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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) kk
(ii) P(X<3)P(X < 3)
(iii) P(X>6)P(X > 6)
(iv) P(0<X<3)P(0 < X < 3)

Explanation

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,.......,xn{x_1},{x_2},{x_3},.......,{x_n}together with respective probabilities p1,p2,......,pn{p_1},{p_2},......,{p_n} satisfying i=1npi=1\sum\limits_{i = 1}^n {{p_i} = 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+9k1=0  \Rightarrow P(X = 0) + P(X = 1) + P(X = 2) + ............. + P(X = 7) = 1 \\\ \Rightarrow 0 + k + 2k + 2k + 3k + {k^2} + 2{k^2} + 7{k^2} + k = 1 \\\ \Rightarrow 9k + 10{k^2} = 1 \\\ \Rightarrow 10{k^2} + 9k - 1 = 0 \\\
Now, after solving this equation, we get
(10k1)(k+1)=0\Rightarrow (10k - 1)(k + 1) = 0.
10k1=0\Rightarrow 10k - 1 = 0 or k+1=0k + 1 = 0
10k=1 k=110  \Rightarrow 10k = 1 \\\ \Rightarrow k = \dfrac{1}{{10}} \\\ or k=1k = - 1.
Hence, k = -1 cannot be possible because probability can’t be negative.
Therefore , k=110k = \dfrac{1}{{10}}.
(ii) P(X<3)P(X < 3).
\Rightarrow P(X<3)P(X < 3)= P(X=0)+P(X=1)+P(X=2)P(X = 0) + P(X = 1) + P(X = 2).
\Rightarrow P(X<3)P(X < 3)= 0+k+2k0 + k + 2k
\Rightarrow P(X<3)P(X < 3)= 3k
putting k=110k = \dfrac{1}{{10}}, we get
\Rightarrow P(X<3)P(X < 3) = 3×110=3103 \times \dfrac{1}{{10}} = \dfrac{3}{{10}}.
Hence, P(X<3)P(X < 3) = 310\dfrac{3}{{10}}.
(iii) P(X>6)P(X > 6).
\Rightarrow P(X>6)P(X > 6) = P(X=7)P(X = 7)
\Rightarrow P(X>6)P(X > 6) = 7k2+k=7×(110)2+1107{k^2} + k = 7 \times {\left( {\dfrac{1}{{10}}} \right)^2} + \dfrac{1}{{10}}.
\Rightarrow P(X>6)P(X > 6) = 7100+110=7+10100\dfrac{7}{{100}} + \dfrac{1}{{10}} = \dfrac{{7 + 10}}{{100}}
\Rightarrow P(X>6)P(X > 6) = 17100\dfrac{{17}}{{100}}.
Hence, P(X>6)P(X > 6) = 17100\dfrac{{17}}{{100}}.
(iv) P(0<X<3)P(0 < X < 3).
\Rightarrow P(0<X<3)P(0 < X < 3)= P(X=1)+P(X=2)P(X = 1) + P(X = 2).
\Rightarrow P(0<X<3)P(0 < X < 3)= k+2k=3kk + 2k = 3k
\Rightarrow P(0<X<3)P(0 < X < 3)= 3×110=3103 \times \dfrac{1}{{10}} = \dfrac{3}{{10}}
Hence , P(0<X<3)P(0 < X < 3) = 310\dfrac{3}{{10}}.

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.