Skip to main content
Back to PROBABILITY

Your Progress

0 / 1130 Answered
0
Correct
0
Incorrect
0%
Accuracy
JEE • MATHEMATICS • PROBABILITY

Probability practice · Mixed difficulty

Choose an answer to check it, or open “View answer” to study the explanation. Sign in to save progress to your account.

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.

Take a chapter quiz →
Showing 241-270 of 1130 questions

Preparing practice and saved progress…

241 Easy
241 of 1130
Application +4 | -1
If X follows a binomial distribution with parameters n=8n=8 and p=1/2p=1 / 2, then P(∣X−4∣≤2)P(|X-4| \leq 2) equals
View answer and explanation
Correct option: B
119128\frac{119}{128}
Option B (119128\frac{{119}}{{128}}) is correct because it represents the sum of probabilities for X values between 2 and 6 inclusive, given the binomial distribution parameters.
242 Medium
242 of 1130
Analyzing +4 | -1
The probability of getting a total of at least 6 in the simultaneously throw of three dice is
View answer and explanation
Correct option: D
103108\frac{{103}}{{108}}
The correct answer is D (103/108) because it represents the probability of the complement event (getting at least 6) calculated by subtracting the probability of the event not happening (getting less than 6) from 1.
243 Medium
243 of 1130
Application +4 | -1
A and B throw a die alternatively. If A starts throwing then the probability that he throws 5 before B throws 3 or 6 is
View answer and explanation
Correct option: A
3/83/8
Option A (3/8) is correct because it represents the probability that A throws a 5 before B throws a 3 or 6, calculated using an infinite geometric series.
244 Easy
244 of 1130
Understanding +4 | -1
Three letters are written to different persons and addresses to three envelops are also written. Without looking at the addresses, the probability that the letters go into right envelops is
View answer and explanation
Correct option: B
1/61/6
Option B (16\frac{1}{6}) is correct because there is only 1 way to put all letters in the correct envelopes out of 3! = 6 total permutations.
245 Easy
245 of 1130
Application +4 | -1
If random variable XX has the following probability distribution:
X:012345678P(X=x):a3a5a7a9a11a13a15a17a\begin{array}{cccccccccccccccccccc} X&:&0&1&2&3&4&5&6&7&8 \\ {P(X = x)}&:&a&{3a}&{5a}&{7a}&{9a}&{11a}&{13a}&{15a}&{17a} \end{array}
then the value of aa is
View answer and explanation
Correct option: D
181\frac{1}{81}
Option D (181\frac{1}{81}) is correct because the sum of all probabilities in a probability distribution must equal 1, allowing us to solve for a.
246 Medium
246 of 1130
Application +4 | -1
Probability that Ajay will not go to office is 15\frac{1}{5} and probability that Ajay and Vijay will go to office is 27\frac{2}{7}, if their visits to office is independent of each other then find probability that Ajay will go to the office but Vijay will not go, is
View answer and explanation
Correct option: C
1835\frac{18}{35}
Option C (1835\frac{18}{35}) is correct because it is the probability of Ajay going to the office and Vijay not going, calculated using independent probabilities.
247 Medium
247 of 1130
Application +4 | -1
