June 2023

1. State whether the following statements are True or False. Give reasons in support of your answers :
   2 each
   (a) If a card is drawn from a well shuffled pack of 52 cards, then the probability that it is either club or king is $\frac{9}{26}$$\frac{9}{26}$(9)/(26)\frac{9}{26}$\frac{9}{26}$.

1. Total number of clubs in a deck: There are 13 clubs in a deck (one for each rank: Ace through King).
2. Total number of kings in a deck: There are 4 kings in a deck (one for each suit: clubs, diamonds, hearts, and spades).

3. Adjusting for the overlap (King of Clubs): Since the King of Clubs is already counted in the total number of clubs, we need to subtract it from the total count to avoid double-counting.

4. Probability calculation: The probability of drawing either a club or a king is the number of favorable outcomes divided by the total number of outcomes (52 cards in the deck). Therefore, the probability is $\frac{16}{52}$$\frac{16}{52}$(16)/(52)\frac{16}{52}$\frac{16}{52}$.

• $p=-\frac{1}{2}$$p=-\frac{1}{2}$p=-(1)/(2)p = -\frac{1}{2}$p=-\frac{1}{2}$ (not valid as probability cannot be negative)
• $p=0$$p=0$p=0p = 0$p=0$ (possible, but would imply no chance of success)
• $p=\frac{1}{4}$$p=\frac{1}{4}$p=(1)/(4)p = \frac{1}{4}$p=\frac{1}{4}$ (a valid probability)
• $p=1$$p=1$p=1p = 1$p=1$ (possible, but would imply certainty of success)

1. The mean of $Z$$Z$ZZ$Z$ is the difference of the means of $X$$X$XX$X$ and $Y$$Y$YY$Y$:
   $${\mu }_Z={\mu }_X-{\mu }_Y$$$${\mu }_Z={\mu }_X-{\mu }_Y$$mu _(Z)=mu _(X)-mu _(Y)\mu_Z = \mu_X – \mu_Y$${\mu }_Z={\mu }_X-{\mu }_Y$$
2. The variance of $Z$$Z$ZZ$Z$ is the sum of the variances of $X$$X$XX$X$ and $Y$$Y$YY$Y$ (since variance is always additive, even for the difference of two variables):
   $${\sigma }_Z^2={\sigma }_X^2+{\sigma }_Y^2$$$${\sigma }_Z^2={\sigma }_X^2+{\sigma }_Y^2$$sigma_(Z)^(2)=sigma_(X)^(2)+sigma_(Y)^(2)\sigma^2_Z = \sigma^2_X + \sigma^2_Y$${\sigma }_Z^2={\sigma }_X^2+{\sigma }_Y^2$$

• ${\mu }_X=7$${\mu }_X=7$mu _(X)=7\mu_X = 7${\mu }_X=7$, ${\sigma }_X^2=12$${\sigma }_X^2=12$sigma_(X)^(2)=12\sigma^2_X = 12${\sigma }_X^2=12$
• ${\mu }_Y=4$${\mu }_Y=4$mu _(Y)=4\mu_Y = 4${\mu }_Y=4$, ${\sigma }_Y^2=9$${\sigma }_Y^2=9$sigma_(Y)^(2)=9\sigma^2_Y = 9${\sigma }_Y^2=9$

• The mean of $Z=X-Y$$Z=X-Y$Z=X-YZ = X – Y$Z=X-Y$ is $3$$3$33$3$ (since $7-4=3$$7-4=3$7-4=37 – 4 = 3$7-4=3$).
• The variance of $Z$$Z$ZZ$Z$ is $21$$21$2121$21$ (since $12+9=21$$12+9=21$12+9=2112 + 9 = 21$12+9=21$).

2. (a) The odds that a book will be favourably reviewed by 3 independent critics are 5 to 2,4 to 3 and 3 to 4 , respectively. Find the probability that of the three reviews, a majority will be favourable.

1. For the first critic: $P_1=\frac{5}{7}\approx 0.7143$$P_1=\frac{5}{7}\approx 0.7143$P_(1)=(5)/(7)~~0.7143P_1 = \frac{5}{7} \approx 0.7143$P_1=\frac{5}{7}\approx 0.7143$
2. For the second critic: $P_2=\frac{4}{7}\approx 0.5714$$P_2=\frac{4}{7}\approx 0.5714$P_(2)=(4)/(7)~~0.5714P_2 = \frac{4}{7} \approx 0.5714$P_2=\frac{4}{7}\approx 0.5714$
3. For the third critic: $P_3=\frac{3}{7}\approx 0.4286$$P_3=\frac{3}{7}\approx 0.4286$P_(3)=(3)/(7)~~0.4286P_3 = \frac{3}{7} \approx 0.4286$P_3=\frac{3}{7}\approx 0.4286$

1. All three critics give favorable reviews.
2. Only the first and second critics give favorable reviews.
3. Only the first and third critics give favorable reviews.
4. Only the second and third critics give favorable reviews.

