Chapter
stringclasses 18
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1
|
7518-7521
|
Clearly, six different cases are there as listed below:
SFFFFF, FSFFFF, FFSFFF, FFFSFF, FFFFSF, FFFFFS Similarly, two successes and four failures can have
6 4 2
|
1
|
7519-7522
|
Similarly, two successes and four failures can have
6 4 2 combinations
|
1
|
7520-7523
|
4 2 combinations It will be
lengthy job to list all of these ways
|
1
|
7521-7524
|
2 combinations It will be
lengthy job to list all of these ways Therefore, calculation of probabilities of 0, 1, 2,
|
1
|
7522-7525
|
combinations It will be
lengthy job to list all of these ways Therefore, calculation of probabilities of 0, 1, 2, ,
n number of successes may be lengthy and time consuming
|
1
|
7523-7526
|
It will be
lengthy job to list all of these ways Therefore, calculation of probabilities of 0, 1, 2, ,
n number of successes may be lengthy and time consuming To avoid the lengthy
calculations and listing of all the possible cases, for the probabilities of number of
successes in n-Bernoulli trials, a formula is derived
|
1
|
7524-7527
|
Therefore, calculation of probabilities of 0, 1, 2, ,
n number of successes may be lengthy and time consuming To avoid the lengthy
calculations and listing of all the possible cases, for the probabilities of number of
successes in n-Bernoulli trials, a formula is derived For this purpose, let us take the
experiment made up of three Bernoulli trials with probabilities p and q = 1 – p for
success and failure respectively in each trial
|
1
|
7525-7528
|
,
n number of successes may be lengthy and time consuming To avoid the lengthy
calculations and listing of all the possible cases, for the probabilities of number of
successes in n-Bernoulli trials, a formula is derived For this purpose, let us take the
experiment made up of three Bernoulli trials with probabilities p and q = 1 – p for
success and failure respectively in each trial The sample space of the experiment is
the set
S = {SSS, SSF, SFS, FSS, SFF, FSF, FFS, FFF}
The number of successes is a random variable X and can take values 0, 1, 2, or 3
|
1
|
7526-7529
|
To avoid the lengthy
calculations and listing of all the possible cases, for the probabilities of number of
successes in n-Bernoulli trials, a formula is derived For this purpose, let us take the
experiment made up of three Bernoulli trials with probabilities p and q = 1 – p for
success and failure respectively in each trial The sample space of the experiment is
the set
S = {SSS, SSF, SFS, FSS, SFF, FSF, FFS, FFF}
The number of successes is a random variable X and can take values 0, 1, 2, or 3 The probability distribution of the number of successes is as below :
P(X = 0) = P(no success)
= P({FFF}) = P(F) P(F) P(F)
= q
|
1
|
7527-7530
|
For this purpose, let us take the
experiment made up of three Bernoulli trials with probabilities p and q = 1 – p for
success and failure respectively in each trial The sample space of the experiment is
the set
S = {SSS, SSF, SFS, FSS, SFF, FSF, FFS, FFF}
The number of successes is a random variable X and can take values 0, 1, 2, or 3 The probability distribution of the number of successes is as below :
