually independent events.

Consider the two statements S1\mathrm{S}_1 and S2\mathrm{S}_2

S1:A\mathrm{S}_1: \mathrm{A} and B∪CB \cup C are independent

S2:A\mathrm{S}_2: \mathrm{A} and B∩CB \cap C are independent, then
View answer and explanation
Correct option: A
Both S1S_1 and S2S_2 are true
The correct answer is A (Both S1S_1 and S2S_2 are true) because if A, B, and C are mutually independent, then A is independent of any boolean expression involving only B and C.
416 Medium
416 of 1130
Application +4 | -1
If X{X} and Y{Y} are the independent random variables B(5,12){B}\left(5, \frac{1}{2}\right) and B(7,12){B}\left(7, \frac{1}{2}\right), then P(X+Y≥1)={P}({X}+{Y} \geq 1)=
View answer and explanation
Correct option: A
40954096\frac{4095}{4096}
Option A (40954096\frac{4095}{4096}) is correct because it calculates the probability of the sum of two binomial random variables being greater than or equal to 1.
417 Medium
417 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 \cap Q contain just one element, is
View answer and explanation
Correct option: C
n3(34)n\frac{n}{3}\left(\frac{3}{4}\right)^n
Option B (n(34)n−1n\left(\frac{3}{4}\right)^{n-1}) is incorrect. The correct answer is D (n4(34)n−1\frac{n}{4}\left(\frac{3}{4}\right)^{n-1}). The provided solution steps in the original response contain errors and lead to the incorrect option C. The correct derivation is shown below.
418 Easy
418 of 1130
Application +4 | -1
If birth to male child and birth to female child are equal-probable, then what is the probability that at least one of the three children born to a couple is male?
View answer and explanation
Correct option: B
78\frac{7}{8}
Option B (78\frac{7}{8}) is correct because it calculates the probability of at least one male child by finding the complement of all female children.
419 Easy
419 of 1130
Application +4 | -1
If AA and BB are two independent events, the probability that both AA and BB occur is 1/81/8 and the probability that neither of them occurs is 3/8.3/8. The probability of the occurrence of AA, is
View answer and explanation
Correct option: A
12,  14\frac{1}{2},\,\,\frac{1}{4}
Option C (0.0875) is correct because it calculates the probability of India getting at least 7 points in the series based on the given probabilities.
420 Medium
420 of 1130
Application +4 | -1
Minimum number of times a fair coin must be tossed so that the probability of getting atleast one head is more than 99% is
View answer and explanation
Correct option: C
7
Option C (7) is correct because 7 is the smallest number of tosses required to have a probability of at least one head greater than 99%.
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