Exercise 11.2 - Chapter 11 Probability Distributions 12th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Ex $11.2$
Question 1.
Three fair coins are tossed simultaneously. Find the probability mass function for number of heads occurred.
Solution:
When three coins are tossed, the sample space is
$\mathrm{S}=\{\mathrm{HHH}, \mathrm{HHT}, \mathrm{HTH}, \mathrm{HTT}, \mathrm{THH}, \mathrm{THT}, \mathrm{TTH}, \mathrm{TTT}\}$
' $\mathrm{X}$ ' is the random variable denotes the number of heads.
$\therefore$ ' $\mathrm{X}$ ' can take the values of $0,1,2$ and 3
Hence, the probabilities
$
\begin{aligned}
& P(X=0)=P(\text { No heads })=\frac{1}{8} ; \quad P(X=1)=P(1 \text { head })=\frac{3}{8} \\
& \mathrm{P}(\mathrm{X}=2)=\mathrm{P}(2 \text { heads })=\frac{3}{8} ; \mathrm{P}(\mathrm{X}=3)=\mathrm{P}(3 \text { heads })=\frac{1}{8} \\
&
\end{aligned}
$
Question 2.
A six sided die is marked ' 1 ' on one face, ' 3 ' on two of its faces, and ' 5 ' on remaining three faces. The die is thrown twice. If $\mathrm{X}$ denotes the total score in two throws, find
(i) the probability mass function
(ii) the cumulative distribution function
(iii) $\mathrm{P}(4 \leq \mathrm{X}<10)$
(iv) $P(X \geq 6)$
Solution:
Given that die is marked ' 1 ' on one face, ' 3 ' on two of its faces and ' 5 ' on remaining three faces. i.e., $\{1$, $3,3,5,5,5\}$ in a single die.
When it is thrown twice, the number of sample points is 36 , in which the sum of faces numbers are $2,, 4,6$, 8 and 10 are the value of random variable ' $\mathrm{X}$ '.
$
\begin{array}{ll}
\mathrm{P}(\mathrm{X}=2)=\frac{1}{36} \quad ; \quad \mathrm{P}(\mathrm{X}=4)=\frac{4}{36} ; \quad P(X=6)=\frac{10}{36} \\
\mathrm{P}(\mathrm{X}=8)=\frac{12}{36} \quad ; \quad \mathrm{P}(\mathrm{X}=10)=\frac{9}{36}
\end{array}
$

(i) Probability mass function:

(ii) The Cumulative distribution function:
$
\mathrm{F}(x)=\left\{\begin{array}{ccc}
0 & \text { for } & x<2 \\
1 / 36 & \text { for } & x \leq 2 \\
5 / 36 & \text { for } & x \leq 4 \\
15 / 36 & \text { for } & x \leq 6 \\
27 / 36 & \text { for } & x \leq 8 \\
1 & \text { for } & x \leq 10
\end{array}\right.
$
(iii) $(4 \leq 10)=\mathrm{P}(\mathrm{X}=4)+\mathrm{P}(\mathrm{X}=6)+\mathrm{P}(\mathrm{X}=8)$
$
=\frac{4}{36}+\frac{10}{36}+\frac{12}{36}=\frac{26}{36}=\frac{13}{18}
$
$
\begin{aligned}
& \text { (iv) } \mathrm{P}(\mathrm{X} \geq 6)=\mathrm{P}(\mathrm{X}=6)+\mathrm{P}(\mathrm{X}=8)+\mathrm{P}(\mathrm{X}=10) \\
& =\frac{10}{36}+\frac{12}{36}+\frac{9}{36}=\frac{31}{36}
\end{aligned}
$
Question 3.
Find the probability mass function and cumulative distribution function of number of girl child in families with 4 children, assuming equal probabilities for boys and girls.
Solution:
Let ' $X$ ' be the random variable which denotes the number of girl children in the family of 4 children and $X$ takes the values of $0,1,2,3,4$.
Probability of child being a boy $=P(B)=\frac{1}{2}$
Probability of child being a girl $=P(G)=\frac{1}{2}$