P(X = 0) = P(no success)
= P({FFF}) = P(F) P(F) P(F)
= q q
|
1
|
7528-7531
|
The sample space of the experiment is
the set
S = {SSS, SSF, SFS, FSS, SFF, FSF, FFS, FFF}
The number of successes is a random variable X and can take values 0, 1, 2, or 3 The probability distribution of the number of successes is as below :
P(X = 0) = P(no success)
= P({FFF}) = P(F) P(F) P(F)
= q q q = q3 since the trials are independent
P(X = 1) = P(one successes)
= P({SFF, FSF, FFS})
= P({SFF}) + P({FSF}) + P({FFS})
= P(S) P(F) P(F) + P(F) P(S) P(F) + P(F) P(F) P(S)
= p
|
1
|
7529-7532
|
The probability distribution of the number of successes is as below :
P(X = 0) = P(no success)
= P({FFF}) = P(F) P(F) P(F)
= q q q = q3 since the trials are independent
P(X = 1) = P(one successes)
= P({SFF, FSF, FFS})
= P({SFF}) + P({FSF}) + P({FFS})
= P(S) P(F) P(F) + P(F) P(S) P(F) + P(F) P(F) P(S)
= p q
|
1
|
7530-7533
|
q q = q3 since the trials are independent
P(X = 1) = P(one successes)
= P({SFF, FSF, FFS})
= P({SFF}) + P({FSF}) + P({FFS})
= P(S) P(F) P(F) + P(F) P(S) P(F) + P(F) P(F) P(S)
= p q q + q
|
1
|
7531-7534
|
q = q3 since the trials are independent
P(X = 1) = P(one successes)
= P({SFF, FSF, FFS})
= P({SFF}) + P({FSF}) + P({FFS})
= P(S) P(F) P(F) + P(F) P(S) P(F) + P(F) P(F) P(S)
= p q q + q p
|
1
|
7532-7535
|
q q + q p q + q
|
1
|
7533-7536
|
q + q p q + q q
|
1
|
7534-7537
|
p q + q q p = 3pq2
P(X = 2) = P (two successes)
= P({SSF, SFS, FSS})
= P({SSF}) + P ({SFS}) + P({FSS})
© NCERT
not to be republished
574
MATHEMATICS
= P(S) P(S) P(F) + P(S) P(F) P(S) + P(F) P(S) P(S)
= p
|
1
|
7535-7538
|
q + q q p = 3pq2
P(X = 2) = P (two successes)
= P({SSF, SFS, FSS})
= P({SSF}) + P ({SFS}) + P({FSS})
© NCERT
not to be republished
574
MATHEMATICS
= P(S) P(S) P(F) + P(S) P(F) P(S) + P(F) P(S) P(S)
= p p
|
1
|
7536-7539
|
q p = 3pq2
P(X = 2) = P (two successes)
= P({SSF, SFS, FSS})
= P({SSF}) + P ({SFS}) + P({FSS})
© NCERT
not to be republished
574
MATHEMATICS
= P(S) P(S) P(F) + P(S) P(F) P(S) + P(F) P(S) P(S)
= p p q
|
1
|
7537-7540
|
p = 3pq2
P(X = 2) = P (two successes)
= P({SSF, SFS, FSS})
= P({SSF}) + P ({SFS}) + P({FSS})
© NCERT
not to be republished
574
MATHEMATICS
= P(S) P(S) P(F) + P(S) P(F) P(S) + P(F) P(S) P(S)
= p p q + p
|
1
|
7538-7541
|
p q + p q
|
1
|
7539-7542
|
q + p q p + q
|
1
|
7540-7543
|
+ p q p + q p
|
1
|
7541-7544
|
q p + q p p = 3p2q
and
P(X = 3) = P(three success) = P ({SSS})
= P(S)
|
1
|
7542-7545
|
p + q p p = 3p2q
and
P(X = 3) = P(three success) = P ({SSS})
= P(S) P(S)
|
1
|
7543-7546
|
p p = 3p2q
and
P(X = 3) = P(three success) = P ({SSS})
= P(S) P(S) P(S) = p3
Thus, the probability distribution of X is
X
0
1
2
3
P(X)
q 3
3q2p
3qp2
p 3
Also, the binominal expansion of (q + p)3 is
q
q
p
qp
p
3
3 2
3
2
3
+
+
+
Note that the probabilities of 0, 1, 2 or 3 successes are respectively the 1st, 2nd,
3rd and 4th term in the expansion of (q + p)3
|
1
|
7544-7547
|
p = 3p2q
and
P(X = 3) = P(three success) = P ({SSS})
= P(S) P(S) P(S) = p3
Thus, the probability distribution of X is
X
0
1
2
3
P(X)
q 3
3q2p
3qp2
p 3
Also, the binominal expansion of (q + p)3 is
q
q
p
qp
p
3
3 2
3
2
3
+
+
+
