Question (1)
State which of the following are not the probability distributions of a random variable. Give
reasons for your answer.
(i)
Solution
∑ p(x) = 0.4 + 0.4 + 0.2
= 1.
Yes it is probability distribution function.
(ii)
| x |
0 |
1 |
2 |
3 |
4 |
| P(x) |
0.1 |
0.5 |
0.2 |
-0.1 |
0.3 |
Solution
As one of the probability is negative , it is not probability distribution function.
(iii)
| x |
-1 |
0 |
1 |
| p(x) |
0.6 |
0.1 |
0.2 |
Solution
∑p(x) = 0.6 + 0.1+ 0.2 = 0.9 < 1
As the sum is less than 1, so no it is not probability distribution function.
(iv)
| z |
3 |
2 |
1 |
0 |
-1 |
| p(x) |
0.3 |
0.2 |
0.4 |
0.1 |
0.05 |
Solution
∑ p(x) = 0.3 + 0.2 + 0.4 + 0.1 + 0.05 = 1.05 > 1
As sum of probabilities is greater than 1, so it is not probability distribution function.
Question (2)
An urn contains 5 red and 2 black balls. Two balls are randomly drawn. Let X represent the
number of black balls. What are the possible values of X? Is X a random variable?
Solution
in an urn 5 red + 2 black = 7 balls
Two balls drawn at random. n = 7C
2 = 21.
X ; the number of black balls
X = 0, 1, 2.
\[P\left( {x = 0} \right) = \frac{{5{C_2}}}{{7{C_2}}} = \frac{{5 \times 4}}{{7 \times 6}} = \frac{{10}}{{21}}\]
\[P\left( {x = 1} \right) = \frac{{5{C_1} \times 2{C_1}}}{{21}} = \frac{{5 \times 2}}{{21}} = \frac{{10}}{{21}}\]
\[P\left( {x = 2} \right) = \frac{{2{C_2}}}{{21}} = \frac{1}{{21}}\]
\[\sum {p(x) = \frac{{10}}{{21}} + \frac{{10}}{{21}} + \frac{1}{{21}} = 1} \]
So yes it is probability distribution function.
Question (3)
Let X represent the difference between the number of heads and the number of tails obtained
when a coin is tossed 6 times. What are possible values of X?
Solution
X ; difference between number of heads and number of tails when coin is tossed 6 times.
| No. of heads |
No. of tails |
Difference. |
| 0 |
6 |
6 |
| 1 |
5 |
4 |
| 2 |
4 |
2 |
| 3 |
3 |
0 |
| 4 |
2 |
2 |
| 5 |
1 |
4 |
| 6 |
0 |
6 |
So possible values of X are 0, 2, 4, and 6.
Question (4)
Find the probability distribution of
(i) number of heads in two tosses of a coin
(ii) number of tails in the simultaneous tosses of three coins
(iii) number of heads in four tosses of a coin
Solution
(i) The coin is tossed two times.
So sample space S = { HH, HT, TH, TT}. n = 4
X ; number of heads in two tosses of a coin.
X = 0, 1, 2.
P(X = 0) = 1/4., P(X = 1) = 2/4, P(X = 2) = 1/4.
The probability distribution function is as follows.
| X |
0 |
1 |
2 |
Total |
| p(X) |
1/4 |
2/4 |
1/4 |
1 |
(ii) The coin is tossed three times.
So sample space S = { HHH, HHT,HTH, HTT, THH,THT, TTH TTT}. n = 8
X ; number of heads in three tosses of a coin.
X = 0, 1, 2, 3.
P(X = 0) = 1/8, P(X = 1) = 3/8, P(X = 2) = 3/8, P(X = 3) = 1/8.
The probability distribution function is as follows.
| X |
0 |
1 |
2 |
3 |
Total |
| p(X) |
1/8 |
3/8 |
3/8 |
1/8 |
1 |
(iii) The coin is tossed four times.
So sample space S = { HHHH, HHHT,HHTH, HHTT, HTHH,HTHT, HTTH HTTT, THHH, THHT, THTH, THTT, TTHH, TTHT, TTTH, TTTT}. n = 16
X ; number of heads in four tosses of a coin.
X = 0, 1, 2, 3, 4.
P(X = 0) = 1/16, P(X = 1) = 4/16, P(X = 2) = 6/16, P(X = 3) = 4/16, P(X = 4) = 1/16.
