Question

A random variable X has the following probability distribution. X 0 1 2 3 4 5 6 7 P (X) 0 k 2 k 2 k 3 k k 2 2 k 2 7 k 2 + k Determine (i) k (ii) P (X < 3) (iii) P (X > 6) (iv) P (0 < X < 3)

Answer 1

(i).

Since X is a random a variable and sum of probabilities is equal to 1.

∑ X=0 7 P( X ) =1 P( X 0 )+P( X 1 )+P( X 2 )+P( X 3 )+P( X 4 )+P( X 5 )+P( X 6 )+P( X 7 )=1 0+k+2k+2k+3k+ k 2 +2 k 2 +7 k 2 +k=1 9k+10 k 2 =1

Further, simplify the equation,

9k+10 k 2 =1 10 k 2 +9k−1=0 ( 10k−1 )( k+1 )=0 k= 1 10 , −1

But k is a probability so, it cannot be negative.

Thus, the value of k is 1 10 .

(ii).

For P( X<3 ),

P( X 0 )+P( X 1 )+P( X 2 )=0+k+2k =3k =3× 1 10

Thus, the value of k is 3 10 .

(iii).

For P( X>6 ),

P( X 7 )=7 k 2 +k =7× ( 1 10 ) 2 + 1 10 = 17 100

(iv).

Here, for P( 0<X<3 ),

P( X 1 )+P( X 2 )=k+2k =3k =3× 1 10 = 3 10

Thus, the value of k is 3 10 .