Note that the probabilities of 0, 1, 2 or 3 successes are respectively the 1st, 2nd,
3rd and 4th term in the expansion of (q + p)3 Also, since q + p = 1, it follows that the sum of these probabilities, as expected, is 1
|
1
|
7545-7548
|
P(S) P(S) = p3
Thus, the probability distribution of X is
X
0
1
2
3
P(X)
q 3
3q2p
3qp2
p 3
Also, the binominal expansion of (q + p)3 is
q
q
p
qp
p
3
3 2
3
2
3
+
+
+
Note that the probabilities of 0, 1, 2 or 3 successes are respectively the 1st, 2nd,
3rd and 4th term in the expansion of (q + p)3 Also, since q + p = 1, it follows that the sum of these probabilities, as expected, is 1 Thus, we may conclude that in an experiment of n-Bernoulli trials, the probabilities
of 0, 1, 2,
|
1
|
7546-7549
|
P(S) = p3
Thus, the probability distribution of X is
X
0
1
2
3
P(X)
q 3
3q2p
3qp2
p 3
Also, the binominal expansion of (q + p)3 is
q
q
p
qp
p
3
3 2
3
2
3
+
+
+
Note that the probabilities of 0, 1, 2 or 3 successes are respectively the 1st, 2nd,
3rd and 4th term in the expansion of (q + p)3 Also, since q + p = 1, it follows that the sum of these probabilities, as expected, is 1 Thus, we may conclude that in an experiment of n-Bernoulli trials, the probabilities
of 0, 1, 2, , n successes can be obtained as 1st, 2nd,
|
1
|
7547-7550
|
Also, since q + p = 1, it follows that the sum of these probabilities, as expected, is 1 Thus, we may conclude that in an experiment of n-Bernoulli trials, the probabilities
of 0, 1, 2, , n successes can be obtained as 1st, 2nd, ,(n + 1)th terms in the expansion
of (q + p)n
|
1
|
7548-7551
|
Thus, we may conclude that in an experiment of n-Bernoulli trials, the probabilities
of 0, 1, 2, , n successes can be obtained as 1st, 2nd, ,(n + 1)th terms in the expansion
of (q + p)n To prove this assertion (result), let us find the probability of x-successes in
an experiment of n-Bernoulli trials
|
1
|
7549-7552
|
, n successes can be obtained as 1st, 2nd, ,(n + 1)th terms in the expansion
of (q + p)n To prove this assertion (result), let us find the probability of x-successes in
an experiment of n-Bernoulli trials Clearly, in case of x successes (S), there will be (n – x) failures (F)
|
1
|
7550-7553
|
,(n + 1)th terms in the expansion
of (q + p)n To prove this assertion (result), let us find the probability of x-successes in
an experiment of n-Bernoulli trials Clearly, in case of x successes (S), there will be (n – x) failures (F) Now, x successes (S) and (n – x) failures (F) can be obtained in
|
1
|
7551-7554
|
To prove this assertion (result), let us find the probability of x-successes in
an experiment of n-Bernoulli trials Clearly, in case of x successes (S), there will be (n – x) failures (F) Now, x successes (S) and (n – x) failures (F) can be obtained in (
|
1
|
7552-7555
|
Clearly, in case of x successes (S), there will be (n – x) failures (F) Now, x successes (S) and (n – x) failures (F) can be obtained in ( )
|
1
|
7553-7556
|
Now, x successes (S) and (n – x) failures (F) can be obtained in ( ) n
x n
x
ways
|
1
|
7554-7557
|
( ) n
x n
x
ways In each of these ways, the probability of x successes and (n − x) failures is
= P(x successes)
|
1
|
7555-7558
|
) n
x n
x
