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
1 answer:
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
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