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probability questions and answers
Q1) The random variable X has probability distribution P(X) of the
Following form, where k is a some number
k if k = 0
2k if k = 1
P(X) = 3k if k = 2
0 otherwise
( a ) Determine the value of k
( b ) P(X < 2), P(X = 2), P( X = 2)
Solution: we know p(x1) + p(x2) +p(x3) +…..+p(xn) = 1
k + 2k + 3k + 0 = 1
6 k = , k = 1 / 6
a) k = 1/6
b) P(X < 2) = k + 2k = 3k = 3 * 1 / 6 = 1/2
P( X = 2) = k + 2k + 3k = 6 k = 6 * 1 / 6 = 1
P( X = 2) = 3k + 0 = 3k = 3 * 1 / 6 = 1 / 2 Ans.
Q3) Find the mean number of heads in 3 tosses of a fair coin
Solution: S = { HHH, HHT, HTH, HTT, TTT,TTH, THT, THH }
3 tosses of a fair coin X = 0, 1, 2, 3
Mean for heads
X = 0, P(0) = 1/8
X = 1, P(1) = 3/8
X = 2, P(2) = 3/8
X = 3, P(3) = 1/8
Probability distribution:
X 0 1 2 3
P(X) 1/8 3/8 3/8 1/8
xi * pi (x) 0 3/8 6/8 3/8
Mean = µ = S xi pi (x)
= 0 + 3/8 + 6/8 + 3/8 = 12/8 = 3/2 Ans.
Q4) 2 dice are thrown simultaneously. If X denotes the number of sixes,
find the expectation of X
Solution: S = 36
Success is a 6 X = 0, 1, 2
Mean for sixes
X = 0, P(0) = 25/36
X = 1, P(1) = 10/36
X = 2, P(2) = 1/36
Probability distribution:
X 0 1 2
P(X) 25/36 10/36 1/36
xi * pi (x) 0 10/36 2/36
Mean = µ = S xi pi (x)
= 0 + 10/36 + 2/36 = 12/36 = 1/3 Ans.
Q5) 2 numbers are selected without replacement from the first six positive
integers. Let X denote the larger of the 2 numbers obtained. Find E(X)
Solution:
12, 21, 31, 41, 51, 61
13, 23, 32, 42, 52, 62
S = 14, 24, 34, 43, 53, 63 = 30
15, 25, 35, 45, 54, 64
16, 26, 36, 46, 56, 65
X denote the larger of the 2 numbers. (F is the no. of favorable chances)
xi fi pi pi xi
2 2 2/30 4/30
3 4 4/30 12/30
4 6 6/30 24/30
5 8 8/30 40/30
6 10 10/30 60/30
??????????Ü 30 ??????????Ü 140/30
Expectation of X = Mean
= S xi pi (x)
= 140/30
= 14/3 Ans.
Q6) Let X denote the sum of the numbers obtained when 2 dice are rolled.
Find the variance and S.D. of X.
Solution:
11, 21, 31, 41, 51, 61
12, 22, 32, 42, 52, 62
S = 13, 23, 33, 43, 53, 63
14, 24, 34, 44, 54, 64 = 36
15, 25, 35, 45, 55, 65
16, 26, 36, 46, 56, 66
X denote the sum of the 2 numbers.
xi 2 3 4 5 6 7 8 9 10 11 12 Sum
S
pi 1/36 2/36 3/36 4/36 5/36 6/36 5/36 4/36 3/36 2/36 1/36
pi xi 2
36 6
36 1236 2036 3036 4236 4036 3636 3036 2236 1236 7
pi xi2 4
36 18
36 48
36 10036 180
36 294
36 320
36 324
36 300
36 242
36 144
36 329
6
Find the variance and S.D. of X.
Expectation of X = Mean
= S xi pi (x) = 7
Variance = s 2 = Sxi2pi(x) – {S xi pi (x)} 2
= 329 - 72
6
= 54.83 - 49 = 5.83 Ans.
Standard deviation = s = v s 2
= v 5.83
= 2.4 Ans.
Q7) 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 and his age X
is recorded. What is the probability distribution of X ? Find mean,
variance and S.D.
Solution: X denotes the age of the student.
Probability distribution:
xi
year 14 15 16 17 18 19 20 21 Sum
S
pi 2/15 1/15 2/15 3/15 1/15 2/15 3/15 1/15
pi xi 28
15 15
15 3215 5115 1815 38
15 60
15 21
15 263
15
pi xi2 392
15 225
15 512
15 86715 324
15 722
15 1200
15 441
15 4683
15
Mean = S xi pi (x) = 263 /15
Find the Mean, variance and S.D. of X.