ways In each of these ways, the probability of x successes and (n − x) failures is
= P(x successes) P(n–x) failures is
=
times
(
) times
P(S)
|
1
|
7556-7559
|
n
x n
x
ways In each of these ways, the probability of x successes and (n − x) failures is
= P(x successes) P(n–x) failures is
=
times
(
) times
P(S) P(S)
|
1
|
7557-7560
|
In each of these ways, the probability of x successes and (n − x) failures is
= P(x successes) P(n–x) failures is
=
times
(
) times
P(S) P(S) P(S)
P(F)
|
1
|
7558-7561
|
P(n–x) failures is
=
times
(
) times
P(S) P(S) P(S)
P(F) P(F)
|
1
|
7559-7562
|
P(S) P(S)
P(F) P(F) P(F)
x
n x
1442443
1442443 = px qn–x
Thus, the probability of x successes in n-Bernoulli trials is
|
1
|
7560-7563
|
P(S)
P(F) P(F) P(F)
x
n x
1442443
1442443 = px qn–x
Thus, the probability of x successes in n-Bernoulli trials is (
|
1
|
7561-7564
|
P(F) P(F)
x
n x
1442443
1442443 = px qn–x
Thus, the probability of x successes in n-Bernoulli trials is ( )
|
1
|
7562-7565
|
P(F)
x
n x
1442443
1442443 = px qn–x
Thus, the probability of x successes in n-Bernoulli trials is ( ) n
x n
−x
px qn–x
or nCx
px qn–x
Thus
P(x successes) =
nC
x
n x
x p q − , x = 0, 1, 2,
|
1
|
7563-7566
|
( ) n
x n
−x
px qn–x
or nCx
px qn–x
Thus
P(x successes) =
nC
x
n x
x p q − , x = 0, 1, 2, ,n
|
1
|
7564-7567
|
) n
x n
−x
px qn–x
or nCx
px qn–x
Thus
P(x successes) =
nC
x
n x
x p q − , x = 0, 1, 2, ,n (q = 1 – p)
Clearly, P(x successes), i
|
1
|
7565-7568
|
n
x n
−x
px qn–x
or nCx
px qn–x
Thus
P(x successes) =
nC
x
n x
x p q − , x = 0, 1, 2, ,n (q = 1 – p)
Clearly, P(x successes), i e
|
1
|
7566-7569
|
,n (q = 1 – p)
Clearly, P(x successes), i e C
n
x
n x
x p q − is the (x + 1)th term in the binomial
expansion of (q + p)n
|
1
|
7567-7570
|
(q = 1 – p)
Clearly, P(x successes), i e C
n
x
n x
x p q − is the (x + 1)th term in the binomial
expansion of (q + p)n Thus, the probability distribution of number of successes in an experiment consisting
of n Bernoulli trials may be obtained by the binomial expansion of (q + p)n
|
1
|
7568-7571
|
e C
n
x
n x
x p q − is the (x + 1)th term in the binomial
expansion of (q + p)n Thus, the probability distribution of number of successes in an experiment consisting
of n Bernoulli trials may be obtained by the binomial expansion of (q + p)n Hence, this
© NCERT
not to be republished
PROBABILITY 575
distribution of number of successes X can be written as
X
0
1
2
|
1
|
7569-7572
|
C
n
x
n x
x p q − is the (x + 1)th term in the binomial
expansion of (q + p)n Thus, the probability distribution of number of successes in an experiment consisting
of n Bernoulli trials may be obtained by the binomial expansion of (q + p)n Hence, this
© NCERT
not to be republished
PROBABILITY 575
distribution of number of successes X can be written as
X
0
1
2 x
|
1
|
7570-7573
|
Thus, the probability distribution of number of successes in an experiment consisting
of n Bernoulli trials may be obtained by the binomial expansion of (q + p)n Hence, this
© NCERT
not to be republished
PROBABILITY 575
distribution of number of successes X can be written as
X
0
1
2 x n
P(X)
nC0 qn
nC1 qn–1p1
nC2 qn–2p2
nCx qn–xpx
nCn pn
The above probability distribution is known as binomial distribution with parameters
