So probability of not coming these chosen items are defective
= ( ) ( )
Standard deviation = √v ri nce
= √
X and Y are independent
∴ ( ) ( ) ( ) re true Only (D) is odd one 6. [Ans. A]
Number of favourable cases are given by HHHT
HHTH HTHH THHH
Total number of cases
= 2C1 2C1 2C1 2C1 =16
∴ Probability =
7. [Ans. A]
A uniform distribution and density function
b 2 3
a
x xL
3(b a) 2 b a
3 3 2 2 2
2
b a (b a ) 3(b a) 4 b a
2 2 2 2
2
(b a)(b ab a ) (b a) (b a)
3(b a) 4 b a
2 2 2 2
4b 4ab 4a 3a 3b 6ab 12
2 2
b a 2ab
12
(b a)2
12
Standard deviation = √v ri nce (b a)2
12
(b a)
12
Given: b=1, a=0
Standard deviation = 1 0 1
12 12
8. [Ans. D]
Let probability of getting atleast one head
= P(H) then
P (at least one head) = 1 P(no head)
P(H)=1 P(all tails) But in all cases, 23=8
P (H) = 1 1 7
8 8 Alternately
Probability of getting at least one head ( ) ( )
1 7
1 8 8
Alternately
From Binomial theorem
Probability of getting at least one head p q
( ) ( ) ( )
1 3 7
(3 3 1)
2 8
9. [Ans. C]
Probability of drawing 2 washers, first followed by 3 nuts, and subsequently the 4 bolts
10. [Ans. D]
Required probability = . / . /
11. [Ans. D]
Given 4R and 6B , -
12. [Ans. C]
Below X (X ) is
(X ) has to be less than 0.5 but greater than zero
13. [Ans. D]
A event that he knows the correct answer
B event that student answered correctly the question P(B) = ?
( ) ( )
X=0 X=1
( )
⏟ he knows correct nswer
⏟ e does not know
correct nswer so he guesses
( ) ( ) ( ⁄ ) ( ⁄ ) ( )
( )
⁄ ⁄
14. [Ans. D]
x 1 2 3
P(x) 0.3 0.6 0.1 (x) x (x)
(x)
x (x)
σ V(x) (x ) , x (x) ( x (x)) ( ) ( ) σ √ 15. [Ans. A]
16. [Ans. *] Range 0.25 to 0.27 p
q
Using Binomial distribution
(x ≥ ) ( ) ( ) ( ) ( )
17. [Ans. *] Range 0.64 to 0.66
Let number of men = 100 Number of women = 100
No. of employed men = 80% of men = 80 No. of employed women
= 50% of women = 50
Probability if the selected one person being employed
= probability of one employed women +probability of one employed man
18. [Ans. A]
So from figure Mean value = 1
V ri nce : μ me n x defective pieces σ (x μ)
n(n )
( ) ( ) ( ) ( )
( )
19. [Ans. *](Range 49 to 51)
Given that μ σ z x μ
σ
x
ere x μ , s x gre ter th n - z
ence prob bility (z ) σ √ ∫ e dz
∴ of s ving ccount holder
orm l distribution
20. [Ans. B]
Mean m = np = 5.2 (x ) e me
m e
e (x )
CE
1. [Ans D]
A, B, C are true
(D) is not true. Since in a negatively skewed distribution
mode > median > mean 2. [Ans. D]
Let the mean and standard deviation of the students of batch C be μ and σ respectively and the mean and standard deviation of entire class of first year students be μ and σ respectively
Now given, μ σ and μ σ
In order to normalise batch C to entire class, the normalize score must be equated
since Z = Z = = Now Z = =
Equation these two and solving, we get
=
x = 8.969 ≃ 9.0 3. [Ans. B]
Since population is finite, hypergeometric distribution is applicable
p( defective in c lcul tors)
4. [Ans. C]
σ μ
5. [Ans. B]
Given f(x) = x for x = 0 else where
( x ) ∫ f(x)dx
∫ x dx
= 0 1
The probability expressed in percentage P =
= 2.469% = 2.47%
6. [Ans. A]
Given
P(private car) = 0.45
P(bus 1 public transport) = 0.55
Since a person has a choice between private car and public transport
P(public transport) = 1 – P(private car)
= 1 – 0.45 =0.55 P(bus) = P(bus public transport)
(bus public tr nsport)
(public tr nsport)
= 0.55 × 0.55
= 0.3025 ≃ 0.30 Now P(metro)
= 1 [P(private car) + P(bus)]
= 1 (0.45 + 0.30) = 0.25 25 Calculators
2 Defective
5 Calculators
23 Non-defective
4 Non-defective 1 Defective
∴ P(private car) = 0.45 P(bus) = 0.30
and P(metro) = 0.25 7. [Ans. D]
ere μ cm; σ cm ( x 102)
= P .