1. All three critics give favorable reviews:
   The probability is $P_1\times P_2\times P_3$$P_1\times P_2\times P_3$P_(1)xxP_(2)xxP_(3)P_1 \times P_2 \times P_3$P_1\times P_2\times P_3$.
2. Only the first and second critics give favorable reviews:
   The probability is $P_1\times P_2\times (1-P_3)$$P_1\times P_2\times (1-P_3)$P_(1)xxP_(2)xx(1-P_(3))P_1 \times P_2 \times (1 – P_3)$P_1\times P_2\times (1-P_3)$.
3. Only the first and third critics give favorable reviews:
   The probability is $P_1\times (1-P_2)\times P_3$$P_1\times (1-P_2)\times P_3$P_(1)xx(1-P_(2))xxP_(3)P_1 \times (1 – P_2) \times P_3$P_1\times (1-P_2)\times P_3$.
4. Only the second and third critics give favorable reviews:
   The probability is $(1-P_1)\times P_2\times P_3$$(1-P_1)\times P_2\times P_3$(1-P_(1))xxP_(2)xxP_(3)(1 – P_1) \times P_2 \times P_3$(1-P_1)\times P_2\times P_3$.

1. All three critics give favorable reviews: $0.1749$$0.1749$0.17490.1749$0.1749$
2. Only the first and second critics give favorable reviews: $0.2332$$0.2332$0.23320.2332$0.2332$
3. Only the first and third critics give favorable reviews: $0.1312$$0.1312$0.13120.1312$0.1312$
4. Only the second and third critics give favorable reviews: $0.0700$$0.0700$0.07000.0700$0.0700$

• Green: 0 and 00 (2 numbers)
• Red: Even numbers (excluding 0 and 00)
• Black: Odd numbers

(i) Probability of getting 13 or Green

• Number of favorable outcomes for 13: 1 (just the number 13)
• Number of favorable outcomes for green: 2 (0 and 00)
• Total favorable outcomes: $1+2=3$$1+2=3$1+2=31 + 2 = 3$1+2=3$

(ii) Probability of getting a number greater than 17 and Black

• Total black numbers from 19 to 35: 9 (since every second number is odd)
• Total favorable outcomes: 9

(iii) Probability of getting Red or Black

• Total red and black numbers: $38-2$$38-2$38-238 – 2$38-2$ (total numbers minus green numbers)
• Total favorable outcomes: $38-2=36$$38-2=36$38-2=3638 – 2 = 36$38-2=36$

Sample Space

• Sample Space ($S$$S$SS$S$): This is the set of all possible outcomes of a random experiment. For example, in a coin toss, the sample space is $S=\{\text{heads},\text{tails}\}$$S=\{\text{heads},\text{tails}\}$S={“heads”,”tails”}S = \{\text{heads}, \text{tails}\}$S=\{\text{heads},\text{tails}\}$.

Events

• Event: An event is a subset of the sample space. It represents a specific outcome or a set of outcomes. For example, getting a heads in a coin toss is an event.

Probability Function

• Probability Function ($P$$P$PP$P$): This is a function that assigns a probability to each event in the sample space.

Axioms of Probability

1. Non-negativity: For any event $A$$A$AA$A$ in the sample space $S$$S$SS$S$, the probability of $A$$A$AA$A$ is non-negative. Mathematically, $P(A)\ge 0$$P(A)\ge 0$P(A) >= 0P(A) \geq 0$P(A)\ge 0$.
2. Probability of the Sample Space: The probability of the entire sample space is 1. Mathematically, $P(S)=1$$P(S)=1$P(S)=1P(S) = 1$P(S)=1$.
3. Additivity: For any two mutually exclusive events $A$$A$AA$A$ and $B$$B$BB$B$ (i.e., $A$$A$AA$A$ and $B$$B$BB$B$ cannot both occur at the same time), the probability of the union of $A$$A$AA$A$ and $B$$B$BB$B$ is the sum of their probabilities. Mathematically, if $A\cap B=\mathrm{\varnothing }$$A\cap B=\mathrm{\varnothing }$A nn B=O/A \cap B = \emptyset$A\cap B=\mathrm{\varnothing }$, then $P(A\cup B)=P(A)+P(B)$$P(A\cup B)=P(A)+P(B)$P(A uu B)=P(A)+P(B)P(A \cup B) = P(A) + P(B)$P(A\cup B)=P(A)+P(B)$.

Implications of the Axioms

• Complement Rule: The probability of the complement of an event $A$$A$AA$A$ (denoted as $A^{\prime }$$A^{\prime }$A^(‘)A’$A^{\prime }$) is $P(A^{\prime })=1-P(A)$$P(A^{\prime })=1-P(A)$P(A^(‘))=1-P(A)P(A’) = 1 – P(A)$P(A^{\prime })=1-P(A)$.
• Addition Rule for Non-Mutually Exclusive Events: If $A$$A$AA$A$ and $B$$B$BB$B$ are not mutually exclusive, then $P(A\cup B)=P(A)+P(B)-P(A\cap B)$$P(A\cup B)=P(A)+P(B)-P(A\cap B)$P(A uu B)=P(A)+P(B)-P(A nn B)P(A \cup B) = P(A) + P(B) – P(A \cap B)$P(A\cup B)=P(A)+P(B)-P(A\cap B)$.
• Conditional Probability and Independence: These concepts arise from the axioms and are fundamental to understanding more complex probability scenarios.