n and p, because for given values of n and p, we can find the complete probability
distribution
|
1
|
7571-7574
|
Hence, this
© NCERT
not to be republished
PROBABILITY 575
distribution of number of successes X can be written as
X
0
1
2 x n
P(X)
nC0 qn
nC1 qn–1p1
nC2 qn–2p2
nCx qn–xpx
nCn pn
The above probability distribution is known as binomial distribution with parameters
n and p, because for given values of n and p, we can find the complete probability
distribution The probability of x successes P(X = x) is also denoted by P(x) and is given by
P(x) = nCx qn–xpx, x = 0, 1,
|
1
|
7572-7575
|
x n
P(X)
nC0 qn
nC1 qn–1p1
nC2 qn–2p2
nCx qn–xpx
nCn pn
The above probability distribution is known as binomial distribution with parameters
n and p, because for given values of n and p, we can find the complete probability
distribution The probability of x successes P(X = x) is also denoted by P(x) and is given by
P(x) = nCx qn–xpx, x = 0, 1, , n
|
1
|
7573-7576
|
n
P(X)
nC0 qn
nC1 qn–1p1
nC2 qn–2p2
nCx qn–xpx
nCn pn
The above probability distribution is known as binomial distribution with parameters
n and p, because for given values of n and p, we can find the complete probability
distribution The probability of x successes P(X = x) is also denoted by P(x) and is given by
P(x) = nCx qn–xpx, x = 0, 1, , n (q = 1 – p)
This P(x) is called the probability function of the binomial distribution
|
1
|
7574-7577
|
The probability of x successes P(X = x) is also denoted by P(x) and is given by
P(x) = nCx qn–xpx, x = 0, 1, , n (q = 1 – p)
This P(x) is called the probability function of the binomial distribution A binomial distribution with n-Bernoulli trials and probability of success in each
trial as p, is denoted by B(n, p)
|
1
|
7575-7578
|
, n (q = 1 – p)
This P(x) is called the probability function of the binomial distribution A binomial distribution with n-Bernoulli trials and probability of success in each
trial as p, is denoted by B(n, p) Let us now take up some examples
|
1
|
7576-7579
|
(q = 1 – p)
This P(x) is called the probability function of the binomial distribution A binomial distribution with n-Bernoulli trials and probability of success in each
trial as p, is denoted by B(n, p) Let us now take up some examples Example 31 If a fair coin is tossed 10 times, find the probability of
(i)
exactly six heads
(ii)
at least six heads
(iii)
at most six heads
Solution The repeated tosses of a coin are Bernoulli trials
|
1
|
7577-7580
|
A binomial distribution with n-Bernoulli trials and probability of success in each
trial as p, is denoted by B(n, p) Let us now take up some examples Example 31 If a fair coin is tossed 10 times, find the probability of
(i)
exactly six heads
(ii)
at least six heads
(iii)
at most six heads
Solution The repeated tosses of a coin are Bernoulli trials Let X denote the number
of heads in an experiment of 10 trials
|
1
|
7578-7581
|
Let us now take up some examples Example 31 If a fair coin is tossed 10 times, find the probability of
(i)
exactly six heads
(ii)
at least six heads
(iii)
at most six heads
Solution The repeated tosses of a coin are Bernoulli trials Let X denote the number