Variance = s 2 = Sxi2pi(x) – {S xi pi (x)} 2
= 4683 - ( 263 )2
15 15
= 312.2 – (17.53)2 = 4.78 Ans.
Standard deviation = s = v s 2
= v 4.78
= 2.19 Ans.
Q8) In a meeting 70% of the members favours a certain proposal, 30%
being against. A member is selected at random and we let X = 0 if
he opposed and X = 1 if he is in favour. Find E(X) and Var(X).
Solution: A member is selected
If he opposes X = 0, p(X) = 30/100
If he favours x = 1 p(x) = 70/100
Probability distribution:
Mean = S xi pi (x) = 70/100
= 0. 7 Ans.
Variance =s 2 = Sxi2pi(x) – {S xi pi (x)} 2
= 70 - (0.7)2
100
= 0.21 Ans.
xi
year 0 1 S
pi 30
100 70
100
pi xi 0 70
100 70
100
pi xi2 0 70
100 70
100
Q9) A die is thrown 6 times. If “getting an odd no.” is a success,
what is the Probability of i) 5 successes ii) at least 5 successes
iii) at most 5 successes ?
Solution: Success = odd no. = 3/6 = 1/2 p = 1/2 q = 1 – 1/2 = 1/2
tossed 6 times n = 6
i) 5 successes
P(x ) = C (n,x) px q n – x
P(x = 5) = C (6,5) (1/2)5 (1/2) 6 – 5
= 6 * (1/2)5 * (1/2) = 6 (1/2)6 = 3/32 Ans.
ii) at least 5 successes
P(x = 5) + P( x = 6)
P(x = 6) = C (n,x) px q n – x
= C (6,6) (1/2)6 (1/2) 6 – 6
= 1 * (1/2)6 * (1/2)0 = (1/2)6 = 1/64
P (at least 5 successes) = 3/32 + 1/64 = 7/64 Ans.
Solution: iii) at most 5 successes
At most 5 means, up to 5
Or
1 – P(x = 6)
P (at most 5 successes) = 1 – P(6)
= 1 – 1/64
= 63/64 Ans.
Q10) A pair dice is thrown 4 times. If getting a doublet is a success, what is the Probability of 2 successes
Solution: A pair of dice is thrown. Total = 36
No. of doublets = { (1,1)(2,2)(3,3)(4,4)(5,5)(6,6) } = 6
Success = getting a doublet = 6/36 = 1/6
p = 1/6 q = 1 – 1/6 = 5/6
tossed 4 times n = 4
2 successes
P(x) = C (n,x) px q n – x
P(2) = C (4,2) (1/6)2 (5/6) 4 – 2
P(2) = 6 * 1 * 25
36 36
= 25 Ans.
216
Q11) There are 5% defective items in a large bulk of items. What is
the probability that a sample of 10 items will include not more than
1 defective item ?
Solution: 5% defective items 95% good items.
p = 5/100 = 1/20 q = 1 – 1/20 = 19/20
sample of 10 items n =10
not more than 1 defective
P(x = 0) + P(x = 1)
P(x) = C (n,x) px q n – x
P(0) = C (10,0) (1/20)0 (19/20) 10 – 0
= 1 * 1 * (19/20)10 = (19/20)10
P(1) = C (10,1) (1/20)1 (19/20) 10 – 1
= 10 * (1/20) * (19/20)9 = 10 * ( 19 )9
20 20
P(not more than 1 defective item) = ( 19 )10 + 10 * ( 19 )9
20 20 20
= ( 19 )9 * { 19 + 10 } = 29 ( 19 )9 Ans.
20 20 20 20 20
Q12) 5 cards are drawn from a pack of 52 cards with replacement. What is
the probability that i) all the 5 are spades ii) only 3 are spades
iii) none is a spade ?
Solution: drawing a spade is a success. Spades = 13
p = 13/52 = 1/4 q = 1 – 1/4 = 3/4
5 cards are drawn n = 5
i) all the 5 are spades
P(x) = C (n,x) px q n – x
P(5) = C (5,5) (1/4)5 (3/4) 5 – 5
= 1 * (1/4)5 * 1 = (1/4)5 = 1/1024
ii) only 3 are spades
P(3) = C (5,3) (1/4)3 (3/4) 5 – 3
= 10 * (1/64) * (9/16)9 = 90/1024 = 45/512
iii) none is a spade
P(0) = C (5,0) (1/4)0 (3/4) 5 – 0
= 1 * 1 * 243 = 243 Ans.
