Question

The probability distribution functions which shows the number of hours (X) a student study during

lockdown period in a day, is given by (where C > 0)


X 0 1 2


P(X) 3C^3
4C – 10C^2
5C – 1


On the basis of above information answer the following questions:


(i) The correct equation for C is

(a) 3C3
– 10C2 + C – 2 = 0
(b) 3C3 + 10C2 + C – 2 = 0

(c) 3C3
– 10C2 + 9C – 2 = 0
(d) None of these


(ii) The value of C is

(a) 1
/2
(b) 1
/3

c) 1
/4

(d) 1
/6


(iii) P(X < 2) =

(a) 1
/2

(b) 1
/3

(c) 1
/4

(d) 1
/6



(iv) P(X = 1) =

(a) 2
/9

(b) 1
/9

(c) 2
/3

(d) 1
/3


(v) P(X ≥ 0) =

(a) 1
/2

(b) 1
/3

(c) 1
(d) 0​

Answer 1

$\underline{\textbf{Given:}}$

$\textsf{The probability function of X is}$

$\begin{array}{|c|c|c|c|}\cline{1-4}X&0&1&2\\\cline{1-4}P(X)&3C^3&4C-10C^2&5C-1\\\cline{1-4}\end{array}$

$\underline{\textbf{To find:}}$

$\textsf{(i) correct equation for C}$

$\textsf{(ii) The value of C}$

$\mathsf{(iii)\;P(X < 2)}$

$\textsf{(iv) P(X=1)}$

$\textsf{(iv)\;P(X{\geq}0)}$

$\underline{\textbf{Solution:}}$

$\mathsf{(i)}$

$\textsf{Since P(X) is a probability function,}$

$\mathsf{\displaystyle\sum\limits^{x=2}_{x=0}\;P(X=x)=1}$

$\implies\mathsf{P(X=0)+P(X=1)+P(X=2)=1}$

$\implies\mathsf{3C^3+4C-10C^2+5C-1=1}$

$\implies\mathsf{3C^3-10C^2+9C-1=1}$

$\implies\boxed{\mathsf{3C^3-10C^2+9C-2=0}}$

$\therefore\mathsf{Option\;(c)\;is\;correct}$

$\mathsf{(ii)\;consider,\;3C^3-10C^2+9C-2=0}$

$\implies\mathsf{3C^3-3C^2-7C^2+7C+2C-2=0}$

$\implies\mathsf{3C^2(C-1)-7C(C-1)+2(C-1)=0}$

$\implies\mathsf{(3C^2-7C+2)(C-1)=0}$

$\implies\mathsf{(3C^2-6C-C+2)(C-1)=0}$

$\implies\mathsf{(3C(C-2)-1(C-2))(C-1)=0}$

$\implies\mathsf{(3C-1)(C-2)(C-1)=0}$

$\implies\mathsf{C=1,2,\dfrac{1}{3}}$

$\mathsf{But\;C=1,2\;are\;not\;suitable\;for\;P(X)}$

$\implies\boxed{\mathsf{C=\dfrac{1}{3}}}$

$\therefore\mathsf{Option\;(b)\;is\;correct}$

$\mathsf{(iii)\;P(X<2)}$

$\mathsf{=P(X=0)+P(X=1)}$

$\mathsf{=3C^3+4C-10C^2}$

$\mathsf{=3\left(\dfrac{1}{27}\right)+4\left(\dfrac{1}{3}\right)-10\left(\dfrac{1}{9}\right)}$

$\mathsf{=\dfrac{1}{9}+\dfrac{4}{3}-\dfrac{10}{9}}$

$\mathsf{=\dfrac{1+12-10}{9}}$

$\mathsf{=\dfrac{3}{9}}$

$\mathsf{=\dfrac{1}{3}}$

$\implies\boxed{\mathsf{P(X\,<\,2)=\dfrac{1}{3}}}$

$\therefore\mathsf{Option\;(b)\;is\;correct}$

$\mathsf{(iv)\;P(X=1)}$

$\mathsf{=4C-10C^2}$

$\mathsf{=4\left(\dfrac{1}{3}\right)-10\left(\dfrac{1}{9}\right)}$

$\mathsf{=\dfrac{4}{3}-\dfrac{10}{9}}$

$\mathsf{=\dfrac{12-10}{9}}$

$\mathsf{=\dfrac{2}{9}}$

$\implies\boxed{\mathsf{P(X=1)=\dfrac{2}{9}}}$

$\therefore\mathsf{Option\;(a)\;is\;correct}$

$\mathsf{(v)\;P(x\;\geq\;0)}$

$\mathsf{=P(X=0)+P(X=1)+P(X=2)}$

$\mathsf{=1}$

$\implies\boxed{\mathsf{P(x\;\geq\;0)=1}}$

$\therefore\mathsf{Option\;(c)\;is\;correct}$

$\underline{\textbf{Find more:}}$

Let A and B be two events such that P(A)=0.6, P(B) = 0.2 and P(A/B)=0.5. Then P (A'/B')

equals

(a) 1/10

(b) 3/10

(C) 3/8

(d) 6/7​

https://brainly.in/question/14116335

Answer 2

Step-by-step explanation:

6/7

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