The probability distribution function is as follows.
| X |
0 |
1 |
2 |
3 |
4 |
Total |
| p(X) |
1/16 |
4/16 |
6/16 |
4/16 |
1/16 |
1 |
Question (5)
Find the probability distribution of the number of successes in two tosses of a die, where a
success is defined as
(i) number greater than 4
(ii) six appears on at least one die
Solution
The die is thrown two times. so sample space will contain 36 elements. n = 36.
(i) The success is defined as number greater than 4.
X ; the number of sucess
X = 0, 1, 2.
p = probability of seccess = 2/6
\[P\left( {x = 0} \right) = \frac{4}{6} \times \frac{4}{6} = \frac{{16}}{{36}}\]
\[P\left( {x = 1} \right) = \frac{4}{6} \times \frac{2}{6} + \frac{2}{6} \times \frac{4}{6} = \frac{{8 + 8}}{{36}} = \frac{{16}}{{36}}\]
\[P\left( {x = 2} \right) = \frac{2}{6} \times \frac{2}{6} = \frac{4}{{36}}\]
The probability distribution function is given as follows.
| X |
0 |
1 |
2 |
Total |
| P(X) |
16/36 |
16/36 |
4/36 |
1 |
(ii) Six appears on at least one die.
X = 0, 1.
P(X = 0) = 25/36, P(X = 1) = 11/36.
| X |
0 |
1 |
Total |
| P(X) |
25/36 |
11/36 |
1 |
Question (6)
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
6 bulbs out of 30 bulbs are selected with replacement.
The probability of selected bulb is defective = 6/30 = 1/5.
The probability that selected bulb is not defective = 24/ 30 = 4/5.
X = Number of defective bulbs
X = 0, 1, 2, 3, 4.
\[P\left( {x = 0} \right) = \frac{4}{5} \times \frac{4}{5} \times \frac{4}{5} \times \frac{4}{5} = \frac{{256}}{{625}}\]
\[P\left( {x = 1} \right) = 4C1 \times \frac{4}{5} \times \frac{4}{5} \times \frac{4}{5} \times \frac{1}{5} = 4 \times \frac{{64}}{{125}} \times \frac{1}{5} = \frac{{256}}{{625}}\]
\[P\left( {x = 2} \right) = 4{C_2} \times \frac{4}{5} \times \frac{4}{5} \times \frac{1}{5} \times \frac{1}{5} = 6 \times \frac{{16}}{{25}} \times \frac{1}{{25}} = \frac{{96}}{{625}}\]
\[P\left( {x = 3} \right) = 4{C_3} \times \frac{4}{5} \times \frac{1}{5} \times \frac{1}{5} \times \frac{1}{5} = 4 \times \frac{4}{5} \times \frac{1}{{125}} = \frac{{16}}{{625}}\]
\[P\left( {x = 4} \right) = \frac{1}{5} \times \frac{1}{5} \times \frac{1}{5} \times \frac{1}{5} = \frac{1}{{625}}\]
The probability distribution function is given as follows.
| X |
0 |
1 |
2 |
3 |
4 |
Total |
| p(X) |
256/625 |
256/625 |
96/625 |
16/625 |
1/625 |
1 |
Question (7)
A coin is biased so that the head is 3 times as likely to occur as tail. If the coin is tossed twice,
find the probability distribution of number of tails.
Solution
P(H) = 3P(T)
P(H) + P(T) = 1
3P(T) + P(T) = 1
4P(T) = 1
P(T) = 1/4, P(H) = 3/4.
A coin is tossed twice , so sample space S = {HH, HT, TH, TT}
X ; the number of tails.
X = 0, 1, 2.
P(X = 0 ) = P( HH)
\[P\left( {x = 0} \right) = \frac{3}{4} \times \frac{3}{4} = \frac{9}{{16}}\]
P(X = 1) = P( HT, TH)
\[P\left( {x = 1} \right) = \frac{3}{4} \times \frac{1}{4} + \frac{1}{4} \times \frac{3}{4} = \frac{{3 + 3}}{{16}} = \frac{6}{{16}}\]
P(X = 2) = P(TT)
\[P\left( {x = 2} \right) = \frac{1}{4} \times \frac{1}{4} = \frac{1}{{16}}\]The probability distribution function is as follows
| X |
0 |
1 |
2 |
Total |
| P(X) |
9/16 |
6/16 |
1/16 |
16/16=1 |
Question (8)
. A random variable X has the following probability distribution:
| X |
0 |
1 |
2 |
3 |
4 |
5 |
6 |
7 |
| P(X) |
0 |
k |
2k |
2k |
3k |
k2 |
2k2 |
7k2 + k |
Determine
(i) k
(ii) P(X < 3)
(iii) P(X > 6)
(iv) P(0 < X < 3)
Solution
(i) As it is given the probability distribution function,
∑p(x) = 1
0 + k + 2k + 2k + 3k + k
2 + 2k
2 + 7k
2 + k = 1
10k
2 + 9k = 1
10 k
2 + 9k - 1 = 0
( k + 1)(10k - 1 ) = 0
k + 1 = 0 or 10k - 1 = 0
k = -1 or k = 1/10
As k is the probability k ≠ -1. so k = 1/10.
