1.

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).​

Answer»

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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