Let the sample space S consist of the 3! permutations of letters a, b and c along with three triples of each letter (e. g. aaa) and let each element of S have probability 1/9. Define Ai = {i-th place in thee triple is occupied by a}, i = 1, 2, 3. Verify whether A1, A2 and A3 are independent?

a)P(A1) = P( aaa, abc, acb ) = 1/3

    P(A2) = P(aaa, bac, cab ) = 1/3

      P(A3) = P(aaa, cba, bca ) = 1/3

b)  To verify IF A1, A2 and A3 are independent  we can see that

P( A1 and A2 and A3 ) = P( aaa ) = 1/9  is not equal to P(A1)P(A2)P(A3)

Step-by-step explanation:

Based on the given conditions The sample space S can be written as

aaa, bbb, ccc, abc, acb, bac, bca, cab,and cba  = 9 most likely outcomes

Defining Ai ;which is the event that ith place in the triple is occupied by a

The probabilities of i=1,2,3 can be computed as

P(A1) = P( aaa, abc, acb ) = 1/3

P(A2) = P(aaa, bac, cab ) = 1/3

P(A3) = P(aaa, cba, bca ) = 1/3

P(A1 and A3) = P(A2 and A3 ) = P ( A1 and A3 ) =  1/9

THIS CAN BE EXPRESSED AS

P(A1)P(A2) = P(A2)P(A3) = P(A1)P(A3)

It can observed that

A1, A2, A3 are pair wise independent of each other

to verify this we can see that

P( A1 and A2 and A3 ) = P( aaa ) = 1/9  is not equal to P(A1)P(A2)P(A3)

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step-by-step explanation:

the direct variation says that:

then the equation is of the form is given by:

            where k is the constant variation.

from the given table:

consider any value of x and m(x)=y

x = 2 and m(x)=y = 3

substitute these value in [1]; we have;

divide both sides by 2 we get;

⇒

check:

put x = -1 and m(x) = -1.5

substitute in the equation:

-1.5 = -1.5                   true

therefore, the function m(x) represents a direct variation

graph of this function as shown below: