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Questions:

Generic 5

15 x 5 = 75

Let X be a continuous random variable with \(\text { p.d.f } f(x)= \begin{cases}a x & 0 \leq x \leq 1 \\ a & 1 \leq x \leq 2 \\ -a x+3 a & 2 \leq x \leq 3 \\ 0 & \text { otherwise }\end{cases}\)
(i) Determine the constant a
(ii) Compute \(\mathrm{P}(x \leq 1.5)\)

A continuous random variable has the following p.d.f, \(f(x)= \begin{cases}k x^{2}, & 0 \leq x \leq 10 \\ 0, & \text { otherwise }\end{cases}\) find k and evaluate
\((i)\ \mathrm{P}(0.2 \leq x \leq 0.5) \)
(ii) \( \mathrm{P}(x \leq 3)\)

For the p.d.f \(f(x)=\left\{\begin{array}{lc} \operatorname{Cx}(1-x)^{3}, & 0<x<1 \\ 0, & \text { elsewhere } \end{array}\right.\)
(i) the constant C
\(\text {(ii) } \mathrm{P}\left(x<\frac{1}{2}\right)\)

If the probability density function of a random variable is given by \(f(x)= \begin{cases}k\left(1-x^{2}\right) & 0<x<1 \\ 0 & \text { elsewhere }\end{cases}\)
(i) Find k
(ii) The distribution function of the random variable

A continuous random variable X has p.d.f \(f(x)=3 x^{2}, \ 0 \leq x \leq 1\). Find a and b such that
(i) \(\mathrm{P}(x \leq \mathrm{a})=\mathrm{P}(x>\mathrm{a})\)
(ii) \( \mathrm{P}(x>\mathrm{b})=0.05\)

A random variable X has the following probability mass function

x	0	1	2	3	4	5	6
P(X=x)	k	3k	5k	7k	9k	11k	13k

(i) Find k
(ii) Evaluate \(\mathbf{P}(x<4), P(x \geq 5), \quad \mathrm{P}(3<x \leq 6)\)
(iii) What is the smallest value of x for which \(\mathrm{P}(X \leq x)>\frac{1}{2}\)

A fair coin is tossed until a head or five tails occur. If X denotes the number of tosses of the coin, find mean of X

Find the probability distribution of the number ofsuccess in 2 tosses of a die, where a success in defined as "getting a number greater than 4". Also find the mean and variance of the distribution

Two numbers are selected at random (without replacement) from the first six positive integers. Let X denote the larger of two numbers obtained. Find E(X) and var (X).

10.

Let X denote the number of hours you study during a randomly selected school day. The probability that X can take the value X has the following form, where k is some unknown constant \(p(X=x)= \begin{cases}0.1 & \text { if } x=0 \\ k x & \text { if } x=1 \text { or } 2 \\ k(5-x) & \text { if } x=3 \text { or } 4 \\ 0 & \text { otherwise }\end{cases}\)
(i) Find the value of k
(ii) What is the probability that you study atleast 2 hours?
(iii) Exactly 2 hours
(iv) At most 2 hours

11.

The probability distribution of a random variable X is

X	1	2	4	2A	3A	5A
P(X)	\(\frac{1}{2}\)	\(\frac{1}{5}\)	\(\frac{3}{25}\)	\(\frac{1}{10}\)	\(\frac{1}{25}\)	\(\frac{1}{25}\)

Calculate
(i) A if E(X) = 2.94
(ii) V(X)

12.

The random variable X tan take only the values 0,1,2. Given that P(X = 0) = P(X = 1) = P and E(X²) = E(X), find the value of p.

13.

The probability distribution of the discrete random variables X and Y are given below

X	0	1	2	3
P(X)	\(\frac{1}{5}\)	\(\frac{2}{5}\)	\(\frac{1}{5}\)	\(\frac{1}{5}\)

Y	0	1	2	3
P(Y)	\(\frac{1}{5}\)	\(\frac{3}{10}\)	\(\frac{2}{5}\)	\(\frac{1}{10}\)

Prove that E(Y²) = 2E(X).

14.

The probability distribution of a random variation X is given below.

X	0	1	2	3	4
P(X)	0.1	0.25	0.3	0.2	0.15

Find
(i) V(X)
ii) V\((\frac{X}{2})\)

15.

A discrete random variable X has the following probability distribution.

x	1	2	3	4	5	6	7
P(X)	c	2c	2c	3c	c²	2c²	7c²+c

Find the value of e. Also, find the mean of the distribution.

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Answers Key:

Generic 5

15 x 5 = 75

f is a p.d.f
\( \int_{-\infty}^{\infty} f(x) d x=1 \)
\(a \int_{0}^{1} x d x+\int_{1}^{2} a d x+\int_{2}^{3}(-a x+3 a) d x=1 \)
\( a\left[\frac{x^{2}}{2}\right]_{0}^{1}+a[x]_{1}^{2}+\left[-\frac{a x^{2}}{2}+3 a x\right]_{2}^{3}=1 \)
\( \frac{a}{2}+a(2-1)+\left[-\frac{9 a}{2}+9 a-(-2 \mathrm{a}+6 a)\right]=1 \)
\( \frac{a}{2}+a-\frac{9 a}{2}+9 a+2 a-6 a=1 \)
\( -\frac{8 a}{2}+6 a=1 \)
\( 2 a=1 \)
\( a=\frac{1}{2} \)
\( \mathrm{P}(x \leq 1.5)=\frac{1}{2} \int_{0}^{1} x d x+\frac{1}{2} \int_{1}^{1+5} d x \)
\( =\frac{1}{2}\left[\frac{x^{2}}{2}\right]_{0}^{1}+\frac{1}{2}[x]_{1}^{1.5} \)
\( =\frac{1}{4}+\frac{1}{2}(0.5)=\frac{2}{4}=\frac{1}{2} \)

f is a p.d.f
\( \int_{-\infty}^{\infty} f(x) d x =1 \)
\(k \int_{0}^{10} x^{2} d x =1 \Rightarrow k\left[\frac{x^{3}}{3}\right]_{0}^{10}=1 \)
\(k\left(\frac{1000}{3}\right) \Rightarrow 1 \mathrm{k}=\frac{3}{1000}=0.003 \)
\(\text {(i) } \mathrm{P}(0.2
\( =0.003\left[\frac{x^{3}}{3}\right]_{0.2}^{0.5} \)
\(\text {(ii) } \mathrm{P}(x \leq 3) =\int_{0}^{3} k x^{2} d x \)
= 0.003\(\left[\frac{x^{3}}{3}\right]_{0}^{3} \)
= 0.001(27-0)
= 0.027

f is a p.d.f
\( \int_{-\infty}^{x} f(x) d x=1 \)
\( C \int_{0}^{1} x(1-x)^{3} d x=1\)
\( \int_{0}^{\infty} f(x) d x=\int_{0}^{a} f(\mathrm{a}-x) d x\)
\( C \int_{0}^{1}(1-x)(1-(1-x))^{3} d x=1\)
\( C \int_{0}^{1}(1-x) x^{3} d x=1 \)
\(C \int_{0}^{1}\left(x^{3}-x^{4}\right) d x=1 \)
\( C\left[\frac{x^{4}}{4}-\frac{x^{5}}{5}\right]_{0}^{1}=1 \)
\( \text {C }\left[\frac{1}{4}-\frac{1}{5}\right]=1 \)
\( C\left(\frac{5-4}{20}\right) =1 \)
\(\mathrm{P}\left(x<\frac{1}{2}\right) =C \int_{0}^{1 / 2} x(1-x)^{3} d x \)
\( =C \int_{0}^{1 / 2} x\left(1-3 x+3 x^{2}-x^{3}\right) d x \)
\( =20 \int_{0}^{1 / 2}\left(x-3 x^{2}+3 x^{3}-x^{+}\right) d x \)
\( =20\left[\frac{x^{2}}{2}-\frac{3 x^{3}}{3}+\frac{3 x^{4}}{4}-\frac{x^{5}}{5}\right]_{0}^{1 / 2} \)
\( =20\left[\frac{x^{2}}{2}-\frac{3 x^{3}}{3}+\frac{3 x^{4}}{4}-\frac{x^{5}}{5}\right]_{0}^{1 / 2} \)
\( =20\left[\frac{1}{8}-\frac{1}{8}+\frac{3}{64}-\frac{1}{160}\right] \)
\( =20\left[\frac{15-2}{320}\right]=\frac{13}{16} \)

f is a p.d.f
\( \int_{-\infty}^{\infty} f(x) d x =1 \)
\(k \int_{0}^{1}\left(1-x^{2}\right) d x =1 \)
\(k\left[x-\frac{x^{3}}{3}\right]_{0}^{1} =1 \)
\(k\left(1-\frac{1}{3}\right) =1 \Rightarrow \mathrm{k}=\frac{3}{2}\)
(ii) The distribution function
\( \mathrm{F}(x)=\int_{-\infty}^{x} f(t) d t \)
\((a)\ When\ x\ \in(-\infty, 0] \)
\(\mathrm{F}(x)=\int_{-\infty}^{x} f(\mathrm{t}) \mathrm{dt}=0 \)
(b) when \(x \in(0,1)\)
\( \mathrm{F}(x) =\int_{-\infty}^{1} f(t) d t=\int_{-\infty}^{11} f(t)^{1} d t+\int_{n}^{5} f(t) d t \)
\( =0+\int_{0}^{2} \frac{3}{2}\left(1-t^{2}\right) d t=\frac{3}{2}\left(x-\frac{x^{3}}{3}\right) \)
(c) when \(x \in[1, \infty]\)
\( \mathrm{F}(x)=\int_{-x}^{x} f(t) d t \)
\( =\int_{-\pi}^{0} f(t) d t+\int_{0}^{1} f(t) d t+\int_{1}^{x} f(t) d t \)
\( =0+\int_{0}^{1} \frac{3}{2}\left(1-t^{2}\right) d t+0 \)
\( =\frac{3}{2}\left[t-\frac{t^{3}}{3}\right]_{0}^{1}=1 \)
\(F(x)=\left\{\begin{array}{lc} 0 & -\infty<x \leq 0 \\ \frac{3}{2}\left(x-\frac{x^{3}}{3}\right) & 0<x<1 \\ 1 & 1 \leq x<\infty \end{array}\right.\)

\( \sum p_{i} =1 \)
\(\mathrm{P}(x \leq \mathrm{a})+\mathrm{P}(x>\mathrm{a}) =1 \)
\(\mathrm{P}(x \leq \mathrm{a})+\mathrm{P}(x \leq \mathrm{a}) =1 \)
\(2 \mathrm{P}(x \leq \mathrm{a}) =1 \)
\(\mathrm{P}(x \leq \mathrm{a}) =\frac{1}{2} \)
\(\int_{0}^{a} f(x) d x =\frac{1}{2} \)
\(3 \int_{0}^{a} x^{2} d x =\frac{1}{2} \)
\({\left[\frac{3 x^{3}}{3}\right]_{0}^{a} } =\frac{1}{2} \)
\(a^{3}=\frac{1}{2} =a=\left(\frac{1}{2}\right)^{1 / 3} \)
(ii) \( \mathrm{P}(x>\mathrm{b}) =0.05 \)
\(\int_{\mathrm{b}}^{1} f(x) d x =0.05 \)
\(3 \int_{b}^{1} x^{2} d x =0.05 \)
\(3\left[\frac{x^{5}}{3}\right]_{1}^{1} =0.05 \)
\(1-\mathrm{b}^{3} =0.05 \)
\(\mathrm{b}^{3} =0.95=\frac{95}{100} \)
\(\mathrm{~b} =\left(\frac{19}{20}\right)^{1 / 3} \)

(i) \( \sum p_{i}=1 \)
\( k+3 k+5 k+7 k+9 k+11 k+13 k=1 \)
\( 49 k=1 \)
\( k=\frac{1}{49}\)
(ii) \( P(x<4)= P(X=0)+P(X=1)+P(X=2) +P(X=3) \)
\(= k+3 k+5 \mathrm{k}+7 \mathrm{k} \)
\(= 16 \mathrm{k}=\frac{16}{49}\)
\( \mathrm{P}(x \geq 5) =\mathrm{P}(\mathrm{X}=5)+\mathrm{P}(\mathrm{X}=6) \)
\( =\mathrm{k}+13 \mathrm{k}+24 \mathrm{k}=\frac{24}{49} \)
\(= \frac{9}{49}+\frac{11}{49}+\frac{13}{49}=\frac{33}{49} \)
(iii) The minimum value of x may be determined by trial and error method
\( \mathrm{P}(x \leq 0)=\frac{1}{49}<\frac{1}{2} \)
\( \mathrm{P}(x \leq 1)=\frac{4}{49}<\frac{1}{2} \)
\( \mathrm{P}(x \leq 2)=\frac{9}{49}<\frac{1}{2} \)
\( \mathrm{P}(x \leq 3)=\frac{16}{49}<\frac{1}{2} \)
\( \mathrm{P}(x \leq 4)=\frac{25}{49}>\frac{1}{2}\)
The smallest value of x for which \(\mathrm{P}(X \leq x)>\frac{1}{2} \text { is } 4 .\)

Sample space is = {H, TH, TTH, TTTH, TTTTH, TTTTT}
X takes value 1, 2, 3, 4, 5
\( \mathrm{P}(\mathrm{X}=1)=\mathrm{P}(\mathrm{H})=\frac{1}{2} \)
\( \mathrm{P}(\mathrm{X}=2)=\mathrm{P}(\mathrm{TH})=\frac{1}{2} \times \frac{1}{2}=\frac{1}{4} \)
\( \mathrm{P}(\mathrm{X}=3)=\mathrm{P}(\mathrm{TTH})=\frac{1}{2} \times \frac{1}{2} \times \frac{1}{2}=\frac{1}{8} \)
\( \mathrm{P}(\mathrm{X}=4) =\mathrm{P}(\mathrm{TTTH}) \)
\( =\frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2}=\frac{1}{16} \)
\(\mathrm{P}(\mathrm{X}=5) =\mathrm{P}(\text { TTTTH U TTTTT }) \)
\( =\left(\frac{1}{2}\right)^{5}+\left(\frac{1}{2}\right)^{5}=\frac{2}{32}=\frac{1}{16}\)
Probability distribution of X is

x	1	2	3	4	5
P(X=x)	\(\frac{1}{2}\)	\(\frac{1}{4}\)	\(\frac{1}{8}\)	\(\frac{1}{16}\)	\(\frac{1}{16}\)

\( \text {Mean } =\mathrm{E}(\mathrm{X})=\sum x p_{X}(x) \)
\( =\frac{1}{2}+\frac{2}{4}+\frac{3}{8}+\frac{4}{16}+\frac{5}{16} \)
\( =\frac{31}{16}=1.9\)

Let x denote the number of successes in two tosses of a die. Then X can take values 0, 1, 2.
\( \mathrm{P}(\mathrm{S}) =\frac{2}{6}=\frac{1}{3}, \mathrm{P}(\mathrm{F})=\frac{2}{3} \)
\(\mathrm{P}(\mathrm{X}=0) =\mathrm{P}(\mathrm{FF}) \)
\( =\frac{2}{3} \times \frac{2}{3}=\frac{4}{9} \)
\(\mathrm{P}(\mathrm{X}=1) =2 \mathrm{P}(\mathrm{SF}) \)
\( =2 \times \frac{1}{3} \times \frac{2}{3}=\frac{4}{9} \)
\(\mathrm{P}(\mathrm{X}=2) =\mathrm{P}(\mathrm{SS}) \)
\( =\frac{1}{3} \times \frac{1}{3}=\frac{1}{9} \)

x	0	1	2
p(x)	\(\frac{4}{9}\)	\(\frac{4}{9}\)	\(\frac{1}{9}\)

\( \text {Mean } =\mathrm{E}(\mathrm{X})=\sum x p_{x}(x) \)
\(=0+\frac{4}{9}+\frac{2}{9}=\frac{6}{9}=\frac{2}{3} \)
\(\mathrm{E}\left(\mathrm{X}^{2}\right) =\sum_{x} x^{2} p_{x}(x) \)
\(=0+\frac{4}{9}+\frac{4}{9}=\frac{8}{9} \)
\(\operatorname{Var}(\mathrm{X}) =\mathrm{E}\left(\mathrm{X}^{2}\right)-[\mathrm{E}(\mathrm{X})]^{2} \)
\(=\frac{8}{9}-\frac{4}{9}=\frac{4}{9} \)

X can take values 2, 3, 4, 5, 6
P(X = 2) = Probability of getting 1 in the first selection and 2 in second selection (or) vice versa
\( =2 \times \frac{1}{6} \times \frac{1}{5}=\frac{1}{15} \)
\(P(X=3)=2 \times\left(\frac{2}{6} \times \frac{1}{5}\right)=\frac{2}{15} \)
\(P(X=4)=2 \times\left(\frac{3}{6} \times \frac{1}{5}\right)=\frac{1}{5} \)
\(P(X=5)=2 \times\left(\frac{4}{6} \times \frac{1}{5}\right)=\frac{4}{15} \)
\(\mathrm{P}(\mathrm{X}=6)=2 \times\left(\frac{5}{6} \times \frac{1}{5}\right)=\frac{1}{3}\)
Probability distribution of X is

x	2	3	4	5	6
p(x)	\(\frac{1}{15}\)	\(\frac{2}{15}\)	\(\frac{1}{15}\)	\(\frac{4}{15}\)	\(\frac{1}{3}\)

\( \mathrm{E}(\mathrm{X}) =\sum x p_{x}(x) \)
\(=\frac{2}{15}+\frac{6}{15}+\frac{4}{5}+\frac{4}{3}+2 \)
\( =\frac{70}{15}=\frac{14}{3} \)

10.

(i) The Probability distribution of X

x	0	1	2	3	4
p(x)	0.1	k	2k	2k	k

\( \sum P_{i} =1 \)
\(0.1+\mathrm{k}+2 \mathrm{k}+2 \mathrm{k}+\mathrm{k} =1 \)
\(6 \mathrm{k} =1-0.1 \)
\(6 \mathrm{k} =0.9 \Rightarrow \mathrm{k}=0.15 \)
(ii) \(\mathrm{P}(x \geq 2)=\mathrm{P}(\mathrm{X}=2)+\mathrm{P}(\mathrm{X}=3)+\mathrm{P}(\mathrm{X}=4)\)
2k + 2k +k = 5k
= 5(0.15) = 0.75
(iii) P(X = 2) = 2k = 2(0.15) = 0.3
(iv) \(\mathrm{P}(x \leq 2)=\mathrm{P}(\mathrm{X}=0)+\mathrm{P}(\mathrm{X}=1)+\mathrm{P}(\mathrm{X}=2)\)
0.1 + k + 2k = 0.1 + 3k
= 0.1 + 3(0.15) = 0.55

11.

E(X) = Σx_i²p_i
\(\Rightarrow E(X)=1(\frac { 1 }{ 2 } )+2\left( \frac { 1 }{ 5 } \right) +4\left( \frac { 3 }{ 25 } \right) +2A\left( \frac { 1 }{ 10 } \right) +3A\left( \frac { 1 }{ 25 } \right) +5A\left( \frac { 1 }{ 25 } \right) \)
\(\Rightarrow \frac { 1 }{ 2 } +\frac { 2 }{ 5 } +\frac { 12 }{ 25 } +\frac { A }{ 5 } +\frac { 3A }{ 25 } +\frac { A }{ 5 } \)
\(=\frac { 69 }{ 50 } +\frac { 13A }{ 25 } \)
Since E(X) = 2.94
\(\frac { 69 }{ 50 } +\frac { 13A }{ 25 } =2.94\Rightarrow \frac { 13A }{ 25 } =2.94-\frac { 69 }{ 50 } \)
= 2.94 - 1.38 = 1.56
\(\Rightarrow A=\frac { 1.56\times 25 }{ 13 } =\frac { 39 }{ 13 } =3\)
\(E({ X }^{ 2 })=1\left( \frac { 1 }{ 2 } \right) +4\left( \frac { 1 }{ 5 } \right) +16\left( \frac { 3 }{ 25 } \right) +{ (2A) }^{ 2 }\left( \frac { 1 }{ 10 } \right) { +(3A) }^{ 2 }\left( \frac { 1 }{ 25 } \right) +{ (5A) }^{ 2 }\left( \frac { 1 }{ 25 } \right) [\because A=3]\)
\(=\frac { 1 }{ 2 } +\frac { 4 }{ 5 } +\frac { 48 }{ 25 } +36\left( \frac { 1 }{ 10 } \right) \)
\(E({ X }^{ 2 })=\frac { 25+40+96+180+162+450 }{ 50 } \)
\(=\frac { 953 }{ 50 } =19.06\)
\(V(X)=E({ X }^{ 2 })-{ [E(X)] }^{ 2 }\)
\(=19.06-{ (2.94) }^{ 2 }[\because E(X)=2.94]\)
= 19.06 - 8.6436
V(X) = 10.4164