3. (a) A random variable $\mathrm{X}$$\mathrm{X}$X\mathrm{X}$\mathrm{X}$ has the following probability mass function :

(i) Determining the Value of $a$$a$aa$a$

(ii) Finding $P(X<3),P(X\ge 3)$$P(X<3),P(X\ge 3)$P(X < 3),P(X >= 3)P(X < 3), P(X \geq 3)$P(X<3),P(X\ge 3)$ and $P(0<X<5)$$P(0<X<5)$P(0 < X < 5)P(0 < X < 5)$P(0<X<5)$

• $P(X<3)$$P(X<3)$P(X < 3)P(X < 3)$P(X<3)$ is the probability that $X$$X$XX$X$ takes a value less than 3. This includes the values 0, 1, and 2. Using the PMF:
  $$P(X<3)=P(X=0)+P(X=1)+P(X=2)=a+3a+5a$$$$P(X<3)=P(X=0)+P(X=1)+P(X=2)=a+3a+5a$$P(X < 3)=P(X=0)+P(X=1)+P(X=2)=a+3a+5aP(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) = a + 3a + 5a$$P(X<3)=P(X=0)+P(X=1)+P(X=2)=a+3a+5a$$
  Substituting $a=\frac{1}{81}$$a=\frac{1}{81}$a=(1)/(81)a = \frac{1}{81}$a=\frac{1}{81}$ and calculating:
  $$P(X<3)=\frac{1}{81}+\frac{3}{81}+\frac{5}{81}=0.1111$$$$P(X<3)=\frac{1}{81}+\frac{3}{81}+\frac{5}{81}=0.1111$$P(X < 3)=(1)/(81)+(3)/(81)+(5)/(81)=0.1111P(X < 3) = \frac{1}{81} + \frac{3}{81} + \frac{5}{81}=0.1111$$P(X<3)=\frac{1}{81}+\frac{3}{81}+\frac{5}{81}=0.1111$$
• $P(X\ge 3)$$P(X\ge 3)$P(X >= 3)P(X \geq 3)$P(X\ge 3)$ is the complement of $P(X<3)$$P(X<3)$P(X < 3)P(X < 3)$P(X<3)$. Therefore, $P(X\ge 3)=1-P(X<3)$$P(X\ge 3)=1-P(X<3)$P(X >= 3)=1-P(X < 3)P(X \geq 3) = 1 – P(X < 3)$P(X\ge 3)=1-P(X<3)$.
• $P(0<X<5)$$P(0<X<5)$P(0 < X < 5)P(0 < X < 5)$P(0<X<5)$ includes the values 1, 2, 3, and 4. Using the PMF:
  $$P(0<X<5)=P(X=1)+P(X=2)+P(X=3)+P(X=4)=3a+5a+7a+9a$$$$P(0<X<5)=P(X=1)+P(X=2)+P(X=3)+P(X=4)=3a+5a+7a+9a$$P(0 < X < 5)=P(X=1)+P(X=2)+P(X=3)+P(X=4)=3a+5a+7a+9aP(0 < X < 5) = P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) = 3a + 5a + 7a + 9a$$P(0<X<5)=P(X=1)+P(X=2)+P(X=3)+P(X=4)=3a+5a+7a+9a$$
  Substituting $a=\frac{1}{81}$$a=\frac{1}{81}$a=(1)/(81)a = \frac{1}{81}$a=\frac{1}{81}$ and calculating:
  $$P(0<X<5)=\frac{3}{81}+\frac{5}{81}+\frac{7}{81}+\frac{9}{81}=0.2962$$$$P(0<X<5)=\frac{3}{81}+\frac{5}{81}+\frac{7}{81}+\frac{9}{81}=0.2962$$P(0 < X < 5)=(3)/(81)+(5)/(81)+(7)/(81)+(9)/(81)=0.2962P(0 < X < 5) = \frac{3}{81} + \frac{5}{81} + \frac{7}{81} + \frac{9}{81}=0.2962$$P(0<X<5)=\frac{3}{81}+\frac{5}{81}+\frac{7}{81}+\frac{9}{81}=0.2962$$

(iii) Smallest Value of $x$$x$xx$x$ for $P(X\le x)>0.5$$P(X\le x)>0.5$P(X <= x) > 0.5P(X \leq x) > 0.5$P(X\le x)>0.5$