1024 1024
Q13) The probability that a bulb produced by a factory will fuse after
150 days of use is 0.05. Find the probability that out of 5 bulbs
i) None ii) not more than 1 iii) more than 1 iv) at least 1
Solution: A bulb will fuse after 150 days of use is 0.05
p = 0.05 q = 1 – 0.05 = 0.95
5 bulbs are tested n = 5
i) none
P(x) = C (n,x) px q n – x
P(0) = C (5,0) (0.05)0 (0.95) 5 – 0
= 1 * 1 * (0.95)5 = (0.95)5
ii) not more than 1 (at most one) i.e. P(0) + P(1)
P(1) = C (5,1) (0.05)1 (0.95) 5 – 1
= 5 * (0.05) * (0.95)4 = 0.25 * (0.95)4
P (not more than 1) = P(0) + P(1)
= (0.95)5 + 0.25 (0.95)4
= (0.95)4 (0.05 + 0.95) = 1.2 * (0.95)4 Ans.
iii) more than 1 i.e. P(2) + P(3) + P(4) + P(5)
or 1 – { P(0) + P(1) }
P ( more than 1) = 1 – { P(0) + P(1) }
= 1 – 1.2 * (0.95)4 Ans.
iv) (at least one) i.e. P(1) + P(2) + P(3) + P(4) + P(5)
or 1 – P(0)
P(at least 1 ) = 1 – P(0) = 1 - (0.95)5 Ans.
Q14) A bag consists of 10 balls each marked with the digits 0 to 9.
if 4 balls are drawn with replacement, what is the probability that none
is marked with the digit 0 ?
Solution: S = {0,1,2,3,4,5,6,7,8,9 } = 10
None is marked with the digit 0
p = 1/10 q = 1 – 1/10 = 9/10
4 balls are drawn n = 4
none ( x = 0)
P(x) = C (n,x) px q n – x
P(0) = C (10,0) (1/10)0 (9/10)4 – 0
= 1 * 1 * (9/10)4 = (9/10)4 Ans.
Q15) In an exam. 20 questions are true/false type are asked. A student
Tosses a coin to find the answer to each question. If the coin shows
head, he answers true, if it shows tail, he answers false. Find the
probability that he answers at least 12 questions correctly.
Solution: A coin is tossed. S = 2
Head = true = p = 1/2 q = 1 – 1/2 = 1/2
20 questions n = 20
at least 12 questions are correct
P(12) + P(13) + P(14) + P(15) + P(16) + P(17) + P(18) + P(19) + P(20)
P(x) = C (n,x) px q n – x
P(12) = C (20,12) (1/2)12 (1/2)20 – 12
= C(20,12) (1/2)20
P ( at least 12 questions are correct)
= (1/2)20 {C(20,12) + C(20,13) +………..+ C(20,20) } Ans
Q16)Suppose X has a binomial distribution B(6, 1/ 2). Show that X = 3
is the Most likely to come.
We have to show P(x=3) is the maximum among xi = 0,1,2,3,4,5,6
Solution: X = 0, 1, 2, 3, 4, 5, 6
B(6, 1/ 2) means n = 6 and p = 1/2 , q = 1 – 1/2 = 1/2
P(0) = C (6,0) (1/2)0 (1/2)6 – 0 = 1/ 64
P(1) = C (6,1) (1/2)1 (1/2)6 – 1 = 6 * 1/ 64 = 6/64
P(2) = C (6,2) (1/2)2 (1/2)6 – 2 = 15 * 1/ 64 = 15/64
P(3) = C (6,3) (1/2)3 (1/2)6 – 3 = 20 * 1/ 64 = 20/64
P(4) = C (6,4) (1/2)4 (1/2)6 – 4 = 15 * 1/ 64 = 15/64
P(5) = C (6,5) (1/2)5 (1/2)6 – 5 = 6 * 1/ 64 = 6/64
P(6) = C (6,6) (1/2)6 (1/2)6 – 6 = 1 * 1/ 64 = 1/64
Of all X = 3 is the maximum
So, X = 3 is most likely to come.
Q17)On a multiple choice exam. With 3 possible answers for each of the
5 questions, what is the probability that a candidate would get 4 or
more correct answers just by guessing ?
Solution: 3 possible answers
5 questions n = 5 and (correct answer) = p = 1/3, q = 1 – 1/3 = 2/3
4 or more correct answers i.e P(4) + P(5)
P(4) = C (5,4) (1/3)4 (2/3)5 – 4
= 5 * (1/3)4 * (2/3) = 5* (1/81) * (2/3) = 10 /243
P(5) = C (5,5) (1/3)5 (2/3)5 – 5
= 1 * 1/ 243 * 1 = 1/243
P( 4 or more) = P(4) + P(5)
= 10 + 1 = 11 Ans.
