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Byju's Answer
Standard VIII
Mathematics
Random Experiment
In a simultan...
Question
In a simultaneous throw of a pair of dice, find the probability of getting:
(i) 8 as the sum
(ii) a doublet
(iii) a doublet of prime numbers
(iv) a doublet of odd numbers
(v) a sum greater than 9
(vi) an even number on first
(vii) an even number on one and a multiple of 3 on the other
(viii) neither 9 nor 11 as the sum of the numbers on the faces
(ix) a sum less than 6
(x) a sum less than 7
(xi) a sum more than 7
(xii) at least once
(xiii) a number other than 5 on any dice.
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Solution
When
a
pair
of
dice
is
thrown
simultaneously
,
the
sample
space
will
be
as
follows
:
S
=
1
,
1
,
1
,
2
,
1
,
3
,
1
,
4
,
⋯
6
,
5
,
6
,
6
Hence
,
the
total
number
of
outcomes
is
36
.
(
i
)
Let
A
be
the
event
of
getting
pairs
whose
sum
is
8
.
Now
,
the
pairs
whose
sum
is
8
are
2
,
6
,
3
,
5
,
4
,
4
,
5
,
3
and
6
,
2
.
Therefore
,
the
total
number
of
favourable
outcomes
is
5
.
∴
P
A
=
Number
of
favourable
outcomes
Total
number
of
outcomes
=
5
36
(
ii
)
Let
A
be
the
event
of
getting
doublets
in
the
sample
space
.
The
doublets
in
the
sample
space
are
1
,
1
,
2
,
2
,
3
,
3
,
4
,
4
,
5
,
5
and
6
,
6
.
Hence
,
the
number
of
favourable
outcomes
is
6
.
∴
P
A
=
Number
of
favourable
outcomes
Total
number
of
outcomes
=
6
36
=
1
6
(
iii
)
Let
A
be
the
event
of
getting
doublets
of
prime
numbers
in
the
sample
space
.
The
doublets
of
prime
numbers
in
the
sample
space
are
2
,
2
,
3
,
3
and
5
,
5
.
Hence
,
the
number
of
favourable
outcomes
is
3
.
∴
P
A
=
Number
of
favourable
outcomes
Total
number
of
outcomes
=
3
36
=
1
12
(
iv
)
Let
A
be
the
event
of
getting
doublets
of
odd
numbers
in
the
sample
space
.
The
doublets
of
odd
numbers
in
the
sample
space
are
1
,
1
,
3
,
3
and
5
,
5
.
Hence
,
the
number
of
favourable
outcomes
is
3
.
∴
P
A
=
Number
of
favourable
outcomes
Total
number
of
outcomes
=
3
36
=
1
12
(
v
)
Let
A
be
the
event
of
getting
pairs
whose
sum
is
greater
than
9
.
The
pairs
whose
sum
is
greater
than
9
are
4
,
6
,
5
,
5
,
5
,
6
,
6
,
4
,
6
,
5
and
6
,
6
.
Hence
,
the
number
of
favorable
outcomes
is
6
.
∴
P
A
=
Number
of
favourable
outcomes
Total
number
of
outcomes
=
6
36
=
1
6
(
vi
)
Let
A
be
the
event
of
getting
pairs
who
has
even
numbers
on
first
in
the
sample
space
.
The
pairs
who
has
even
numbers
on
first
are
:
2
,
1
,
2
,
2
,
…
2
,
6
,
4
,
1
,
⋯
,
4
,
6
,
6
,
1
,
⋯
6
,
6
.
Hence
,
the
number
of
favourable
outcomes
is
18
.
∴
P
A
=
Number
of
favourable
outcomes
Total
number
of
outcomes
=
18
36
=
1
3
(
vii
)
Let
A
be
the
event
of
getting
pairs
with
an
even
number
on
one
die
and
a
multiple
of
3
on
the
other
.
The
pairs
with
an
even
number
on
one
die
and
a
multiple
of
3
on
the
other
are
2
,
3
,
2
,
6
,
4
,
3
,
4
,
6
,
6
,
3
and
6
,
6
.
Hence
,
the
number
of
favourable
outcomes
is
6
.