• $P(X\le 0)=a=\frac{1}{81}$$P(X\le 0)=a=\frac{1}{81}$P(X <= 0)=a=(1)/(81)P(X \leq 0) = a = \frac{1}{81}$P(X\le 0)=a=\frac{1}{81}$
• $P(X\le 1)=a+3a=\frac{4}{81}$$P(X\le 1)=a+3a=\frac{4}{81}$P(X <= 1)=a+3a=(4)/(81)P(X \leq 1) = a + 3a = \frac{4}{81}$P(X\le 1)=a+3a=\frac{4}{81}$
• $P(X\le 2)=a+3a+5a=\frac{9}{81}$$P(X\le 2)=a+3a+5a=\frac{9}{81}$P(X <= 2)=a+3a+5a=(9)/(81)P(X \leq 2) = a + 3a + 5a = \frac{9}{81}$P(X\le 2)=a+3a+5a=\frac{9}{81}$
• $P(X\le 3)=a+3a+5a+7a=\frac{16}{81}$$P(X\le 3)=a+3a+5a+7a=\frac{16}{81}$P(X <= 3)=a+3a+5a+7a=(16)/(81)P(X \leq 3) = a + 3a + 5a + 7a = \frac{16}{81}$P(X\le 3)=a+3a+5a+7a=\frac{16}{81}$
• $P(X\le 4)=a+3a+5a+7a+9a=\frac{25}{81}$$P(X\le 4)=a+3a+5a+7a+9a=\frac{25}{81}$P(X <= 4)=a+3a+5a+7a+9a=(25)/(81)P(X \leq 4) = a + 3a + 5a + 7a + 9a = \frac{25}{81}$P(X\le 4)=a+3a+5a+7a+9a=\frac{25}{81}$
• $P(X\le 5)=a+3a+5a+7a+9a+11a=\frac{36}{81}$$P(X\le 5)=a+3a+5a+7a+9a+11a=\frac{36}{81}$P(X <= 5)=a+3a+5a+7a+9a+11 a=(36)/(81)P(X \leq 5) = a + 3a + 5a + 7a + 9a + 11a = \frac{36}{81}$P(X\le 5)=a+3a+5a+7a+9a+11a=\frac{36}{81}$
• $P(X\le 6)=a+3a+5a+7a+9a+11a+13a=\frac{49}{81}$$P(X\le 6)=a+3a+5a+7a+9a+11a+13a=\frac{49}{81}$P(X <= 6)=a+3a+5a+7a+9a+11 a+13 a=(49)/(81)P(X \leq 6) = a + 3a + 5a + 7a + 9a + 11a + 13a = \frac{49}{81}$P(X\le 6)=a+3a+5a+7a+9a+11a+13a=\frac{49}{81}$

• $P(X\le 0)=\frac{1}{81}\approx 0.0123$$P(X\le 0)=\frac{1}{81}\approx 0.0123$P(X <= 0)=(1)/(81)~~0.0123P(X \leq 0) = \frac{1}{81} \approx 0.0123$P(X\le 0)=\frac{1}{81}\approx 0.0123$
• $P(X\le 1)=\frac{4}{81}\approx 0.0494$$P(X\le 1)=\frac{4}{81}\approx 0.0494$P(X <= 1)=(4)/(81)~~0.0494P(X \leq 1) = \frac{4}{81} \approx 0.0494$P(X\le 1)=\frac{4}{81}\approx 0.0494$
• $P(X\le 2)=\frac{9}{81}=\frac{1}{9}\approx 0.1111$$P(X\le 2)=\frac{9}{81}=\frac{1}{9}\approx 0.1111$P(X <= 2)=(9)/(81)=(1)/(9)~~0.1111P(X \leq 2) = \frac{9}{81} = \frac{1}{9} \approx 0.1111$P(X\le 2)=\frac{9}{81}=\frac{1}{9}\approx 0.1111$
• $P(X\le 3)=\frac{16}{81}\approx 0.1975$$P(X\le 3)=\frac{16}{81}\approx 0.1975$P(X <= 3)=(16)/(81)~~0.1975P(X \leq 3) = \frac{16}{81} \approx 0.1975$P(X\le 3)=\frac{16}{81}\approx 0.1975$
• $P(X\le 4)=\frac{25}{81}\approx 0.3086$$P(X\le 4)=\frac{25}{81}\approx 0.3086$P(X <= 4)=(25)/(81)~~0.3086P(X \leq 4) = \frac{25}{81} \approx 0.3086$P(X\le 4)=\frac{25}{81}\approx 0.3086$
• $P(X\le 5)=\frac{36}{81}=\frac{4}{9}\approx 0.4444$$P(X\le 5)=\frac{36}{81}=\frac{4}{9}\approx 0.4444$P(X <= 5)=(36)/(81)=(4)/(9)~~0.4444P(X \leq 5) = \frac{36}{81} = \frac{4}{9} \approx 0.4444$P(X\le 5)=\frac{36}{81}=\frac{4}{9}\approx 0.4444$
• $P(X\le 6)=\frac{49}{81}\approx 0.6049$$P(X\le 6)=\frac{49}{81}\approx 0.6049$P(X <= 6)=(49)/(81)~~0.6049P(X \leq 6) = \frac{49}{81} \approx 0.6049$P(X\le 6)=\frac{49}{81}\approx 0.6049$