243 243 243
Q18) A person buys a lottery ticket in 50 lotteries, in each of which his
Chance of winning a prize is 1/100. what is the probability that he will
win a prize a) at least once b) exactly once c) at least twice
Solution: 50 lottery tickets n = 50
(win a prize ) = p = 1/100, q = 1 – 1/100 = 99/100
a) At least once i.e 1 – P(0)
P(0) = C (50,0) (1/100)0 (99/100)50 – 0
= 1 * 1 * (99/100)50 = (99/100)50
P(at least once) = 1 – (99/100)50 Ans.
b) exactly once
P(1) = C (50,1) (1/100)1 (99/100)50 – 1
= 50 * 1/ 100 * (99/100)49 = 1 * ( 99 )49 Ans
2 100
c) at least twice i.e 1 – { P(0) + P(1) }
P(at least twice) = 1 – { P(0) + P(1) }
= 1 – { (99/100)50 + ½ * (99/100)49
= 1 – (99/100)49 { (99/100) + 1/2} = 1 – 149 * ( 99 )49 Ans
100 100
Q19) Find the probability of getting 5 exactly twice in 7 throws of a die.
Solution:
7 throws of a die n = 7
(getting 5 ) = p = 1/6 q = 1 – 1/6 = 5/6
getting 5 twice
P(x = 2) = C(7,2) (1/6)2 (5/6) 7 – 2
= 21 * 1/36 * (5/6) 5
= 21 * 1 * ( 5 )5
36 6
= 7 * ( 5 )5 Ans.
12 6
Q20) Find the probability of throwing at most 2 sixes in 6 throws of a die
Solution:
6 throws of a die n = 6
(getting 6 ) = p = 1/6 q = 1 – 1/6 = 5/6
getting 2 sixes i.e P(0) + P(1) + P(2)
P(0) = C(6,0) (1/6)0 (5/6) 6 – 0 = (5/6)6
P(1) = C(6,1) (1/6)1 (5/6) 6 – 1
= 6 * (1/6) * (5/6)5 = (5/6)5
P(2) = C(6,2) (1/6)2(5/6) 6 – 2
= 15 * (1/6)2 * (5/6)4 = (5/12) (5/6)4
P( at most 2 sixes) = (5/6)6 + (5/6)5 + (5/36) (5/6)4
= ( 5 )4 { 25 + 5 + 5 } = 35 ( 5 )4 Ans.
6 36 6 12 18 6
Q21) It is known that 10% of certain articles manufactured are defective.
what is the probability that in a random sample of 12 such articles
9 are defective ?
Solution:
sample of 12 such articles n =12
(defective) = p = 10% = 10/100 = 1/10 q = 1 – 1/10 = 9/10
9 are defective
P(9) = C(12,9) (1/10)9(9/10) 12 – 9
= 220 * (1/10)9 (9/10)3
= 220 * 1 * 93
109 103
= 22 * 93
1011 Ans.
Q22) If A and B are 2 events such that P(A) = 1/ 4 and P(B) = 1/ 2 and
P(A n B) = 1/ 8 find P (not A and not B)
Solution: 1st we check if A and B are independent.
P (A nB) = P (A) . P(B)
1/8 = 1 /4 * 1/ 2 ( True)
A and B are Independent events.
So, not A and not B are also Independent events.
P (not A and not B) = P (not A n not B)
= P (not A) . P (not B)
= P(1 – P(A) ) . P( 1 – P (B) )
= { 1 – ¼} . { 1 – ½ }
= 3 * 1
4 2
= 3/8 Ans.
Q23) Find P(E/F) if 2 coins are tossed once
E: no tail appears F: no head appears
Solution: P(F) = { TT } = ¼
P(E) = { HH } = ¼
P(E U F) = 0
P(E/F) = 0 = 0 Ans.
1/4
Q24) A die is tossed 180 times. Find the expected value (µ ) of times the face with 3 will appear. Find also standard deviation(s) and variance (s 2 )
Solution: n = 180
The appearance of 3 is a success.
p = 1/6
q = 1 – p = 1 – 1/6 = 5/6
Mean µ = np = 180 * 1 = 30
6
Variance s 2 = n p q = 180 * 1 * 5 = 25
6 6
Standard Deviation s = v25 = 5 Ans.
Q25) The mean and variance of the binomial distribution are 4 and 4/3.
Find P(X = 1)
Solution: Mean = np = 4 …………(1)
Variance = npq = 4/3 ……….(2)
Dividing (2) by (1)
npq = 4/3
np 4
q = 1/3
p = 1 – q = 1 – 1/3 = 2/3
np = 4, n = 4/p = 4 * 3/2 = 6
Thus, P(X = 1) = 1 – P(X = 0) = 1 – qn
= 1 – (1/3)6 = 1 – 0.0014
= 0.9984 Ans.
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