Three people each flip two fair coins. The probability that exactly two of the people flipped one head and one tail, is
View answer and explanation
Correct option: B
3/83 / 8
Option B (3 / 8) is correct because it is the probability calculated using the binomial probability formula with n=3, r=2, and p=1/2.
248 Easy
248 of 1130
Application +4 | -1
If any four numbers are selected and they are multiplied, then the probability that the last digit will 1, 3, 5 or 7, is
View answer and explanation
Correct option: C
16/625
Option C (16/625) is correct because the probability of the last digit being 1, 3, 5, or 7 is (2/5) for each number, and since four numbers are selected, the probability is (2/5)4.
249 Easy
249 of 1130
Application +4 | -1
Let U1U_1 and U2U_2 be two urns such that U1U_1 contains 3 white and 2 red balls, and U2U_2 contains only 1 white ball. A fair coin is tossed. If head appears then 1 ball is drawn at random from urnU1u r n U_1 and put into U2⋅U_2 \cdot However, if tail appears then 2 balls are drawn at random from U1U_1 and put into U2U_2. Now, 1 ball is dravn at randnm from U2U_2. Then the probability of the drawn ball from U2U_2 being white is
View answer and explanation
Correct option: B
2330\frac{{23}}{{30}}
Option C (19/30) is incorrect. Option B (23/30) is correct because it calculates the probability of drawing a white ball from U2 after transferring balls from U1, considering both head and tail outcomes of the coin toss.
250 Easy
250 of 1130
Application +4 | -1
A fair coin is tossed a fixed number of times. If the probability of getting seven heads is equal to that of getting nine heads, the probability of getting two heads, is
View answer and explanation
Correct option: C
15/21315 / 2^{13}
Option A (15/21315 / 2^{13}) is correct because it calculates the probability of getting two heads given that the probability of getting seven heads is equal to that of getting nine heads.
251 Medium
251 of 1130
Application +4 | -1
Of a total of 600 bolts, 20% are too large and 10% are too small. The remainder are considered to be suitable. If a bolt is selected at random, the probability that it will be suitable is
View answer and explanation
Correct option: B
710\frac{7}{{10}}
Option B (7/10) is correct because it calculates the probability of a bolt being suitable by dividing the number of suitable bolts by the total number of bolts.
252 Easy
252 of 1130
Application +4 | -1
Let XX be a set containing 10 elements and P(X)P(X) be its power set. If AA and BB are picked up at random from P(X)P(X), with replacement, then the probability that AA and BB have equal number elements, is :
View answer and explanation
Correct option: D
20C10220\frac{{ }^{20} \mathrm{C}_{10}}{2^{20}}
Option D (20C10220\frac{{ }^{{20}} \mathrm{{C}}_{{10}}}{{2^{{20}}}}) is correct because it represents the probability of selecting two subsets with the same number of elements from the power set.
253 Easy
253 of 1130
Understanding +4 | -1
Let S{S} be a sample space containing outcomes ω1,ω2,ω3,…,ωn\omega_1, \omega_2, \omega_3, \ldots, \omega_{{n}} i.e., S={ω1,ω2,….,ωn}{S}=\left\{\omega_1, \omega_2, \ldots ., \omega_{{n}}\right\}.