x
/
= P ( x ) This area is shown below:
The shades area in above figure is given by F(0) –F ( 0.44)
= ( ) ( ( )( ) )
= 0.5 – 0.3345
= 1.1655 ≃ 16.55%
Closest answer is 16.7%
8. [Ans. C]
P(2 heads) = 9. [Ans. C]
P(one ball is Red & another is blue)
= P(first is Red and second is Blue)
=
10. [Ans. A]
Given μ = 1000, σ = 200 We know that Z
When X= 1200, Z Req. Prob = P (X )
(Z )
( Z ) Less than 50%
11. [Ans. D]
(X ) (X ) (X ) ( ) ( )
12. [Ans. 6]
∫ f(x)dx
f(x) { ( x x ) x otherwise
∴ ∫ ( x x )dx 6 x
x
x 7
[ ( ) ( ) ( )]
[ ] [
]
13. [Ans. 0.4]
( ) ∫ f( )d ∫ d
( )| ( )
14. [Ans. *] Range 0.26 to 0.27 Avg= 5
Let x denote penalty
(x ) (x ) (x ) (x ) (x )
ew (x n) e x p(x ) e
e
e
e e [
]
15. [Ans. B]
S * T+
( ) n( ) n(S)
16. [Ans. *] Range 0.25 to 0.28 (n t) e ( t)
n -0.44
no of vehicles veh km ( ) e . /
= 2.e
= 0.2707 CS
1. [Ans. A]
P: Event of selecting Box P, Q: Event of selecting Box P P(P)=1/3, P(Q)=2/3 P(R/P)=2/5, P(R/Q)=3/4
P(R/P).P(P) P(P/R)=
P(R/P).P(P) P(R/Q)P(Q)
2/5 1/3 4/19
2/5 1/3 3/4 2/3
2. [Ans. C]
If f (x) is the continuous probability density function of a random variable X then,
( x b ) P ( x b) =b
a
f x dx
3. [Ans. A]
The probability that exactly n elements are chosen
=The probability of getting n heads out of 2n tosses
= ( ) . / (Binomial formula)
= ( ) ( )
= 4. [Ans. A]
Consider the following diagram
The robot can move only right or up as defined in problem. Let us denote right move by ‘ ’ nd up move by ‘U’ ow to reach (3, 3), from (0,0) , the robot has to
m ke ex ctly ‘ ’ moves nd ‘U’ moves in any order.
Similarly to reach (10, 10) from (0,0) the robot h s to m ke ‘ ’ moves nd ‘U’
moves in any order. The number of ways this can be done is same as number of permutations of a word consisting of 10
‘ ’ s nd ’U’s
Applying formula of permutation with limited repetitions we get the answer as
5. [Ans. D]
The robot can reach (4,4) from (0,0) in
8C4 ways as argued in previous problem.
Now after reaching (4,4) robot is not allowed to go to (5,4)
Let us count how many paths are there from (0,0) to (10,10) if robot goes from (4,4) to (5,4) and then we can subtract this from total number of ways to get the answer.
Now there are 8C4 ways for robot to reach (4,4) from (0,0) and then robot takes the
‘U’ move from ( ) to ( ) ow from (5,4) to (10,10) the robot has to make 5
‘U’ moves nd ‘ ’ moves in ny order which can be done in 11! ways
= 11C5 ways
Therefore, the number of ways robot can move from (0,0) (10,10) via (4,4) – (5,4) move is
8C4 11C5 = 8 11
4 5
No. of ways robot can move from (0,0) to (10,10) without using (4,4) to (5,4) move is
20 8 11
10 4 5
ways which is choice (D) (3,3)
(0,0)
6. [Ans. D]
umber of permut tions with ‘ ’ in the first position =19!
Number of permutations with ‘ ’ in the second position = 10 18!
(Fill the first space with any of the 10 odd numbers and the 18 spaces after the 2 with 18 of the remaining numbers in 18!
ways)
umber of permut tions with ‘ ’ in rd position =10 9 17!
(Fill the first 2 place with 2 of the 10 odd numbers and then the remaining 17 places with remaining 17 numbers) nd so on until ‘ ’ is in th place. After that it is not possible to satisfy the given condition, since there are only 10 odd numbers v il ble to fill before the ‘ ’ So the desired number of permutations which satisfies the given condition is …
Now the probability of this happening is given by =
( … )
Which are clearly not choices (A), (B) or (C) 7. [Ans. A]
Given μ = 1, σ = 4 σ = 2 and μ = 1, σ is unknown Given, P(X ) = P (Y ≥ 2 )
Converting into standard normal variates, .z / = P (z ≥ )
.z / = P (z ≥ ( ))
(z ) = P (z ≥ ) _____(i) Now since we know that in standard normal distribution
P (z ) = P (z ≥ 1) _____(ii) Comparing (i) and (ii) we can say that
= 1 σ = 3
8. [Ans. C]
Let C denote computes science study and M denotes maths study.