(ii) P(X < 3) = P(0) + P(1) + P(2)
= 0 + k + 2k
=3k
= 3/10
(iii) P( X > 6) = P(7)
= 7k
2 + k
= 7 (1/10)
2 + 1/10
= 7/100 + 1/10
= 17/100
(iv) P(0 < X < 3) = P(1) + P(2)
= k + 2k
= 3k
= 3/10.
Question (9)
The random variable X has probability distribution P(X) of the following form, where k is
some number:
P(X) = {
𝑘, if x = 0
2𝑘, if x = 1
3𝑘, if x = 2
0, otherwise
(a) Determine the value of k.
(b) Find P(X < 2),P( X≤2) , P(X ≥ 2),
Solution
( a ) The probability distribution function for the random variale is given.
∑p(x) = 1
k + 2k + 3k +0 = 1
6k = 1
k = 1/6
( b ) P( X < 2 ) = P( x = 0) + P( x = 1)
= k + 2k
= 3k
= 3(1/6) = 1/2
P( X ≤ 2) = P(X = 0) + P(X = 1) + P(X = 2)
= k + 2k + 3k
= 6k
=6(1/6) = 1
P(X ≥ 2) = P(X = 2)
= 3k
=3(1/6) = 1/2.
Question (10)
Find the mean number of heads in three tosses of a fair coin.
Solution
As coin is toss three times , n = 8
X ; the number of heads.
X = 0, 1, 2, 3.
P(X = 0) = 1/8, P(X = 1) = P(X = 2) = 3/8, P(X = 3) = 1/8.
| X |
P(X) |
X P(X) |
| 0 |
1/8 |
0 |
| 1 |
3/8 |
3/8 |
| 2 |
3/8 |
6/8; |
| 3 |
1/8 |
3/8 |
| Total |
1 |
12/8 |
Mean of X = E(X)
= ∑ XP(X)
= 12/8
= 1.5
Question (11)
Two dice are thrown simultaneously. If X denotes the number of sixes,
find the expectation of X.
Solution
Two dics are thrown , so n = 36.
X ; the number of sixes.
X = 0, 1, 2.
P(X = 0) = 25/36, P(X = 1) = 10/36, P(X = 2) = 1/36
| X |
P(X) |
X P(X) |
| 0 |
25/36 |
0 |
| 1 |
10/36 |
10/36 |
| 2 |
1/36 |
2/36 |
| Total |
36/36 = 1 |
12/36; |
E(X) = ∑XP(X)
= 12/36
= 1/3.
Question (12)
Two numbers are selected at random (without replacement) from the first six positive
integers. Let X denotes the larger of the two numbers obtained. Find E(X).
Solution
There are 6 positive numbers , Two number are selected with out replacement.
S = {(1,2), (1,3), (1,4), (1,5), (1,6), (2,3), (2,4), (2,5), (2,6), (3,4), (3,5), (3,6), (4,5), (4,6), (5,6)}
n = 15.
X ; the large number between them
X = 2, 3, 4, 5, 6.
| X |
P(X) |
X P(X) |
| 2 |
1/15 |
2/15 |
| 3 |
2/15 |
6/15 |
| 4 |
3/15 |
12/15 |
| 5 |
4/15 |
20/15 |
| 6 |
5/15 |
30/15 |
| Total |
15/15=1 |
70/15 |
E(X) = ∑XP(X)
= 70/15
= 14/3
Question (13)
Let X denote the sum of the numbers obtained when two fair dice are rolled. Find the variance
and standard deviation of X
Solution
When two dice are rolled then n = 36.
X ; Sum og the numbers on dics.
