JEE Main Maths Test- 4 Free Online Test 2026

Del=(1-w^3n)+w^n(0)+w^2n(w^4n-w^n) =1-(w^3)^n+(w^6)^n-(w^3)^n =1-1+1-1 =0

For infinitely many solutions the two equation become identical.

To find the probability that both cards drawn are black:

• The first card has a 26/52 (1/2) chance of being black.
• After removing one black card, the second draw has a 25/51 chance.
• Multiply these probabilities: (1/2) * (25/51) = 25/102.

Thus, the correct answer is A.

The probability that the least number on any selected chit is 5 means all three numbers drawn must be 5, 6, or 7. The process can be summarised as follows:

• There are 3 favourable outcomes: 5, 6, 7.
• There are 7 possible outcomes for each draw.
• Each draw is independent and with replacement.
• The probability for each draw is: (3/7).

Therefore, the combined probability for three draws is: (3/7)³. Thus, the correct answer is option C.

Let A, B, and C be the events of passing tests I, II, and III respectively. The student's success is defined as (A ∩ B) ∪ (A ∩ C). Using the principle of inclusion-exclusion:

P(success) = P(A ∩ B) + P(A ∩ C) - P(A ∩ B ∩ C)

Given independence:

• P(success) = p*q + p*(1/2) - p*q*(1/2)
• = pq + p/2 - (pq/2)
• = (pq/2) + p/2

Set equal to 1/2:

• (pq)/2 + p/2 = 1/2
• p(q + 1)/2 = 1/2
• p(q + 1) = 1

Testing option C (p=1, q=0):

• 1*(0 + 1) = 1, which satisfies the equation.

Thus, the correct answer is C.

To solve the problem, we first calculate the total number of distinct permutations of the letters in 'PROBABILITY', which has 11 letters with two B's and two I's. The total permutations are given by:

Total permutations = ¹¹! / (2! × 2!)

Next, we treat each pair (BB and II) as single blocks, reducing the problem to arranging 9 items (the two blocks and the seven unique letters). The number of favourable permutations is:

Favourable permutations = ⁹!

The probability is then calculated by dividing the favourable permutations by the total permutations:

Probability = ⁹! / (11! / (2! × 2!)) = ⁹! / (11! / 4) = 4 / 110 = 2 / 55.

Thus, the correct answer is: B.

To find the probability of hitting the target at least once in 10 shots, we follow these steps:

• The probability of missing one shot is 0.7.
• Since the shots are independent, the probability of missing all 10 shots is (0.7)¹⁰.
• The probability of hitting at least once is given by:

P(at least one hit) = 1 - (0.7)¹⁰

This calculation yields approximately 0.0282475249.

1. Calculate the probability each is dead:

• P(Krishna dead) = 1 - 7/15 = 8/15
• P(Hari dead) = 1 - 7/10 = 3/10

2. Since these are independent events, multiply their probabilities:

• P(both dead) = (8/15) * (3/10) = 24/150

Value of odd order skew symmetric determinant

P(boy and his sister both are selected)

∴  Required probability
= 1 – 1/15 = 14/15 = 0.93

Probability of selecting a month = 1/12.
13^th day of a randomly selected month is Friday if its first day is Sunday.
Probability of this event = 1/7.
Hence required probability  = 0.01

Since m and n are selected between 1 and 100, hence sample space = 100 × 100
Also, 7¹ = 7,  7² = 49, 7³ = 343, 7⁴ = 2401,
7⁵ = 16807 etc. Hence 1, 3, 7 and 9 will be the last digits in the powers of 7.
n, m →
↓
1,1       1,2       1,3 ...........       1, 100
2,1       2,2       2,3 ...........       2, 100
...........................................
100, 1  100,2   100, 3.........     100, 100
For m = 1 ; n = 3, 7, 11, ...... 97
∴ Favourable cases =  25
For m = 2, n = 4, 8, 12, ......., 100
∴  Favourable cases =  25
Similarly for every m, favourable n are 25.
∴  Total favourable cases  = 100 × 25
Hence required probability =