Then, which of the following is true?

I. 0≤P(ωi)≤10 \leq {P}\left(\omega_{{i}}\right) \leq 1 for each ωi∈S\omega_{{i}} \in {S}

II. P(ω1)+P(ω2)+….P(ωn)=1{P}\left(\omega_1\right)+{P}\left(\omega_2\right)+\ldots . {P}\left(\omega_{{n}}\right)=1

III. For any event A,P(A)=∑P(ωi),ωi∈A{A}, {P}({A})=\sum {P}\left(\omega_{{i}}\right), \omega_{{i}} \in{A}
View answer and explanation
Correct option: D
All of these
Option D (All of these) is correct because all three statements are fundamental axioms and properties of probability in a sample space.
254 Medium
254 of 1130
Application +4 | -1
A sample of a 4 times is drawn at a random without replacement from a lot of 10 items containing 3 defectives. If xx denotes the number of defective items in the sample, then P(0<x<3)P(0 < x < 3) is equal to
View answer and explanation
Correct option: B
45\frac{4}{5}
Option B (4/5) is correct because it calculates the probability that the number of defective items in the sample is between 0 and 3 by considering the possible combinations.
255 Easy
255 of 1130
Application +4 | -1
Ten unbiased coins are thrown simultaneously. The probability of getting at least seven heads is
View answer and explanation
Correct option: D
1164\frac{11}{64}
Option D (1164\frac{{11}}{{64}}) is correct because it represents the probability of getting at least seven heads when ten unbiased coins are thrown simultaneously, calculated using the binomial distribution.
256 Medium
256 of 1130
Application +4 | -1
A five digit number if formed by writing the digits 1, 2, 3, 4, 5 in a random order without repetitions. Then, the probability that the number is divisible by 4 is
View answer and explanation
Correct option: C
15\frac{1}{5}
Option C (15\frac{1}{5}) is correct because it is the probability that the number is divisible by 4, calculated by dividing the number of favorable outcomes by the total number of outcomes.
257 Easy
257 of 1130
Analyzing +4 | -1
10 different vectors are lying on a plane out of which four are parallel with respect to each other. Probability that three vectors chosen from them will satisfy the equation λ1a+λ2b+λ3c=0\lambda_1 \mathbf{a}+\lambda_2 \mathbf{b}+\lambda_3 \mathbf{c}=0, where λ1,λ2\lambda_1, \lambda_2 and λ3≠0\lambda_3 \neq 0 is
View answer and explanation
Correct option: D
(6C2+4C1)+(6C2×4C1)10C3\frac{\left({ }^6 C_2+{ }^4 C_1\right)+\left({ }^6 C_2 \times{ }^4 C_1\right)}{{ }^{10} C_3}
Option D ((6C2+4C1)+(6C2×4C1)10C3\frac{\left({ }^6 C_2+{ }^4 C_1\right)+\left({ }^6 C_2 \times{ }^4 C_1\right)}{{ }^{10} C_3}) is correct because it considers all possible combinations of vectors from the parallel and non-parallel sets that satisfy the given equation.
258 Easy
258 of 1130
Application +4 | -1
If mm is a natural such that m≤5m \leq 5, then the probability that the quadratic equation x2+mx+12+m2=0x^2+m x+\frac{1}{2}+\frac{m}{2}=0 has real roots is
View answer and explanation
Correct option: C
35\frac{3}{5}
Option C (35\frac{3}{5}) is correct because it is the probability that the quadratic equation has real roots, given that mm is a natural number such that m≤5m \leq 5.
259 Easy
259 of 1130
Application +4 | -1
In a certain town, 40%40 \% of the people have brown hair, 25%25 \% have brown eyes and 15%15 \% have both brown hair and brown eyes. If a person selected at random from the town, has brown hair, the probability that he also has brown eyes is
View answer and explanation
Correct option: B
38\frac{3}{8}
Option B (3/8) is correct because it is the conditional probability of having brown eyes given that the person has brown hair, calculated using the formula P(B|A) = P(A ∩ B) / P(A).
260 Easy
260 of 1130
Application +4 | -1
A problem in Mathematics is given to three students A,B,CA, B, C and their respective probability of solving the problem are 12,13\frac{1}{2}, \frac{1}{3} and 14\frac{1}{4}. Probability that the problem is solved, is
View answer and explanation
Correct option: A
34\frac{3}{4}
Option A (34\frac{3}{4}) is correct because it calculates the probability that at least one of the three students solves the problem.
261 Easy
261 of 1130
Application +4 | -1
Bag AA contains 7 white balls and 3 red balls. Bag BB contains 3 white balls and 2 red balls. A ball is chosen randomly and found to be red then find the probability that it is taken from bag AA.
View answer and explanation
Correct option: C
37\frac{3}{7}
Option C (37\frac{{3}}{{7}}) is correct because it is calculated using Bayes' theorem to find the conditional probability.
262 Easy
262 of 1130
Application +4 | -1
The probability that an event AA happens in one trial of an experiment is 0.4. Three independent trials of the experiment are formed. The probability that the event A happens at least once is