$\begin{aligned}
& P(X=0)=P(4 \text { Boys })=\frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2}=\frac{1}{16} \\
& P(X=1)=P(1 \mathrm{G}, 3 \mathrm{~B})=\left(\frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2}\right) 4 \mathrm{C}_1=\frac{4}{16}=\frac{1}{4} \\
& \mathrm{P}(\mathrm{X}=2)=\mathrm{P}(2 \mathrm{G}, 2 \mathrm{~B})=\left(\frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2}\right) 4 \mathrm{C}_2=\frac{6}{16}=\frac{3}{8} \\
& \mathrm{P}(\mathrm{X}=3)=\mathrm{P}(3 \mathrm{G}, 1 \mathrm{~B})=\left(\frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2}\right) 4 \mathrm{C}_3=\frac{4}{16}=\frac{1}{4} \\
& P(X=4)=P(4 \text { Girls })=\frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2}=\frac{1}{16} \\
& \text { The probability mass function is } f(x)=\left\{\begin{array}{lll}
\frac{1}{16} & \text { for } & x=0,4 \\
\frac{1}{4} & \text { for } & x=1,3 \\
\frac{3}{8} & \text { for } & x=2
\end{array}\right. \\
& \text { Cumulative distribution function } \mathrm{F}(x)=\left\{\begin{array}{ccc}
0 & \text { for } & x<0 \\
1 / 16 & \text { for } & x \leq 0 \\
5 / 16 & \text { for } & x \leq 1 \\
11 / 16 & \text { for } & x \leq 2 \\
15 / 16 & \text { for } & x \leq 3 \\
1 & \text { for } & x \leq 4
\end{array}\right. \\
&
\end{aligned}$

Question 4.
Suppose a discrete random variable can only take the values 0,1 , and 2 . The probability mass function is defined by $f(x)=\left\{\begin{array}{cc}\frac{x^2+1}{k}, & \text { for } x=0,1,2 \\ 0 & \text { otherwise }\end{array}\right.$
Find
(i) the value of $\mathrm{k}$
(ii) cumulative distribution function
(iii) $\mathrm{P}(\mathrm{X} \geq 1)$.

Solution:
Given probability mass function $f(x)=\left\{\begin{array}{cc}\frac{x^2+1}{k}, & \text { for } x=0,1,2 \\ 0, & \text { otherwise }\end{array}\right.$
(i) Value of $k: \Sigma \mathrm{P}_i=1 \Rightarrow f(0)+f(1)+f(2)=1$
i.e., $\frac{1}{k}+\frac{2}{k}+\frac{5}{k}=1$
(ii) Cumulative distribution function: $\mathrm{F}(x)=\left\{\begin{array}{ccc}0 & \text { for } & x<0 \\ 1 / 8 & \text { for } & x \leq 0 \\ 3 / 8 & \text { for } & x \leq 1 \\ 1 & \text { for } & x \leq 2\end{array}\right.$

$\text { (iii) } \begin{aligned}
\mathrm{P}(\mathrm{X} \geq 1) & =\mathrm{P}(\mathrm{X}=1)+\mathrm{P}(\mathrm{X}=2) \\
& =\frac{2}{8}+\frac{5}{8}=\frac{7}{8}
\end{aligned}$

Question 5.
The cumulative distribution function of a discrete random variable is given by
$
F(x)=\left\{\begin{array}{cr}
0 & -\infty<x<-1 \\
0.15 & -1 \leq x<0 \\
0.35 & 0 \leq x<1 \\
0.60 & 1 \leq x<2 \\
0.85 & 2 \leq x<3 \\
1 & 3 \leq x<\infty
\end{array}\right.
$
Find the (i) the probability mass function
(ii) $\mathrm{P}(\mathrm{X}<1)$
(iii) $P(X \geq 2)$
Solution:

$\text { Given cumulative distribution function is } \mathrm{F}(x)=\left\{\begin{array}{cc}
0 & -\infty<x<-1 \\
0.15 & -1 \leq x<0 \\
0.35 & 0 \leq x<1 \\
0.60 & 1 \leq x<2 \\
0.85 & 2 \leq x<3 \\
1 & 3 \leq x<\infty
\end{array}\right.$
The values of ' $x$ ' are $-1,0,1,2,3$
nis $\mathrm{F}(x)=\left\{\begin{array}{cr}0 & -\infty For $\mathrm{x}=-1, \mathrm{f}(\mathrm{x})=0.15-0=0.15$
For $\mathrm{x}=0, \mathrm{f}(\mathrm{x})=0.35-0.15=0.20$
For $x=1, f(x)=0.60-0.35=0.25$
For $x=2, f(x)=0.85-0.60=0.25$
For $\mathrm{x}=3, \mathrm{f}(\mathrm{x})=1-0.85=0.15$
(i) Probability mass function table

(ii) $\mathrm{P}(\mathrm{X}<1)=\mathrm{P}(\mathrm{X}=-1)+\mathrm{P}(\mathrm{X}=0)=0.15+0.20=0.35$
(iii) $P(X \geq 2)=P(X=2)+P(X=3)=0.25+0.15=0.40$

Question 6.
A random variable $\mathrm{X}$ has the following probability mass function.