1. For $0\le x<2$$0\le x<2$0 <= x < 20 \leq x < 2$0\le x<2$:
   $$f(x)=F^{\prime }(x)={\left(\frac{x}{8}\right)}^{\prime }=\frac{1}{8}$$$$f(x)=F^{\prime }(x)={\left(\frac{x}{8}\right)}^{\prime }=\frac{1}{8}$$f(x)=F^(‘)(x)=((x)/(8))^(‘)=(1)/(8)f(x) = F'(x) = \left( \frac{x}{8} \right)’ = \frac{1}{8}$$f(x)=F^{\prime }(x)={\left(\frac{x}{8}\right)}^{\prime }=\frac{1}{8}$$
2. For $2\le x<4$$2\le x<4$2 <= x < 42 \leq x < 4$2\le x<4$:
   $$f(x)=F^{\prime }(x)={\left(\frac{x^2}{16}\right)}^{\prime }=\frac{x}{8}$$$$f(x)=F^{\prime }(x)={\left(\frac{x^2}{16}\right)}^{\prime }=\frac{x}{8}$$f(x)=F^(‘)(x)=((x^(2))/(16))^(‘)=(x)/(8)f(x) = F'(x) = \left( \frac{x^2}{16} \right)’ = \frac{x}{8}$$f(x)=F^{\prime }(x)={\left(\frac{x^2}{16}\right)}^{\prime }=\frac{x}{8}$$
3. For $x<0$$x<0$x < 0x < 0$x<0$ and $x\ge 4$$x\ge 4$x >= 4x \geq 4$x\ge 4$, the pdf $f(x)=0$$f(x)=0$f(x)=0f(x) = 0$f(x)=0$ since the CDF is constant in these intervals.

Expected Value $\mathrm{E}(X)$$\mathrm{E}(X)$E(X)\mathrm{E}(X)$\mathrm{E}(X)$

1. ${\int }_0^{2}x\cdot \frac{1}{8}dx$${\int }_0^2 x\cdot \frac{1}{8}dx$int_(0)^(2)x*(1)/(8)dx\int_{0}^{2} x \cdot \frac{1}{8} \, dx${\int }_0^{2}x\cdot \frac{1}{8}dx$
2. ${\int }_2^{4}x\cdot \frac{x}{8}dx$${\int }_2^4 x\cdot \frac{x}{8}dx$int_(2)^(4)x*(x)/(8)dx\int_{2}^{4} x \cdot \frac{x}{8} \, dx${\int }_2^{4}x\cdot \frac{x}{8}dx$

Variance $\text{Var}(X)$$\text{Var}(X)$“Var”(X)\text{Var}(X)$\text{Var}(X)$

1. For the first part of the expected value $\mathrm{E}(X)$$\mathrm{E}(X)$E(X)\mathrm{E}(X)$\mathrm{E}(X)$:
   $${\int }_0^{2}x\cdot \frac{1}{8}dx=\frac{1}{4}$$$${\int }_0^2 x\cdot \frac{1}{8}dx=\frac{1}{4}$$int_(0)^(2)x*(1)/(8)dx=(1)/(4)\int_{0}^{2} x \cdot \frac{1}{8} \, dx = \frac{1}{4}$${\int }_0^{2}x\cdot \frac{1}{8}dx=\frac{1}{4}$$
2. For the second part of $\mathrm{E}(X)$$\mathrm{E}(X)$E(X)\mathrm{E}(X)$\mathrm{E}(X)$:
   $${\int }_2^{4}x\cdot \frac{x}{8}dx=\frac{7}{3}$$$${\int }_2^4 x\cdot \frac{x}{8}dx=\frac{7}{3}$$int_(2)^(4)x*(x)/(8)dx=(7)/(3)\int_{2}^{4} x \cdot \frac{x}{8} \, dx = \frac{7}{3}$${\int }_2^{4}x\cdot \frac{x}{8}dx=\frac{7}{3}$$

• The expected value $\mathrm{E}(X)$$\mathrm{E}(X)$E(X)\mathrm{E}(X)$\mathrm{E}(X)$ of the random variable $X$$X$XX$X$ is $\frac{31}{12}$$\frac{31}{12}$(31)/(12)\frac{31}{12}$\frac{31}{12}$.
• The variance $\text{Var}(X)$$\text{Var}(X)$“Var”(X)\text{Var}(X)$\text{Var}(X)$ of $X$$X$XX$X$ is $\frac{167}{144}$$\frac{167}{144}$(167)/(144)\frac{167}{144}$\frac{167}{144}$.

Advantages of Random Variables

1. Quantitative Analysis of Random Phenomena: Random variables allow for the quantitative analysis of random phenomena. By assigning numerical values to outcomes, we can apply mathematical tools and statistical techniques to analyze and interpret these phenomena. This is crucial in fields like economics, engineering, physics, and biology, where understanding and predicting outcomes is essential.
2. Simplification and Modeling of Complex Systems: Random variables enable the simplification and modeling of complex systems. By abstracting real-world processes into random variables and their distributions, we can create models that are easier to analyze and understand. These models can be used to make predictions, test hypotheses, and inform decision-making processes in various fields, from finance to meteorology.

4. (a) A taxi cab company has 12 Ambassadors and 8 Fiats. If 5 of these taxi cabs are in the workshop for repairs and an Ambassador is as likely to be in for repairs as a Fiat, what is the probability that (i) 3 of them are Ambassadors and 2 are Fiats, (ii) at least 3 of them are Ambassadors, and (iii) all 5 are of the same make?