of heads in an experiment of 10 trials Clearly, X has the binomial distribution with n = 10 and p = 1
2
Therefore
P(X = x) = nCxqn–xpx, x = 0, 1, 2,
|
1
|
7579-7582
|
Example 31 If a fair coin is tossed 10 times, find the probability of
(i)
exactly six heads
(ii)
at least six heads
(iii)
at most six heads
Solution The repeated tosses of a coin are Bernoulli trials Let X denote the number
of heads in an experiment of 10 trials Clearly, X has the binomial distribution with n = 10 and p = 1
2
Therefore
P(X = x) = nCxqn–xpx, x = 0, 1, 2, ,n
Here
n = 10,
21
p
, q = 1 – p = 1
2
Therefore
P(X = x) =
10
10
10
10
1
1
1
C
C
2
2
2
x
x
x
x
−
⎛
⎞
⎛
⎞
⎛
⎞
=
⎜
⎟
⎜
⎟
⎜
⎟
⎝
⎠
⎝
⎠
⎝
⎠
Now
(i) P(X = 6) =
10
10
6
10
1
10
|
1
|
7580-7583
|
Let X denote the number
of heads in an experiment of 10 trials Clearly, X has the binomial distribution with n = 10 and p = 1
2
Therefore
P(X = x) = nCxqn–xpx, x = 0, 1, 2, ,n
Here
n = 10,
21
p
, q = 1 – p = 1
2
Therefore
P(X = x) =
10
10
10
10
1
1
1
C
C
2
2
2
x
x
x
x
−
⎛
⎞
⎛
⎞
⎛
⎞
=
⎜
⎟
⎜
⎟
⎜
⎟
⎝
⎠
⎝
⎠
⎝
⎠
Now
(i) P(X = 6) =
10
10
6
10
1
10 1
105
C
2
6
|
1
|
7581-7584
|
Clearly, X has the binomial distribution with n = 10 and p = 1
2
Therefore
P(X = x) = nCxqn–xpx, x = 0, 1, 2, ,n
Here
n = 10,
21
p
, q = 1 – p = 1
2
Therefore
P(X = x) =
10
10
10
10
1
1
1
C
C
2
2
2
x
x
x
x
−
⎛
⎞
⎛
⎞
⎛
⎞
=
⎜
⎟
⎜
⎟
⎜
⎟
⎝
⎠
⎝
⎠
⎝
⎠
Now
(i) P(X = 6) =
10
10
6
10
1
10 1
105
C
2
6 4
|
1
|
7582-7585
|
,n
Here
n = 10,
21
p
, q = 1 – p = 1
2
Therefore
P(X = x) =
10
10
10
10
1
1
1
C
C
2
2
2
x
x
x
x
−
⎛
⎞
⎛
⎞
⎛
⎞
=
⎜
⎟
⎜
⎟
⎜
⎟
⎝
⎠
⎝
⎠
⎝
⎠
Now
(i) P(X = 6) =
10
10
6
10
1
10 1
105
C
2
6 4 512
2
⎛
⎞
=
=
⎜
⎟
×
⎝
⎠
(ii) P(at least six heads) = P(X ≥ 6)
= P (X = 6) + P (X = 7) + P (X = 8) + P(X = 9) + P (X = 10)
© NCERT
not to be republished
576
MATHEMATICS
=
10
10
10
10
10
10
10
10
10
10
6
7
8
9
10
1
1
1
1
1
C
C
C
C
C
2
2
2
2
2
⎛
⎞
⎛
⎞
⎛
⎞
⎛
⎞
⎛
⎞
+
+
+
+
⎜
⎟
⎜
⎟
⎜
⎟
⎜
⎟
⎜
⎟
⎝
⎠
⎝
⎠
⎝
⎠
⎝
⎠
⎝
⎠
=
10
10
|
1
|
7583-7586
|
1
105
C
2
6 4 512
2
⎛
⎞
=
=
⎜
⎟
×
⎝
⎠
(ii) P(at least six heads) = P(X ≥ 6)
= P (X = 6) + P (X = 7) + P (X = 8) + P(X = 9) + P (X = 10)
© NCERT
not to be republished
576
MATHEMATICS
=
10
10
10
10
10
10
10
10
10
10
6
7
8
9
10
1
1
1
1
1
C
C
C
C
C
2
2
2
2
2
⎛
⎞
⎛
⎞
⎛
⎞
⎛
⎞
⎛
⎞
+
+
+
+
⎜
⎟
⎜
⎟
⎜
⎟
⎜
⎟
⎜
⎟
⎝
⎠
⎝
⎠
⎝
⎠
⎝
⎠
⎝
⎠
=
10
10 10
|
1
|
7584-7587
|
4 512
2
⎛
⎞
=
=
⎜
⎟
×
⎝
⎠
(ii) P(at least six heads) = P(X ≥ 6)
= P (X = 6) + P (X = 7) + P (X = 8) + P(X = 9) + P (X = 10)
© NCERT
not to be republished
576
MATHEMATICS
=
10
10
10
10
10
10
10
10
10
10
6
7
8
9
10
1
1
1
1
1
C
C
C
C
C
2
2
2
2
2
⎛
⎞
⎛
⎞
⎛
⎞
⎛
⎞
⎛
⎞
+
+
+
+
⎜
⎟
⎜
⎟
⎜
⎟
⎜
⎟
⎜
⎟
⎝
⎠
⎝
⎠
⎝
⎠
⎝
⎠
⎝
⎠
=
10
10 10 10
|
1
|
7585-7588
|
512
2
⎛
⎞
=
=
⎜
⎟
×
⎝
⎠
(ii) P(at least six heads) = P(X ≥ 6)
= P (X = 6) + P (X = 7) + P (X = 8) + P(X = 9) + P (X = 10)
© NCERT
not to be republished
576
MATHEMATICS
=
10
10
10
10
10
10
10
10
10
10
6
7
8
9
10
1
1
1
1
1
C
C
C
C
C
2
2
2
2
2
⎛
⎞
⎛
⎞
⎛
⎞
⎛
⎞
⎛
⎞
+
+
+
+
⎜
⎟
⎜
⎟
⎜
⎟
⎜
⎟
⎜
⎟
⎝
⎠
⎝
⎠
⎝
⎠
⎝
⎠
⎝
⎠
=
10
10 10 10 10
|
1
|
7586-7589
|
10 10 10 10
|
1
|
7587-7590
|
10 10 10 1
6
|
1
|
7588-7591
|
10 10 1
6 4
|
1
|
7589-7592
|
10 1
6 4 7
|
1
|
7590-7593
|
1
6 4 7 3
|
1
|
7591-7594
|
4 7 3 8
|
1
|
7592-7595
|
7 3 8 2
|
1
|