X = 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12
| X |
P(X) |
XP(X) |
X2P(X) |
| 2 |
1/36 |
2/36 |
4/36 |
| 3 |
2/36 |
6/36 |
18/36 |
| 4 |
3/36/td>
| 12/36 |
48/36 |
| 5 |
4/36 |
20/36 |
100/36 |
| 6 |
5/36 |
30/36 |
180/36 |
| 7 |
6/36 |
42/36 |
294/36 |
| 8 |
5/36 |
40/36; |
320/36 |
| 9 |
4/36 |
36/36 |
324/36 |
| 10 |
3/36 |
30/36 |
300/36 |
| 11 |
2/36 |
22/36 |
242/36 |
| 12 |
1/36 |
12/36 |
144/36 |
| Total |
36/36 = 1 |
/252/36 |
1974/36 |
E(X) = ∑XP(X)
= 252 / 36
= 7
E(X
2) = ∑X
2P(X)
= 1974/ 36
=54.833
V(X) = E(X
2) - [E(X)]
2
= 54.833 - (7)
2
= 54.833 - 49
= 5.833
S.D = √V(X)
= √5.833
= 2.415.
Question (14)
A class has 15 students whose ages are 14, 17, 15, 14, 21, 17, 19, 20, 16, 18, 20, 17,
16, 19 and 20 years. One student is selected in such a manner that each has the same chance
of being chosen and the age X of the selected student is recorded. What is the probability
distribution of the random variable X? Find mean, variance and standard deviation of X.
Solution
X ; age of the student
X = 14,15, 16, 17, 18, 19, 20, 21.
| X |
frquency |
P(X) |
XP(X) |
X2P(X) |
| 14 |
2 |
2/15 |
28/15 |
392/15 |
| 15 |
1 |
1/15 |
15/15 |
225/15 |
| 16 |
2 |
2/15 |
32/15 |
512/15 |
| 17 |
3 |
3/15 |
51/15 |
867/15 |
| 18 |
1 |
1/15 |
18/15 |
324/15 |
| 19 |
2 |
2/15 |
38/15 |
722/15 |
| 20 |
3 |
3/15 |
60/15 |
1200/15 |
| 21 |
1 |
1/15 |
21/15 |
441/15 |
| Total |
15 |
1 |
263/15 |
4683/15 |
E(X) = ∑XP(X)
= 263/15
=17.53
E(X
2) = ∑X
2P(X)
= 4683/15
V(X) = E(X
2) - [E(X)]
2
= 4683/ 15 - (263/15)
2
= (70245 - 69169)/225
= 1076 / 225
=4.78
S.D.(X) = √V(X)
= √4.78
=2.186 = 2.19.
Question (15)
. In a meeting, 70% of the members favour and 30% oppose a certain proposal. A member is
selected at random and we take X = 0 if he opposed, and X = 1 if he is in favour. Find E(X)
and Var(X).
Solution
IN a meeting 70% favors and 30% oppose a proposal.
X = 0 for oppose and X = 1 for favour.
| X |
P(X) |
XP(X) |
X2P(X) |
| 0 |
0.3 |
0 |
0 |
| 1 |
0.7 |
0.7 |
0.7 |
| Total |
1 |
0.7 |
0.7 |
E(X) = ∑XP(X)
= 0.7
E(X
2) = ∑X
2P(X)
= 0.7
V(X) = E(X
2) -[ E(X)]
2
= 0.7 - (0.7)
2
= 0.7 - 0.49
= 0.21
Question (16)
The mean of the numbers obtained on throwing a die having written 1 on three faces, 2 on
two faces and 5 on one face is
(A) 1
(B) 2
(C) 5
(D) 8
Solution
X = 1, 2, 5.
| X |
P(X) |
XP(X) |
| 1 |
3/6 |
3/6 |
| 2 |
2/6 |
4/6 |
| 5 |
1/6 |
5/6 |
| Total |
1 |
12/6 |
Mean E(X) = ∑XP(X)
=12/6
= 2
So B is the correct answer.
Question (17)
Suppose that two cards are drawn at random from a deck of cards. Let X be the number of
aces obtained. Then the value of E(X) is
(A) 37/221
(B) 5/13
(C) 1/13
(D) 2/13
Solution
Two cards are drawn from a pack of cards.
the probability of ace card = 4/52 = 1/13
X ; the number of ace card
X = 0, 1, 2.
| X |
P(X) |
XP(X) |
| 0 |
144/169 |
0 |
| 1 |
24/169 |
24/169 |
| 2 |
1/169 |
2/169 |
| Total |
1 |
26/169 |
E(X) = ∑XP(X)
= 26/169
= 2/13
D is the correct answer.