(i) Probability that 3 are Ambassadors and 2 are Fiats

(ii) Probability that at least 3 are Ambassadors

• Probability of exactly 3 Ambassadors: $\frac{(\genfrac{}{}{0ex}{}{12}{3})\times (\genfrac{}{}{0ex}{}{8}{2})}{(\genfrac{}{}{0ex}{}{20}{5})}$$\frac{(\genfrac{}{}{0ex}{}{12}{3})\times (\genfrac{}{}{0ex}{}{8}{2})}{(\genfrac{}{}{0ex}{}{20}{5})}$(((12)/(3))xx((8)/(2)))/(((20)/(5)))\frac{\binom{12}{3} \times \binom{8}{2}}{\binom{20}{5}}$\frac{(\genfrac{}{}{0ex}{}{12}{3})\times (\genfrac{}{}{0ex}{}{8}{2})}{(\genfrac{}{}{0ex}{}{20}{5})}$ (as calculated in part (i)).
• Probability of exactly 4 Ambassadors: $\frac{(\genfrac{}{}{0ex}{}{12}{4})\times (\genfrac{}{}{0ex}{}{8}{1})}{(\genfrac{}{}{0ex}{}{20}{5})}$$\frac{(\genfrac{}{}{0ex}{}{12}{4})\times (\genfrac{}{}{0ex}{}{8}{1})}{(\genfrac{}{}{0ex}{}{20}{5})}$(((12)/(4))xx((8)/(1)))/(((20)/(5)))\frac{\binom{12}{4} \times \binom{8}{1}}{\binom{20}{5}}$\frac{(\genfrac{}{}{0ex}{}{12}{4})\times (\genfrac{}{}{0ex}{}{8}{1})}{(\genfrac{}{}{0ex}{}{20}{5})}$.
• Probability of exactly 5 Ambassadors: $\frac{(\genfrac{}{}{0ex}{}{12}{5})\times (\genfrac{}{}{0ex}{}{8}{0})}{(\genfrac{}{}{0ex}{}{20}{5})}$$\frac{(\genfrac{}{}{0ex}{}{12}{5})\times (\genfrac{}{}{0ex}{}{8}{0})}{(\genfrac{}{}{0ex}{}{20}{5})}$(((12)/(5))xx((8)/(0)))/(((20)/(5)))\frac{\binom{12}{5} \times \binom{8}{0}}{\binom{20}{5}}$\frac{(\genfrac{}{}{0ex}{}{12}{5})\times (\genfrac{}{}{0ex}{}{8}{0})}{(\genfrac{}{}{0ex}{}{20}{5})}$.

(iii) Probability that all 5 are of the same make

• Probability of all 5 being Ambassadors: $\frac{(\genfrac{}{}{0ex}{}{12}{5})}{(\genfrac{}{}{0ex}{}{20}{5})}$$\frac{(\genfrac{}{}{0ex}{}{12}{5})}{(\genfrac{}{}{0ex}{}{20}{5})}$(((12)/(5)))/(((20)/(5)))\frac{\binom{12}{5}}{\binom{20}{5}}$\frac{(\genfrac{}{}{0ex}{}{12}{5})}{(\genfrac{}{}{0ex}{}{20}{5})}$.
• Probability of all 5 being Fiats: $\frac{(\genfrac{}{}{0ex}{}{8}{5})}{(\genfrac{}{}{0ex}{}{20}{5})}$$\frac{(\genfrac{}{}{0ex}{}{8}{5})}{(\genfrac{}{}{0ex}{}{20}{5})}$(((8)/(5)))/(((20)/(5)))\frac{\binom{8}{5}}{\binom{20}{5}}$\frac{(\genfrac{}{}{0ex}{}{8}{5})}{(\genfrac{}{}{0ex}{}{20}{5})}$.

