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```TutorBreeze.com
13. Probability
(Random Variables and Binomial Distributions)
QUESTION 1: Let X denote a random variable. The probability that X can take the values x, has the
following form, where k is some unknown constant. Find the value of k.
SOLUTION:
x=0
 0.2,
 kx,
x = 1, 2

P(X = x) = 
x = 3, 4
k (6 − x),
 0,
otherwise
P(X = 0) = 0.2 , P( X=1) = k , P ( X= 2) = 2k P (X = 3) =3k , P ( X= 4) =2 k
, So
0.2 + k + 2k + 2k + 3k =1
i.e.8 k+0.2 = 1
8k = 0.8
k= 0.1
QUESTION 2: From a lot of 30 bulbs which include 6 defectives, a sample of 4 bulbs is drawn
at random with replacement. Find the probability distribution of the number of defective bulbs.
SOLUTION:
Let X = number of defective bulbs So, X = 0,1,2,3,4
Then
p = 1/5 , q = 4/5
4
4
 24   4  256
P( X = 0) = C0 p q =   =   =
 30   5  625
4
0 4
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3
256
 1  4 
P ( X = 1) = 4C1 p1 q 3 = 4     =
625
 5  5 
2
2
96
1 4
P ( X = 2) = 4C2 p 2 q 2 = 4     =
5
5
625
   
3
1
16
1 4
P ( X = 3) = C3 p q = 4     =
 5   5  625
4
3 1
4
0
1
1 4
P ( X = 4) = 4C4 p 4 q 0 = 4     =
625
5  5
X
0
1
P(X)
256 256
625 625
2
3
4
96
625
16
625
1
625
QUESTION 3: Suppose X follows the Binomial distribution B(6,1/2)Which is the most likely
outcome.
SOLUTION:
Here , n = 6 , p = ½ , q = ½
6
1
1
1
P ( X = 0) = 6C0 p 0 q 6 = 6C0   = 6 =
64
2 2
1
5
1
6
1 1
P ( X = 1) = 6C1 p1q 5 = 6C1     = 6 6 =
2
64
2 2
2
4
2
4
 1   1  6 × 5  1   1  15
P ( X = 2) = 6C2 p 2 q 4 = 6C2     =
×    =
 2   2  1× 2  2   2  64
3
3
3
3
 1   1  6 × 5 × 4  1   1  20
P ( X = 3) = 6C3 p 3q 3 = 6C3     =
×    =
 2   2  1× 2 × 3  2   2  64
4
2
4
2
 1   1  6 × 5  1   1  15
P ( X = 4) = 6C4 p 4 q 2 = 6C4     =
×    =
 2   2  1× 2  2   2  64
5
1
5
1
6
1 1
1 1
P ( X = 5) = 6C5 p 5 q1 = 6C5     = 6 ×     =
2 2
 2   2  64
6
6
1
1
1
P ( X = 6) = 6C6 p 6 q 0 = 6C6   = 1×   =
2
 2  64
20
Maximum Value =
64
∴X = 3 is the most likely outcome.
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QUESTION 4: A pair of dice is thrown 4 times. If getting a doublet is considered a success, find the
probability distribution of number of successes.
SOLUTION:
S = “throwing a Doublet in a single throw of a pair of dice”.
F= “not throwing a Doublet in a single throw of a pair of dice”.
p=
6 1
=
36 6
q = 1− p = 1−
1 5
=
6 6
3
125
5
P ( X = 0) = 3C0 p 0 q 3 = 1.q 3 =   =
216
6
2
75
 1  5 
P ( X = 1) = 3C1 p1 q 2 = 3 p.q 2 = 3     =
216
 6  6 
2
 1   5  15
P ( X = 2) = 3C2 p 2 q1 = 3 p 2 .q = 3     =
 6   6  216
3
1
1
P ( X = 3) = 3C3 p 3 q 0 = .1. p 3 =   =
216
6
X
0
1
2
3
P(X
)
125
216
75
216
15
216
1
216
QUESTION 5: The mean and variance of a Binomial distribution for a random variable are 10
and 5/3 respectively. Find P ( X≥1)
SOLUTION:
(q + p )n.
np = 10 , …(1)
npq = 5/3 ….(2)
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q=
5
1
=
3 × 10 6
p = 1− q = 1−
n=
1 5
=
6 6
10 10 × 6
=
= 12
p
5
12
1
P ( X ≥ 1) = 1 − P ( X = 0) = 1 − 12C0 p 0 q12 = 1 − q12 = 1 −  
6
QUESTION 6(i): You are taking a 10 question multiple choice test. If each question has four
choices and you guess on each question, what is the probability of getting exactly 7
questions correct?
SOLUTION:
QUESTION 6(ii): In a box containing 100 bulbs, 10 are defective. The what is the probability that
out of a sample of 5 bulbs, none is defective .
SOLUTION:
Here n= 5, p = 10/100 = 1/10, q = 9/10 P(0) = q 5 = (9/10)5
QUESTION 7: Find the mean of the Binomial distribution B(3, ¼)
SOLUTION:
: Let X be the random variable that follows the given binomial distribution.
X can take the values 0,1,2,3
Here n = 3 , p = ¼ and q = 1- p = 1- ¼ = ¾
.P(x) = nCx px qn-x
Substituting in this X = 0,1,2,3, we can get the respective probabilities.
They work out to be P(X=0) = 27/64 , P(X=1) also equal to the same 27/64, P(X=2) =9/64 and P(X=3) =
1/64.
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In the tabular form the Binomial distribution is :
X
P(X)
0
1
2
3
27/64
27/64
9/64
1/64
We know that the mean of a distribution is given by :
µ = Σxi pi,
Let us add one more row to our table and find the product of each x with its corresponding probability.
