Question
Question: A random variable X has the following probability distribution: Determine (i) \[k\] (ii) \(P(X...
A random variable X has the following probability distribution:
Determine
(i) k
(ii) P(X<3)
(iii) P(X>6)
(iv) P(0<X<3)
Solution
Hint: In this question we will use the concept of probability distribution of a random variable. The probability distribution of random variable X is defined only when we have the various values of the random variable e.g. x1,x2,x3,.......,xntogether with respective probabilities p1,p2,......,pn satisfying i=1∑npi=1
Complete step-by-step answer:
(i) we know that the sum of all probabilities in a probability distribution is always unity. Therefore,
⇒P(X=0)+P(X=1)+P(X=2)+.............+P(X=7)=1 ⇒0+k+2k+2k+3k+k2+2k2+7k2+k=1 ⇒9k+10k2=1 ⇒10k2+9k−1=0
Now, after solving this equation, we get
⇒(10k−1)(k+1)=0.
⇒10k−1=0 or k+1=0
⇒10k=1 ⇒k=101 or k=−1.
Hence, k = -1 cannot be possible because probability can’t be negative.
Therefore , k=101.
(ii) P(X<3).
⇒ P(X<3)= P(X=0)+P(X=1)+P(X=2).
⇒ P(X<3)= 0+k+2k
⇒ P(X<3)= 3k
putting k=101, we get
⇒ P(X<3) = 3×101=103.
Hence, P(X<3) = 103.
(iii) P(X>6).
⇒ P(X>6) = P(X=7)
⇒ P(X>6) = 7k2+k=7×(101)2+101.
⇒ P(X>6) = 1007+101=1007+10
⇒ P(X>6) = 10017.
Hence, P(X>6) = 10017.
(iv) P(0<X<3).
⇒ P(0<X<3)= P(X=1)+P(X=2).
⇒ P(0<X<3)= k+2k=3k
⇒ P(0<X<3)= 3×101=103
Hence , P(0<X<3) = 103.
Note: Whenever we are asked this type of question, first we have to remember the basic points of the probability distribution of a random variable. As according to the questions asked we will use the properties and formulae of probability distribution, we can easily solve them and we will get the required answers.