View answer and explanation
Correct option: B
0.7840.784
Option B (0.784) is correct because it's the probability of event A happening at least once, calculated as 1 minus the probability of it not happening in any trial.
263 Medium
263 of 1130
Application +4 | -1
If two events AA and BB are such that P(Ac)=0.3,P(B)=0.4,P(A∩Bc)=0.5,P\left( {{A^c}} \right) = 0.3,P\left( B \right) = 0.4,P\left( {A \cap {B^c}} \right) = 0.5, then P(B/A∪Bc)=P\left( {B/A \cup {B^c}} \right) =
View answer and explanation
Correct option: B
0.25
Option B (0.25) is correct because it is calculated using the conditional probability formula and the given probabilities.
264 Medium
264 of 1130
Application +4 | -1
A party is thrown for families having boys, girls, ladies and gentleman. When one member is randomly selected, odds against being a boy is 275\frac{27}{5}, odds in favour of being a girl is 725\frac{7}{25} and probability of being a lady is 1132\frac{11}{32}, then the odds in favour of the chosen person being a gentleman, is
View answer and explanation
Correct option: A
923\frac{9}{23}
Option A (923\frac{9}{23}) is correct because it calculates the probability of selecting a gentleman and then converts it to odds in favor.
265 Easy
265 of 1130
Application +4 | -1
A bag contains (n+1)(n+1) coins. It is known that one of these coins shows heads on both sides, whereas the other coins are fair. One coin is selected at random and tossed. If the probability that toss results in heads is 7/12, then the value of nn is
View answer and explanation
Correct option: C
5
Option C (5) is correct because it sets up and solves the equation for the probability of getting heads, given one coin has heads on both sides.
266 Easy
266 of 1130
Understanding +4 | -1
A card is drawn at random from a pack of 52 cards. The probability that the drawn card is a court card i.e. a jack, a queen or a king, is
View answer and explanation
Correct option: B
313\frac{3}{13}
Option B (313\frac{3}{13}) is correct because there are 12 court cards (3 per suit) in a standard 52-card deck, so the probability is 12/52, which simplifies to 3/13.
267 Medium
267 of 1130
Application +4 | -1
If from each of the three boxes containing 3 white and 1 black, 2 white and 2 black, 1 white and 3 black balls, one ball is drawn at random, then the probability that 2 white and 1 black ball will be drawn is
View answer and explanation
Correct option: A
1332\frac{13}{32}
Option A (1332\frac{13}{32}) is correct because it calculates the probability of drawing 2 white and 1 black ball from the three boxes.
268 Medium
268 of 1130
Analyzing +4 | -1
Let two fair six faced dice AA and BB be thrown simultaneously. If E1E_1 is the event that die A shows up four, E2E_2 is the event that BB shows up two and E3E_3 is the event that the sum of the numbers on both dice is odd, then which the following statements is not true?
View answer and explanation
Correct option: D
E1,E2E_1, E_2 and E3E_3 are independent
Option D (E1,E2E_1, E_2 and E3E_3 are independent) is the correct answer because while E1E_1, E2E_2, and E3E_3 are pairwise independent, they are not mutually independent.
269 Easy
269 of 1130
Application +4 | -1
If the probability of A\mathrm{A} to pass in an examination is 0.3 and for B\mathrm{B} is 0.5 . Then the probability that either A or B pass in the examination is
View answer and explanation
Correct option: A
0.650.65
Option A (0.65) is correct because the probability of either A or B passing is calculated using the formula P(A∪B)=P(A)+P(B)−P(A)P(B)P(A \cup B) = P(A) + P(B) - P(A)P(B) with independent events.
270 Medium
270 of 1130
Application +4 | -1
3%3 \% of the electric bulbs manufactured by a company are defective. Using Poisson distribution (approximation), the probability that a sample of 100 bulbs will contain exactly one defective, is

(Given antilog (.1742)=1.5)(.1742)=1.5)
View answer and explanation
Correct option: B
0.15
Option B (0.15) is correct because it approximates the probability of exactly one defective bulb using the Poisson distribution.
Revision notes and key takeaways

Review PROBABILITY in your textbook or class notes, then use this set to check your understanding.

Showing 241-270 of 1130 questions

Your practice

0 / 1130 answered · 1130 remaining · 0% accuracy

Progress is stored on this device. Sign in to save to your account.

✓ Correct · ✗ Incorrect · ○ Unanswered

Practice settings

Type at least 2 characters or jump directly:
🔓

Save your learning progress

Sign in with a free account to save your practice answers and quiz results, then return to them when you revise.