Solution:
Given probability mass function

(i) We know that $\Sigma P_i=1$
i.e., $\mathrm{k}^2+2 \mathrm{k}^2+3 \mathrm{k}^2+2 \mathrm{k}+3 \mathrm{k}=1$
$6 k^2+5 k=1$
$6 \mathrm{k}^2+5 \mathrm{k}-1=0$
$(k+1)(6 k-1)=0$

$
\begin{aligned}
& \text { (ii) } \mathrm{P}(2 \leq \mathrm{X}<5) \\
& =\mathrm{P}(\mathrm{X}=2)+\mathrm{P}(\mathrm{X}=3)+\mathrm{P}(\mathrm{X}=4) \\
& =2 \mathrm{k}^2+3 \mathrm{k}^2+2 \mathrm{k}=5 \mathrm{k}^2+2 \mathrm{k} \\
& =5\left(\frac{1}{36}\right)+2\left(\frac{1}{6}\right)=\frac{5}{36}+\frac{1}{3}=\frac{17}{36}
\end{aligned}
$
(iii) $\mathrm{P}(3<\mathrm{X})=\mathrm{P}(\mathrm{X}>3)$
$
=P(X=4)+P(X=5)=2 \mathrm{k}+3 \mathrm{k}=5 \mathrm{k}=5 / 6
$
Question 7.
The cumulative distribution function of a discrete random variable is given by.

$
F(x)=\left\{\begin{array}{lll}
0 & \text { for } & -\infty 1 / 2 & \text { for } & 0 \leq x<1 \\
3 / 5 & \text { for } & 1 \leq x<2 \\
4 / 5 & \text { for } & 2 \leq x<3 \\
9 / 10 & \text { for } & 3 \leq x<4 \\
1 & \text { for } & 4 \leq x<\infty
\end{array}\right.
$
Find (i) the probability mass function
(ii) $\mathrm{P}(\mathrm{X}<3)$ and
(iii) $P(X \geq 2)$.
Solution:

$\text { Given cumulative distribution function is } \mathrm{F}(x)=\left\{\begin{array}{lll}
0 & \text { for } & -\infty<x<0 \\
1 / 2 & \text { for } & 0 \leq x<1 \\
3 / 5 & \text { for } & 1 \leq x<2 \\
4 / 5 & \text { for } & 2 \leq x<3 \\
9 / 10 & \text { for } & 3 \leq x<4 \\
10 & \text { for } & 4 \leq x<\infty
\end{array}\right.$

(i) The value of ' $x$ ' are $0,1,2,3$ and 4
For $x=0, f(x)=\frac{1}{2}-0=\frac{1}{2} \quad ; \quad$ For $x=1, f(x)=\frac{3}{5}-\frac{1}{2}=\frac{1}{10}$
For $x=2, f(x)=\frac{4}{5}-\frac{3}{5}=\frac{1}{5} \quad ; \quad$ For $x=3, f(x)=\frac{9}{10}-\frac{4}{5}=\frac{1}{10}$
For $x=4, f(x)=1-\frac{9}{10}=\frac{1}{10}$
Probability mass function is

$\text { (ii) } \begin{aligned}
\mathrm{P}(\mathrm{X}<3) & =\mathrm{P}(\mathrm{X}=0)+\mathrm{P}(\mathrm{X}=1)+\mathrm{P}(\mathrm{X}=2) \\
& =\frac{1}{2}+\frac{1}{10}+\frac{1}{5}=\frac{5+1+2}{10}=\frac{8}{10}=\frac{4}{5}
\end{aligned}$

$\text { (iii) } \begin{aligned}
\mathrm{P}(\mathrm{X} \geq 2) & =P(X=2)+P(X=3)+P(X=4) \\
& =\frac{1}{5}+\frac{1}{10}+\frac{1}{10}=\frac{2+2}{10}=\frac{4}{10}=\frac{2}{5}
\end{aligned}$