Question

Q1 (a) A random variable X has the following probability function: Х O 1 2 3 4 5 and P(x) 0 k 2k 2k 3k 5k

Answer 1

Answer:

(i) It is known that the sum of probabilities of a probability distribution of random variables is one.

∴0+k+2k+3k+k

2

+2k

2

+(7k

2

+k)=1

⇒10k

2

+9k−1=0

⇒(10k−1)(k+1)=0

⇒k=−1,

10

1

k=−1 is not possible as the probability of an event is never negative.

∴k=

10

1

(ii) P(X<3)=P(X=0)+P(X=1)+P(X=2)

=0+k+2k

=3k

=3×

10

1

=

10

3

(iii) P(X>6)=P(X=7)

=7k

2

+k

=7×

10

1

2

+

10

1

=

100

7

+

10

1

=

100

17

(iv) P(0<X<3)=P(X=1)+P(X=2)

=k+2k

=3k

=3×

10

1

=

10

3