1. Probability that 3 of them are Ambassadors and 2 are Fiats:
   $$P(\text{3 Ambassadors, 2 Fiats})=\frac{(\genfrac{}{}{0ex}{}{12}{3})\times (\genfrac{}{}{0ex}{}{8}{2})}{(\genfrac{}{}{0ex}{}{20}{5})}=\frac{385}{969}$$$$P(\text{3 Ambassadors, 2 Fiats})=\frac{(\genfrac{}{}{0ex}{}{12}{3})\times (\genfrac{}{}{0ex}{}{8}{2})}{(\genfrac{}{}{0ex}{}{20}{5})}=\frac{385}{969}$$P(“3 Ambassadors, 2 Fiats”)=(((12)/(3))xx((8)/(2)))/(((20)/(5)))=(385)/(969)P(\text{3 Ambassadors, 2 Fiats}) = \frac{\binom{12}{3} \times \binom{8}{2}}{\binom{20}{5}} = \frac{385}{969}$$P(\text{3 Ambassadors, 2 Fiats})=\frac{(\genfrac{}{}{0ex}{}{12}{3})\times (\genfrac{}{}{0ex}{}{8}{2})}{(\genfrac{}{}{0ex}{}{20}{5})}=\frac{385}{969}$$
2. Probability that at least 3 of them are Ambassadors:
   $$P(\text{At least 3 Ambassadors})=\frac{(\genfrac{}{}{0ex}{}{12}{3})\times (\genfrac{}{}{0ex}{}{8}{2})+(\genfrac{}{}{0ex}{}{12}{4})\times (\genfrac{}{}{0ex}{}{8}{1})+(\genfrac{}{}{0ex}{}{12}{5})\times (\genfrac{}{}{0ex}{}{8}{0})}{(\genfrac{}{}{0ex}{}{20}{5})}=\frac{682}{969}$$$$P(\text{At least 3 Ambassadors})=\frac{(\genfrac{}{}{0ex}{}{12}{3})\times (\genfrac{}{}{0ex}{}{8}{2})+(\genfrac{}{}{0ex}{}{12}{4})\times (\genfrac{}{}{0ex}{}{8}{1})+(\genfrac{}{}{0ex}{}{12}{5})\times (\genfrac{}{}{0ex}{}{8}{0})}{(\genfrac{}{}{0ex}{}{20}{5})}=\frac{682}{969}$$P(“At least 3 Ambassadors”)=(((12)/(3))xx((8)/(2))+((12)/(4))xx((8)/(1))+((12)/(5))xx((8)/(0)))/(((20)/(5)))=(682)/(969)P(\text{At least 3 Ambassadors}) = \frac{\binom{12}{3} \times \binom{8}{2} + \binom{12}{4} \times \binom{8}{1} + \binom{12}{5} \times \binom{8}{0}}{\binom{20}{5}} = \frac{682}{969}$$P(\text{At least 3 Ambassadors})=\frac{(\genfrac{}{}{0ex}{}{12}{3})\times (\genfrac{}{}{0ex}{}{8}{2})+(\genfrac{}{}{0ex}{}{12}{4})\times (\genfrac{}{}{0ex}{}{8}{1})+(\genfrac{}{}{0ex}{}{12}{5})\times (\genfrac{}{}{0ex}{}{8}{0})}{(\genfrac{}{}{0ex}{}{20}{5})}=\frac{682}{969}$$
3. Probability that all 5 are of the same make:
   $$P(\text{All 5 same make})=\frac{(\genfrac{}{}{0ex}{}{12}{5})+(\genfrac{}{}{0ex}{}{8}{5})}{(\genfrac{}{}{0ex}{}{20}{5})}=\frac{53}{969}$$$$P(\text{All 5 same make})=\frac{(\genfrac{}{}{0ex}{}{12}{5})+(\genfrac{}{}{0ex}{}{8}{5})}{(\genfrac{}{}{0ex}{}{20}{5})}=\frac{53}{969}$$P(“All 5 same make”)=(((12)/(5))+((8)/(5)))/(((20)/(5)))=(53)/(969)P(\text{All 5 same make}) = \frac{\binom{12}{5} + \binom{8}{5}}{\binom{20}{5}} = \frac{53}{969}$$P(\text{All 5 same make})=\frac{(\genfrac{}{}{0ex}{}{12}{5})+(\genfrac{}{}{0ex}{}{8}{5})}{(\genfrac{}{}{0ex}{}{20}{5})}=\frac{53}{969}$$

• The probability of getting exactly 3 even numbers in 5 throws is twice the probability of getting exactly 2 even numbers in 5 throws.

Understanding the Scenario

• Buses arrive at 15-minute intervals: 7:00, 7:15, 7:30, etc.
• A passenger arrives randomly between 7:00 and 7:30.

Probability Calculations

(i) Probability of Waiting Less Than 5 Minutes

• Between 7:10 and 7:15.
• Between 7:25 and 7:30.

(ii) Probability of Waiting At Least 12 Minutes

• Between 7:00 and 7:03.
• Between 7:15 and 7:18.

Final Probabilities

1. Probability of waiting less than 5 minutes:
   $$P(\text{wait}<5\text{min})=\frac{10}{30}=\frac{1}{3}$$$$P(\text{wait}<5\text{ min})=\frac{10}{30}=\frac{1}{3}$$P(“wait” < 5″ min”)=(10)/(30)=(1)/(3)P(\text{wait} < 5 \text{ min}) = \frac{10}{30} = \frac{1}{3}$$P(\text{wait}<5\text{ min})=\frac{10}{30}=\frac{1}{3}$$
2. Probability of waiting at least 12 minutes:
   $$P(\text{wait}\ge 12\text{min})=\frac{6}{30}=\frac{1}{5}$$$$P(\text{wait}\ge 12\text{ min})=\frac{6}{30}=\frac{1}{5}$$P(“wait” >= 12″ min”)=(6)/(30)=(1)/(5)P(\text{wait} \geq 12 \text{ min}) = \frac{6}{30} = \frac{1}{5}$$P(\text{wait}\ge 12\text{ min})=\frac{6}{30}=\frac{1}{5}$$

1. $₹100$$₹100$₹100₹100$₹100$ for $30<X<35$$30<X<35$30 < X < 3530 < X < 35$30<X<35$
2. $₹50$$₹50$₹50₹50$₹50$ for $25<X\le 30$$25<X\le 30$25 < X <= 3025 < X \leq 30$25<X\le 30$ or $35\le X<40$$35\le X<40$35 <= X < 4035 \leq X < 40$35\le X<40$
3. $-₹60$$-₹60$-₹60-₹60$-₹60$ (a loss) for $X\le 25$$X\le 25$X <= 25X \leq 25$X\le 25$ or $X\ge 40$$X\ge 40$X >= 40X \geq 40$X\ge 40$