7593-7596
|
3 8 2 9
|
1
|
7594-7597
|
8 2 9 1
|
1
|
7595-7598
|
2 9 1 10
|
1
|
7596-7599
|
9 1 10 2
193
512
=
(iii) P(at most six heads) = P(X ≤ 6)
= P (X = 0) + P (X = 1) + P (X = 2) + P (X = 3)
+ P (X = 4) + P (X = 5) + P (X = 6)
=
10
10
10
10
10
10
10
1
2
3
1
1
1
1
C
C
C
2
2
2
2
⎛
⎞
⎛
⎞
⎛
⎞
⎛
⎞
+
+
+
⎜
⎟
⎜
⎟
⎜
⎟
⎜
⎟
⎝
⎠
⎝
⎠
⎝
⎠
⎝
⎠
+
10
10
10
10
10
10
4
5
6
1
1
1
C
C
C
2
2
2
⎛
⎞
⎛
⎞
⎛
⎞
+
+
⎜
⎟
⎜
⎟
⎜
⎟
⎝
⎠
⎝
⎠
⎝
⎠
= 848
53
1024
64
=
Example 32 Ten eggs are drawn successively with replacement from a lot containing
10% defective eggs
|
1
|
7597-7600
|
1 10 2
193
512
=
(iii) P(at most six heads) = P(X ≤ 6)
= P (X = 0) + P (X = 1) + P (X = 2) + P (X = 3)
+ P (X = 4) + P (X = 5) + P (X = 6)
=
10
10
10
10
10
10
10
1
2
3
1
1
1
1
C
C
C
2
2
2
2
⎛
⎞
⎛
⎞
⎛
⎞
⎛
⎞
+
+
+
⎜
⎟
⎜
⎟
⎜
⎟
⎜
⎟
⎝
⎠
⎝
⎠
⎝
⎠
⎝
⎠
+
10
10
10
10
10
10
4
5
6
1
1
1
C
C
C
2
2
2
⎛
⎞
⎛
⎞
⎛
⎞
+
+
⎜
⎟
⎜
⎟
⎜
⎟
⎝
⎠
⎝
⎠
⎝
⎠
= 848
53
1024
64
=
Example 32 Ten eggs are drawn successively with replacement from a lot containing
10% defective eggs Find the probability that there is at least one defective egg
|
1
|
7598-7601
|
10 2
193
512
=
(iii) P(at most six heads) = P(X ≤ 6)
= P (X = 0) + P (X = 1) + P (X = 2) + P (X = 3)
+ P (X = 4) + P (X = 5) + P (X = 6)
=
10
10
10
10
10
10
10
1
2
3
1
1
1
1
C
C
C
2
2
2
2
⎛
⎞
⎛
⎞
⎛
⎞
⎛
⎞
+
+
+
⎜
⎟
⎜
⎟
⎜
⎟
⎜
⎟
⎝
⎠
⎝
⎠
⎝
⎠
⎝
⎠
+
10
10
10
10
10
10
4
5
6
1
1
1
C
C
C
2
2
2
⎛
⎞
⎛
⎞
⎛
⎞
+
+
⎜
⎟
⎜
⎟
⎜
⎟
⎝
⎠
⎝
⎠
⎝
⎠
= 848
53
1024
64
=
Example 32 Ten eggs are drawn successively with replacement from a lot containing
10% defective eggs Find the probability that there is at least one defective egg Solution Let X denote the number of defective eggs in the 10 eggs drawn
|
1
|
7599-7602
|
2
193
512
=
(iii) P(at most six heads) = P(X ≤ 6)
= P (X = 0) + P (X = 1) + P (X = 2) + P (X = 3)
+ P (X = 4) + P (X = 5) + P (X = 6)
=
10
10
10
10
10
10
10
1
2
3
1
1
1
1
C
C
C
2
2
2
2
⎛
⎞
⎛
⎞
⎛
⎞
⎛
⎞
+
+
+
⎜
⎟
⎜
⎟
⎜
⎟
⎜
⎟
⎝
⎠
⎝
⎠
⎝
⎠
⎝
⎠
+
10
10
10
10
10
10
4
5
6
1
1
1
C
C
C
2
2
2
⎛
⎞
⎛
⎞
⎛
⎞
+
+
⎜
⎟
⎜
⎟
⎜
⎟
⎝
⎠
⎝
⎠
⎝
⎠
= 848
53
1024
64
=
Example 32 Ten eggs are drawn successively with replacement from a lot containing
10% defective eggs Find the probability that there is at least one defective egg Solution Let X denote the number of defective eggs in the 10 eggs drawn Since the
drawing is done with replacement, the trials are Bernoulli trials
|
1
|
7600-7603
|
Find the probability that there is at least one defective egg Solution Let X denote the number of defective eggs in the 10 eggs drawn Since the
drawing is done with replacement, the trials are Bernoulli trials Clearly, X has the
binomial distribution with n = 10 and
10
1
100
10
p
|
1
|
7601-7604
|
Solution Let X denote the number of defective eggs in the 10 eggs drawn Since the
drawing is done with replacement, the trials are Bernoulli trials Clearly, X has the
binomial distribution with n = 10 and
10
1
100
10
p Therefore
q =
9
1
−p10
=
Now
P(at least one defective egg) = P(X ≥ 1) = 1 – P (X = 0)
=
10
10
0
9
1
C
⎛10
⎞
−
⎜
⎟
⎝
⎠ =
10
910
1
10
−
EXERCISE 13
|
1
|
7602-7605
|
Since the
drawing is done with replacement, the trials are Bernoulli trials Clearly, X has the
binomial distribution with n = 10 and
10
