Out of 3n consecutive natural numbers, 3 natural numbers are chosen at random without replacement. Show that the probability that the sum of these numbers is divisible by 3 is :
(3n^2 - 3n + 2)/(3n - 1) (3n - 2).
Answers
Answer:
Let the sequence of 3n consecutive integers begin with m. Then
3n consecutive integers are
m,m+1,m+2,....m+(3n−1)
3 integers from 3n can be selected in
3n
C
3
ways
∴ Total no. of outcomes =
3n
C
3
Now 3n integers can be divided into 3 groups
G
1
:n numbers of form 3p
G
2
:n numbers of form 3p+1
G
3
:n numbers of form 3p+2
The sum of 3 integers chosen from 3n integers will be divisible by 3 if either all the three are from same group or one integer from each group. The no. of ways that the three integers are from same group is
n
C
3
+
n
C
3
+
n
C
3
n
C
1
and no. of ways that the integers are from different group is
n
C
1
×
n
C
1
×
n
C
1
∴ favourable cases =(
n
C
3
+
n
C
3
+
n
C
3
)+(
n
C
1
×
n
C
1
×
n
C
1
)
∴ Required probability =
3n
C
3
3.
n
C
3
+(
n
C
1
)
3
=
(3n−1)(3n−2)
3n
2
−3n+2
Step-by-step explanation:
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Step-by-step explanation:
Let the sequence of 3n consecutive integers begin with m. Then
3n consecutive integers are
m,m+1,m+2,....m+(3n−1)
3 integers from 3n can be selected in
3n
C
3
ways
∴ Total no. of outcomes =
3n
C
3
Now 3n integers can be divided into 3 groups
G
1
:n numbers of form 3p
G
2
:n numbers of form 3p+1
G
3
:n numbers of form 3p+2
The sum of 3 integers chosen from 3n integers will be divisible by 3 if either all the three are from same group or one integer from each group. The no. of ways that the three integers are from same group is
n
C
3
+
n
C
3
+
n
C
3
n
C
1
and no. of ways that the integers are from different group is
n
C
1
×
n
C
1
×
n
C
1
∴ favourable cases =(
n
C
3
+
n
C
3
+
n
C
3
)+(
n
C
1
×
n
C
1
×
n
C
1
)
∴ Required probability =
3n
C
3
3.
n
C
3
+(
n
C
1
)
3
=
(3n−1)(3n−2)
3n
2
−3n+2