12.

Clearly P(X = 0) + P(X = 1) + P(X = 2)= 1
p + P + P(X = 2) =1
2p + P(X = 2) = 1
P(X = 2) = 1 - 2p
so, probability distribution of X is

X	0	1	2
P(X)	p	p	1-2p

∴ E(X) = 0xp+1xp+2(1-2p)
= p+2-4p = 2-3p
and E(X²) = 0xp+1(p)+4(1-2p)
= p+4-8p = 4-7p
Since E(X²) = E(X)we get,
4-7p = 2-3p ⇒ 4-2 = -3p+7p
⇒2 = 4p ⇒ p =\(\frac{2}{4}=\frac{1}{2}\)
∴ p = \(\frac{1}{2}\)

13.

\(E(X)=0\times \frac { 1 }{ 5 } +1(\frac { 2 }{ 5 } )+2\left( \frac { 1 }{ 5 } \right) +3\times \frac { 1 }{ 5 } \)
\(=\frac { 2 }{ 5 } +\frac { 2 }{ 5 } +\frac { 3 }{ 25 } =\frac { 7 }{ 5 } \)
\(\\ \therefore 2E(X)=\frac { 14 }{ 5 } ...(1)\)
\(E({ Y }^{ 2 })=0\times \frac { 1 }{ 5 } +{ 1 }^{ 2 }(\frac { 3 }{ 10 } )+{ 2 }^{ 2 }(\frac { 2 }{ 5 } )+{ 3 }^{ 2 }(\frac { 1 }{ 10 } )\)
\(=\frac { 3 }{ 10 } +\frac { 8 }{ 5 } +\frac { 9 }{ 10 } =\frac { 3+16+9 }{ 10 } \)
\(=\frac { 28 }{ 10 } =\frac { 14 }{ 5 } ..(2)\)
From (1) and (2), E(Y²) = 2 E(X).

14.

i) E(X) = Σx_ip_i
= 0(0.1)+1(0.25)+2(0.3)+3(0.2)+4.(0.15)
= 2.05
E(X²) = ∑x_i²p_i
= 0(0.1)+1(0.25)+4(0.3)+9(0.2)+16(0.15)
= 5.65
Now, V(X) = E(X²)-[E(X)]²
= 5.65 - (2.05)²= 1.4475
ii) \(V\left( \frac { X }{ 2 } \right) =\frac { 1 }{ 4 } v(X)\quad [\because V(aX)={ a }^{ 2 }V(X)]\)
\(=\frac { 1 }{ 4 } (1.4475)\)
\(V\left( \frac { X }{ 2 } \right) =0.361875\)

15.

Since X is the random variable taking values
1, 2,....7
P(X=1)+P(X=2)+...P(X=7) = 1
⇒ c+2c+2c+3c+c²+2c²+7c²+c = 1
⇒ 10c²+ 9c +1 = 0
⇒ (c+1)(10c-1)= 0
⇒ c = \(\frac{1}{10}\)[∵ c-1 is not possible]
Now, E(X) = Σxipi
= 1(c) + 2(2c) + 3(2c) + 4(3c) + 5c²+ 12c²+ 7(7c²+ c)
= 66c²+ 30c = 66(\(\frac{1}{10}\))²+ 30(\(\frac{1}{10}\))
= \(\frac{66}{100}+3=\frac{366}{100}=3.66\)
∴ E(X) = 3.66