P(C on Monday and C on Wednesday)
= P(C on Monday, M on Tuesday and C on Wednesday)
+ P(C on Monday, C on Tuesday and C on Wednesday)
=1 0.6 0.4+ 1 0.4 0.4
= 0.24 + 0.16 = 0.40 9. [Ans. B]
It is given that
P (odd) = 0.9 P (even) Now since 𝜮P(x) = 1
∴ P (odd) + P (even) = 1 0.9 P (even) + P (even) = 1 P(even) = = 0.5263
Now, it is given that P(any even face) is same i.e. P(2) = P(4) = P(6)
Now since,
P(even) = P(2) or P(4) or P(6)
= P(2) + P(4) + P(6)
∴ P(2) = P(4) = P(6) = P(even)
= (0.5263)
= 0.1754 It is given that
P(even | face > 3) = 0.75 ( )
( ) = 0.75 ( )
( ) = 0.75
P(f ce ) ( ) ( ) = =0.468 10. [Ans. A]
The tree diagram of probabilities is shown below
From above tree
(decl red f ulty) pq ( q)( p)
p
p f ulty
not f ulty
q
q q
q
decl red f ulty
decl red not f ulty decl red f ulty
decl red not f ulty
11. [Ans. A]
If b c … Then, no. of divisors of (x )(y )(z ) … iven
∴ o of ivisors of
( )( ) ( )( )
No. of divisors of which are multiples of
o of divisors of ( )( )
∴ equired rob bility
12. [Ans. C]
V(x) (x ) , (x)- Where V(x) is the variance of x,
Since variance is σ and hence never negative, ≥
13. [Ans. A]
The five cards are * + Sample space ordered pairs P (1st card = 2nd card + 1)
*( )( )( )( )+
14. [Ans. D]
𝛔y = a 𝛔x is the correct expression Since variance of constant is zero.
15. [Ans. A]
Let A be the event of head in one coin. B be the event of head in second coin.
The required probability is
( )| ∪ ) * ) ( ∪ )+
( ∪ ) ( )
( ∪ )
( ) (both coin he ds)
( )
( ∪ ) ( t le st one he d) ( T T )
So required prob bility
16. [Ans. B]
Required Probability
= P (getting 6 in the first time)
+ P (getting 1 in the first time and getting 5 or 6 in the second time)
+ P (getting 2 in the first time and getting 4 or 5 or 6 in the second time)
+ P (getting 3 in the first time and getting 3 or 4 or 5 or 6 in the second time)
( ) ( ) ( ) 17. [Ans. C]
The p.d.f of the random variable is
x +1
P(x) 0.5 0.5
The cumulative distribution function F(x) is the probability upto x as given below
x +1
F(x) 0.5 1.0 So correct option is (C) 18. [Ans. C]
(k) e k
P is no. of cars per minute travelling.
For no cars. (i.e. k = 0) For no cars. P(0) e
So P can be either 0,1,2. (i.e. k = 0,1,2) For k = 1, p(1)=e
For k = 2 , P(2)=
Hence
( ) ( ) ( ) e e e
e 4 5
e ( ⁄ ) ( )
e
e 19. [Ans. *] Range 0.24 to 0.27
The smaller sticks, therefore, will range in length from almost 0 meters upto a maximum of 0.5 meters, with each length equally possible.
Thus, the average length will be about 0.25 meters, or about a quarter of the stick.
20. [Ans. 10]
22 occurs in following ways 6 6 6 4 w ys 6 6 5 5 w ys equired prob bility
x
21. [Ans. *] Range 11.85 to 11.95
For functioning 3 need to be working (function)
p
22. [Ans. *] Range 3.8 to 3.9
Expected length = Average length of all words
23. [Ans. *] Range 0.259 to 0.261
Let A = divisible by 2, B = divisible by 3 and C = divisible by 5, then
n(A) = 50, n(B) = 33, n(C) = 20 n( ) n( ) n( ) n( )
P( ∪ ∪ )
( ) ( ) ( ) ( ) ( ) ( ) ( )
∴ equired prob bility is
( ) ( ∪ ∪ ) 24. [Ans. 0.25]
( ∪ ) P(S) = 1
( ) ( ) ( ) utu lly exclusive ( ) ( ) ( )
et ( ) x; ( ) x P(A) P(B) = x( x)
Maximum value of y = x ( x) dy
dx ( x) x
= 2x = 1 x
(max)
ximum v lue of y ( )
ECE
1. [Ans. D]
P(Odd number) 3 1
6 2 P(even number ) 3 1
6 2
Since events are independent, therefore
P(odd/even) 1 1 1 2 2 4
2. [Ans. A]
I √ ∫ e(
)
dx omp ring with
σ√ ∫ e
( )
dx ut μ
∫
√ σe dx … rom nd
x σ
x σ
Put σ in equ tion
Using relation , E[X+Y]= E[X]+E[Y]
And E[CX]=CE[X]
On var[x]= σ =E[(x x)2] σ = ,X- x2
= E[X2] [ ,X-]
6. [Ans. C]
Probability of failing in paper 1, P (A) = 0.3
Probability of failing in paper 2, P (B) = 0.2 Hence the CDF is, shown in the figure (A).
8. [Ans. A]