Test: Probability (Competition Level) - 2


30 Questions MCQ Test Mathematics For JEE | Test: Probability (Competition Level) - 2


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QUESTION: 1

The equation of line of intersection of the planes x + 2y + z = 3 and 6x + 8y + 3z = 13 can be written as

Solution:

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).

QUESTION: 2

The direction cosines of two lines are such that l + m + n = 0, l2 + m2 – n2 = 0, then angle between them is

Solution:

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

QUESTION: 3

The point of intersection of the lines  and 

Solution:

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 

QUESTION: 4

If the straight lines x = 1 + s, y = - 3 - λs, z = 1 + λs and x = t/2, y = 1 + t, z = -t + 2 with parameters s and t respectively are coplanar, then λ equals

Solution:

Given line can be rewritten as

and 

since two line are coplanar, then

λ = -2

QUESTION: 5

Shortest distance between the line  and 

Solution:

Given, lines can be written as

and 
∴ Requires SD =

Where 

∴ Requires SD 

QUESTION: 6

If P(B) = 3/4,  = 1/3  = 1/3, then 

Solution:


QUESTION: 7

If events are independent and P(A) = 1/3, P(B) = 1/3, P(C) = 1/4 then   is equal to

Solution:



 

QUESTION: 8

A fair die is tossed eight times. The probability that a third six is observed on the eight throw is

Solution:

Required probability 

QUESTION: 9

For k = 1, 2, 3, the box Bk contains k red balls and (k + 1) white balls. Let P(B1) = 1/2, P(B2) = 1/3, P(B3) = 1/6. A box is selected at random and a ball is drawn from it. If a red ball is drawn, then probability that it has come from box B2 is

Solution:

Use Baye’s Theorem formula


A : Red ball is drawn






= 14/39

QUESTION: 10

Let A and B be two event such that  stands for complement of event A. Then events A and B are

Solution:





Independent event but not equally likely

QUESTION: 11

Three of the six vertices of a regular hexagon are chosen at random. The probability that triangle with three chosen vertices is an equilateral triangle is

Solution:

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

QUESTION: 12

If A and B are two events such that P(A∪B) = 3/4, P(A∩B) 1/4,  = 2/3, then  

Solution:


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

QUESTION: 13

Three critics review a book. Odds in favour of book are 5 : 2, 4 : 3 and 3 : 4, respectively for the three critics. The probability that majority are in favour of the book is

Solution:

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 

QUESTION: 14

Out of a set of integers given by {1, 2, 3, …. 30}, three numbers are selected at random. Find the probability the sum of the number chosen is divisible by 3.

Solution:

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

QUESTION: 15

A is a set containing n element. A subset P of A is chosen at random. The set A is reconstructed by replacing the element of P. A set ‘Q’ is again chosen at random. Find probability such that P∩Q = φ.

Solution:

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

QUESTION: 16

If the integers m and n are chosen at random from 1 to 100, then probability that a no. of form 7m + 7n is divisible by 5 equals 

Solution:

Consider last digit of power of 7 
last digit
74k → 1
74k+1 →​ 7
74K+2 →​ 9
74k+3 →​ 3
7m + 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 

QUESTION: 17

If  are the probability of three mutually exclusive and exhaustive events, then the set of all value of p is

Solution:

Since events are mutually exclusive and exhaustive






Hence the set of value satisfying all the above inequalities are 

QUESTION: 18

A man alternately tosses a coin and throws a dice beginning with the coin. The probability that he gets a head in the coin before he gets a 5 or 6 in the dice is

Solution:

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.


QUESTION: 19

Three distinct numbers are selected from first 100 natural numbers. The probability that all the three numbers are divisible by 2 and 3 is

Solution:

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 

QUESTION: 20

A student appears for test I, II and III. The student is successful if he passes either in test I and II or test I and III. The probability of the student passing in test I, II, III are p, q and 1/2 respectively. If the probability that the student is successful is 1/2, then  

Solution:

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. 

QUESTION: 21

It is given that events A and B are such that P(A) = 1/4, P(A/B) = 1/2 and P(B/A) = 2/3, then P (B) is 

Solution:

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

QUESTION: 22

Out of a set of integers given by {1, 2, 3, …. 30}, three numbers are selected at random. Find the probability the sum of the number chosen is divisible by 3.

Solution:

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

QUESTION: 23

The equation of plane passing through the point (0, 7, -7) and containing the line 

Solution:

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 

QUESTION: 24

The angle between line  and plane 3x – 2y + 6z = 0 is (μ is scalar)

Solution:

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

QUESTION: 25

Let P(3, 2, 6) be a point in space and Q be a point on line  Then value of m for which the vector  is parallel to the plane x-4y + 3z = 1 is

Solution:

 = 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

QUESTION: 26

The equation of the plane through the point (-1, 2, 0) and parallel to the line  and 

Solution:


QUESTION: 27

Let A (1,1,1) , B (2, 3, 5) and C (-1, 0, 2) be three points, then equation of a plane parallel to the plane ABC which is at a distance 2 from origin is 

Solution:


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 

QUESTION: 28

The distance between the line  and the plane 

Solution:


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

QUESTION: 29

If the distance between the plane x - 2y + z = d and the plane containing the line  and  then |d| is

Solution:

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

QUESTION: 30

A plane π passes through the point (1, 1, 1). If b, c, a are the direction ratios of a normal to the plane, where a, b, c (a < b < c) are the prime factors of 2001, then the equation of the plane π is

Solution:

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

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