∴
P
A
=
Number
of
favourable
outcomes
Total
number
of
outcomes
=
6
36
=
1
6
(
viii
)
Let
A
be
the
event
of
getting
pairs
whose
sum
is
9
or
11
.
The
pairs
whose
sum
is
9
are
3
,
6
,
4
,
5
,
5
,
4
and
6
,
3
.
And
,
the
pairs
whose
sum
is
11
are
5
,
6
and
6
,
5
.
Hence
,
the
number
of
favourable
outcomes
is
6
.
∴
P
A
=
Number
of
favourable
outcomes
Total
number
of
outcomes
=
6
36
=
1
6
∴
P
sum
of
the
pairs
with
neither
9
nor
11
=
1
-
P
sum
of
the
pairs
having
9
or
11
=
1
-
1
6
=
5
6
(
ix
)
Let
A
be
the
event
of
getting
pairs
whose
sum
is
less
than
6
.
The
pairs
whose
sum
is
less
than
6
are
1
,
1
,
1
,
2
,
1
,
3
,
1
,
4
,
2
,
1
,
2
,
2
,
2
,
3
,
3
,
1
,
3
,
2
and
4
,
1
.
Hence
,
the
number
of
favourable
outcomes
is
10
.
∴
P
A
=
Number
of
favourable
outcomes
Total
number
of
outcomes
=
10
36
=
5
18
(
x
)
Let
A
be
the
event
of
getting
pairs
whose
sum
is
less
than
7
.
The
pairs
whose
sum
is
less
than
7
are
1
,
1
,
1
,
2
,
1
,
3
,
1
,
4
,
1
,
5
,
2
,
1
,
2
,
2
,
2
,
3
,
2
,
4
,
3
,
1
,
3
,
2
,
3
,
3
,
4
,
1
,
4
,
2
and
5
,
1
.
Hence
,
the
number
of
favourable
outcomes
is
15
.
∴
P
A
=
Number
of
favourable
outcomes
Total
number
of
outcomes
=
15
36
=
5
12
(
xi
)
Let
A
be
the
event
of
getting
pairs
whose
sum
is
more
than
7
.
The
pairs
whose
sum
is
more
than
7
are
2
,
6
,
3
,
5
,
3
,
6
,
4
,
4
,
4
,
5
,
4
,
6
,
5
,
3
,
5
,
4
,
5
,
5
,
5
,
6
,
6
,
2
,
6
,
3
,
6
,
4
,
6
,
5
and
6
,
6
.
Hence
,
the
number
of
favourable
outcomes
is
15
.
∴
P
A
=
Number
of
favourable
outcomes
Total
number
of
outcomes
=
15
36
=
5
12
(
xii
)
Incomplete
question
(
xiii
)
Let
A
be
the
event
of
getting
pairs
that
has
the
number
5
.
The
pairs
that
has
the
number
5
are
1
,
5
,
2
,
5
,
3
,
5
,
4
,
5
,
5
,
1
,
5
,
2
,
5
,
3
,
5
,
4
,
5
,
5
,
5
,
6
,
6
,
1
,
6
,
2
,
6
,
3
,
6
,
4
and
6
,
6
.
Hence
,
the
number
of
favourable
outcomes
is
11
.
∴
P
A
=
Number
of
favourable
outcomes
Total
number
of
outcomes
=
11
36
∴
P
A
=
1
-
P
A
=
1
-
11
36
=
25
36
Suggest Corrections
0
Similar questions
Q.
In a simultaneous throw of pair of dice, find the probability of getting:
(i) 8 as the sum
(ii) a doublet
(iii) a doublet of prime numbers
(iv) a doublet of odd numbers
(v) a sum greater than 9
(vi) an even number on first
(vii) an even number on one and a multiple of 3 on the other
(viii) neither 9 nor 11 as the sum of the numbers on the faces
(ix) a sum less than 6
(x) a sum less than 7
(xi) a sum more than 7
(xii) at least once
(xiii) a number other than 5 on any dice.
(xiv) even number on each die [CBSE 2014, 2015]
(xv) 5 as the sum [CBSE 2014, 2015]
(xvi) 2 will come up at least once [CBSE 2015]
(xvii) 2 will not come either time [CBSE 2015]
