JEE Advanced Level Test: Probability- 2 - EmSAT Achieve MCQ

Let the d.r’s of a required line be a, b and c. Since, the normal to the given planes x + 2y + z = 3 and 6x + 8y + 3z = 13 are perpendicular to the line.
∴ a +2b+c = 0 and 6a + 8b + 3c = 0
⇒ 
⇒ 
or 
Also, line passes through (2, -1, 3).

l + m + n = 0
and  l² + m² - n² = 0
l² + m² (- l - m)² = 0
2lm = 0  (l = 0 or m = 0)
If l = 0 then n = - m
l : m : n = 0 : l : -1 
If m = 0 then n = - l 
l : m : n = l : 0 : -1 

⇒ θ = π/3

Find general pt. of both line and equate them

⇒ (3λ - 1, 5λ - 3, 7λ - 5 ) .......(i)

⇒ (μ +2, 3μ + 4, 5μ + 6) .......(ii)
on solving λ = 1/2,  & μ = -3/2
on putting λ = 1/2, & μ = -3/2 we get 

Given line can be rewritten as

and 

since two line are coplanar, then

λ = -2

Given, lines can be written as

and 
∴ Requires SD =

Where 

∴ Requires SD 

Required probability 

Use Baye’s Theorem formula


A : Red ball is drawn






= 14/39

Independent event but not equally likely

Total No. of ways = ⁶C₃
No. of favorable ways = 2

P(A∪B) = P(A) = P(A) + P(B) - P(A∩B)
P(B) = 2/3
 = P(B) - P(A∩B) = 

Probability that first critic favour is P(ε₁) = 5/7 
Similarly, P(ε₂) = 4/7  P(ε₃) = 3/7 
Majority are in favour if at least two favour 

number of type 3k → 10
3k + 1 → 10
3k + 2 → 10
Either all the no. are of the same type or one No. from each type

For 1 element not to be present in P∩Q possibilities for n element ⇒ Total
No. of fovourable cases = 3ⁿ

Consider last digit of power of 7 
last digit
7^4k → 1
7^4k+1 →​ 7
7^4K+2 →​ 9
7^4k+3 →​ 3
7^m + 7ⁿ is divided by 5 if
(i) m → 4k + 1, n → 4k + 3
(ii) m→ 4k + 2 and n → 4k
(iii) m→ 4k+3 andn→ 4k+1
(iv) m → 4k and n → 4k + 2 

Since events are mutually exclusive and exhaustive






Hence the set of value satisfying all the above inequalities are 

Probability of getting head = 1/2 and probability of throwing 5 or 6 with a dice = 2/6 1/3 . He starts with a coin and alternately tosses the coin and throws the dice and he will if he get a head before he get 5 or 6.

The numbers should be divisible by 6. Thus, the number of favourable ways is ¹⁶C₃ (as there are 16 numbers in first 100 natural numbers, divisible by 6).
Required probability is 

Let A, B and C be the events that the student is successful in test I, II and III respectively, then P(the student is successful)

P[A∩B∩C')∪(A∩B'∩C)∪(A∩B∩C)]
= P(A∩B∩C') + P(A∩B'∩C) + P(A∩B∩C) 
P(A).P(B).P(C') + P(A)P(B')P(C) + P(A)P(B)P(C)   
{∴ A, B, C are independent}



This equation has infinitely many values of p and q. 

P(A/B) = P(A⋂B)/P(B)
P(B/A) = P(B⋂A)/P(A)
2/3 = P(B⋂A)/(1/4)
P(A⋂B) = 2/3*1/4
= 1/6
Therefore, P(A/B) = P(A⋂B)/P(B)
1/2 = 1/6P(B)
P(B) = 1/3

number of type 3k → 10
3k + 1 → 10
3k + 2 → 10
Either all the no. are of the same type or one No. from each type

The equation of plane passing through (0, 7, -7) is a(x – 0) + b(y – 7) + c(z + 7) = 0
Plane contains line and passes through (-1, 3, -2)
∴ a(-1) + b(3 – 7) + c(-2 + 7) = 0
-3a + 2b + c = 0
∴ a : b : c = 1 : 1 : 1
x + y + z = 0 

D. R’s of line are 2 : 1 : 2
D.R’s of Normal to plane is :3 : -2 : 6
Angle between line and plane

 = i(- 2 - 3μ) + j(μ - 3) + k(5μ - 4)
 is parallel to x - 4y + 3z = 1
1(- 2 - 3μ) + 4(μ - 3) + 3(5μ - 4) = 0
μ = 1/4

A (1,1,1), B (2,3,5), C(-1,0,2) direction ratios of AB are < 1,2,4 >
Therefore, direction ratios of normal to plane ABC are < 2, -3,1 >
As a result, equation of the required plane is 2x – 3y +z = k then 

Hence, equation of the required plane is 

Therefore, the line and the plane are parallel. A point on the line is (2, -2, 3). Required distance = distance of (2, -2, 3) from the given plane x + 5y + z - 5 = 0

Equation of plane containing the given lines is 
⇒ (x - 1) (-1) - (y - 2) (-2) + (z - 3) (-1) = 0
⇒ -x  + 1 + 2y - 4 - z + 3
⇒ -x + 2y - z = 0
Given plane is
x - 2y + z = d .....(ii)
Eqs. (i) and (ii) are parallel
Now, distance between planes

⇒ |d| = 6

2001 = 3 x 23 x 29 and (3+23+29)
= 55 ⇒ a = 3, b = 23, c = 29