1
100
10
p Therefore
q =
9
1
−p10
=
Now
P(at least one defective egg) = P(X ≥ 1) = 1 – P (X = 0)
=
10
10
0
9
1
C
⎛10
⎞
−
⎜
⎟
⎝
⎠ =
10
910
1
10
−
EXERCISE 13 5
1
|
1
|
7603-7606
|
Clearly, X has the
binomial distribution with n = 10 and
10
1
100
10
p Therefore
q =
9
1
−p10
=
Now
P(at least one defective egg) = P(X ≥ 1) = 1 – P (X = 0)
=
10
10
0
9
1
C
⎛10
⎞
−
⎜
⎟
⎝
⎠ =
10
910
1
10
−
EXERCISE 13 5
1 A die is thrown 6 times
|
1
|
7604-7607
|
Therefore
q =
9
1
−p10
=
Now
P(at least one defective egg) = P(X ≥ 1) = 1 – P (X = 0)
=
10
10
0
9
1
C
⎛10
⎞
−
⎜
⎟
⎝
⎠ =
10
910
1
10
−
EXERCISE 13 5
1 A die is thrown 6 times If ‘getting an odd number’ is a success, what is the
probability of
(i) 5 successes
|
1
|
7605-7608
|
5
1 A die is thrown 6 times If ‘getting an odd number’ is a success, what is the
probability of
(i) 5 successes (ii) at least 5 successes
|
1
|
7606-7609
|
A die is thrown 6 times If ‘getting an odd number’ is a success, what is the
probability of
(i) 5 successes (ii) at least 5 successes (iii) at most 5 successes
|
1
|
7607-7610
|
If ‘getting an odd number’ is a success, what is the
probability of
(i) 5 successes (ii) at least 5 successes (iii) at most 5 successes © NCERT
not to be republished
PROBABILITY 577
2
|
1
|
7608-7611
|
(ii) at least 5 successes (iii) at most 5 successes © NCERT
not to be republished
PROBABILITY 577
2 A pair of dice is thrown 4 times
|
1
|
7609-7612
|
(iii) at most 5 successes © NCERT
not to be republished
PROBABILITY 577
2 A pair of dice is thrown 4 times If getting a doublet is considered a success, find
the probability of two successes
|
1
|
7610-7613
|
© NCERT
not to be republished
PROBABILITY 577
2 A pair of dice is thrown 4 times If getting a doublet is considered a success, find
the probability of two successes 3
|
1
|
7611-7614
|
A pair of dice is thrown 4 times If getting a doublet is considered a success, find
the probability of two successes 3 There are 5% defective items in a large bulk of items
|
1
|
7612-7615
|
If getting a doublet is considered a success, find
the probability of two successes 3 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 one defective item
|
1
|
7613-7616
|
3 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 one defective item 4
|
1
|
7614-7617
|
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 one defective item 4 Five cards are drawn successively with replacement from a well-shuffled deck
of 52 cards
|
1
|
7615-7618
|
What is the probability
that a sample of 10 items will include not more than one defective item 4 Five cards are drawn successively with replacement from a well-shuffled deck
of 52 cards What is the probability that
(i) all the five cards are spades
|
1
|
7616-7619
|
4 Five cards are drawn successively with replacement from a well-shuffled deck
of 52 cards What is the probability that
(i) all the five cards are spades (ii) only 3 cards are spades
|
1
|
7617-7620
|
Five cards are drawn successively with replacement from a well-shuffled deck
of 52 cards What is the probability that
(i) all the five cards are spades (ii) only 3 cards are spades (iii) none is a spade
|
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