X
P(X)
XP(X)
0
1
2
3
27/64
27/64
27/64
9/64
18/64
1/64
3/64
0
µ = Σxi pi,
=48/ 64 =3/4,which is the mean
QUESTION 8:
State True or False with justification : The above distribution can not be the probability distribution of a
random variable
SOLUTION:False, because
, the sum of all probabilities is one .
QUESTION 9: Find the mean of the variable Y whose probability distribution is given graphically
SOLUTION:
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QUESTION 10: Two cards are drawn simultaneously (or successively without replacement)
from a well shuffled pack of 52 cards. Find the mean and standard deviation of the number of kings.
SOLUTION:
the probability distribution of X is
MEAN
VARIANCE
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STANDARD DEVIATION
QUESTION 11: .
The chances of a patient having a heart attack is 40%. According to latest research
Drug A reduces the risk of heart attack by 30% and drug B reduces its chances by 25%. At a time a
patient can choose any one of the two options with equal probabilities. It is given that after going
through one of the two options the patient selected at random
suffers a heart attack. Find the probability that the patient had been prescribed Drug A
SOLUTION:
Let E1 be the event that a patient used Drug A∴P(E1) = ½
Let E2 be the event that a patient used Drug B∴P(E2 ) = ½
Let E be the event that a patient had a heart attack
Required probability :P(E1/E)
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P ( E / E1 ) =
40 
30  28
1 −
=
100  100  100
40 
25  30
1 −
=
100  100  100
P ( E / E1 ) P ( E1 )
P ( E1 / E ) =
P ( E / E1 ) P ( E1 ) + P ( E / E2 ) P ( E2 )
P ( E / E1 ) =
28 1
×
14
100
2
=
=
28 1 30 1 29
× +
×
100 2 100 2
QUESTION 12:
A random variable X has the following probability distribution , find
(i) The value of K (ii) P( X ≤ 1)
X
P(X)
0
0.1
1
K
0
0.1
1
K
2
0.2
3
2K
4
0.3
5
K
SOLUTION:
X
P(X)
i)
2
0.2
3
2K
The value of K:0.1 + K + 0.2 +2K + 0.3 + K = 1
⇒ 4K = 0.4 ⇒K = 0.1
(ii) P( X≤ 1) = P(0) + P1) = 0.1 + 0. 1 = 0.2
4
0.3
5
K
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QUESTION 13: 27.A bag contains 25 balls of which 10 are purple and the remaining are
pink . A ball is drawn at random , its colour is noted and it is replaced . 6 balls are drawn in this
way, find the probability that
(i) All balls were purple
(ii) Not more than 2 were Pink
(iii) Atleast one ball was pink
(iv) An equal number of purple and pink balls were drawn.
SOLUTION:
This is a case of Bernoulli’s trials .
Let Success : Getting a purple ball on a draw
Failure : Getting a pink ball on a draw
p = P( success ) =
10 2
3
= ⇒q=
25 5
5
6
6
2
2
(i ) P(6 success ) = 6C6 p 6 q 0 = 1×   ×1 =  
5
5
(ii ) P( not more than 2 failures ) = P( not less than 4 success) = P(4) + P(5) + P(6)
4
2
5
 2 3
 2 3
2
= 6C4 p 4 q 2 + 6C5 p 5 q1 + 6C6 p 6 q 0 = 15 ×     + 6 ×     + 1×  
 5 5
5 5
5
4
2
2
2 
 3
 2  3   2  
=   15 ×   + 6 ×    +   
 5  
5
 5  5   5  
6
4
 2  135 36 4 
=  
+ +
 5   25 25 25 
4
 2  175 
2
=  
= 7× 
 5   25 
5
4
3
3
864
 2 3
(iii )P(3success7failures) = P(3) = 6C3 p 3 q 3 = 15     =
 5   5  3125
(iv) P (atleast 1 failure)=P( at most 5 success) = P(0) + P(1) + P(2) + P(3) + P(4) + P(5) = 1 - P(6)=1- 6C6 p 6 q 0
6
2
2
= 1 − 1×   × 1 = 1 −  
5
5
6
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Four cards are drawn successively with replacement from a well shuffled deck
of 52 cards . What is the probability that
(i)
All the four cards are spades ?
(ii)
QUESTION 14:
SOLUTION:
This is a case of bernoullis trial.
p = P(Success) = P(getting a spade in a single draw ) =
q = P( Failure) = 1 − p = 1 −
13 1
=
52 4
1 3
=
4 4
4
1
1
All the four cards are spades =P(X =4) = C4 p q =   =
256
4
12
3
Only 3 cards are spades=P(X=3)= 4C3 p 3 q1 =
=
256 64
4
(i)
(ii)
4
0
4
81
3
None is a spade=P(X=0) = C0 p q =   =
256
4
4
(iii)
0
4
QUESTION 15: Find the probability distribution of doublets in three throws of a pair of dice
SOLUTION:. S =
“throwing a Doublet in a single throw of a pair of dice”.
F= “not throwing a Doublet in a single throw of a pair of dice”
p=
6 1
=
36 6
q = 1− p = 1−
1 5
=
6 6
3
 5  125
P ( X = 0) = 3C0 p 0 q 3 = 1.q 3 =   =
 6  216
2
75
 1  5 
P ( X = 1) = C1 p q = 3 p.q = 3    =
216
 6  6 
3
1 2
2
2
 1   5  15
P ( X = 2) = C2 p q = 3 p .q = 3     =
 6   6  216
3
2 1
2
3
1
1
P ( X = 3) = 3C3 p 3 q 0 = .1. p 3 =   =
 6  216
X
0
1
2
3
P(X
)
125
216
75
216
15
216
1
216