Calculating Probabilities

1. Probability for $30<X<35$$30<X<35$30 < X < 3530 < X < 35$30<X<35$:
   $$P(30<X<35)=P(X<35)-P(X\le 30)$$$$P(30<X<35)=P(X<35)-P(X\le 30)$$P(30 < X < 35)=P(X < 35)-P(X <= 30)P(30 < X < 35) = P(X < 35) – P(X \leq 30)$$P(30<X<35)=P(X<35)-P(X\le 30)$$
2. Probability for $25<X\le 30$$25<X\le 30$25 < X <= 3025 < X \leq 30$25<X\le 30$ or $35\le X<40$$35\le X<40$35 <= X < 4035 \leq X < 40$35\le X<40$:
   $$P(25<X\le 30)=P(X\le 30)-P(X\le 25)$$$$P(25<X\le 30)=P(X\le 30)-P(X\le 25)$$P(25 < X <= 30)=P(X <= 30)-P(X <= 25)P(25 < X \leq 30) = P(X \leq 30) – P(X \leq 25)$$P(25<X\le 30)=P(X\le 30)-P(X\le 25)$$
   $$P(35\le X<40)=P(X<40)-P(X\le 35)$$$$P(35\le X<40)=P(X<40)-P(X\le 35)$$P(35 <= X < 40)=P(X < 40)-P(X <= 35)P(35 \leq X < 40) = P(X < 40) – P(X \leq 35)$$P(35\le X<40)=P(X<40)-P(X\le 35)$$
   The total probability for this range is the sum of these two probabilities.
3. Probability for $X\le 25$$X\le 25$X <= 25X \leq 25$X\le 25$ or $X\ge 40$$X\ge 40$X >= 40X \geq 40$X\ge 40$:
   $$P(X\le 25)+P(X\ge 40)=P(X\le 25)+(1-P(X<40))$$$$P(X\le 25)+P(X\ge 40)=P(X\le 25)+(1-P(X<40))$$P(X <= 25)+P(X >= 40)=P(X <= 25)+(1-P(X < 40))P(X \leq 25) + P(X \geq 40) = P(X \leq 25) + (1 – P(X < 40))$$P(X\le 25)+P(X\ge 40)=P(X\le 25)+(1-P(X<40))$$

1. Probability for $30<X<35$$30<X<35$30 < X < 3530 < X < 35$30<X<35$: $P(30<X<35)\approx 0.588852$$P(30<X<35)\approx 0.588852$P(30 < X < 35)~~0.588852P(30 < X < 35) \approx 0.588852$P(30<X<35)\approx 0.588852$
2. Probability for $25<X\le 30$$25<X\le 30$25 < X <= 3025 < X \leq 30$25<X\le 30$: $P(25<X\le 30)\approx 0.154825$$P(25<X\le 30)\approx 0.154825$P(25 < X <= 30)~~0.154825P(25 < X \leq 30) \approx 0.154825$P(25<X\le 30)\approx 0.154825$
3. Probability for $35\le X<40$$35\le X<40$35 <= X < 4035 \leq X < 40$35\le X<40$: $P(35\le X<40)\approx 0.242677$$P(35\le X<40)\approx 0.242677$P(35 <= X < 40)~~0.242677P(35 \leq X < 40) \approx 0.242677$P(35\le X<40)\approx 0.242677$
4. Probability for $X\le 25$$X\le 25$X <= 25X \leq 25$X\le 25$ or $X\ge 40$$X\ge 40$X >= 40X \geq 40$X\ge 40$: $P(X\le 25)+P(X\ge 40)\approx 0.00981533+0$$P(X\le 25)+P(X\ge 40)\approx 0.00981533+0$P(X <= 25)+P(X >= 40)~~0.00981533+0P(X \leq 25) + P(X \geq 40) \approx 0.00981533 + 0$P(X\le 25)+P(X\ge 40)\approx 0.00981533+0$ (since $P(X\le 25)$$P(X\le 25)$P(X <= 25)P(X \leq 25)$P(X\le 25)$ is negligible)

1. Traffic Flow Analysis

2. Call Center Operations

3. Defects in Manufacturing

4. Radioactive Decay

6. (a) A can hit a target in 4 out of 5 shots and B can hit the target in 3 out of 4 shots. Find the probability that (i) the target being hit when both try (ii) the target being hit by exactly one person.

• A can hit the target in 4 out of 5 shots, so $P(A)=\frac{4}{5}$$P(A)=\frac{4}{5}$P(A)=(4)/(5)P(A) = \frac{4}{5}$P(A)=\frac{4}{5}$.
• B can hit the target in 3 out of 4 shots, so $P(B)=\frac{3}{4}$$P(B)=\frac{3}{4}$P(B)=(3)/(4)P(B) = \frac{3}{4}$P(B)=\frac{3}{4}$.

(i) Probability of the Target Being Hit When Both Try

(ii) Probability of the Target Being Hit by Exactly One Person

1. Probability of the target being hit when both try:
   $$P(\text{Target hit})=\frac{4}{5}+\frac{3}{4}-(\frac{4}{5}\times \frac{3}{4})=\frac{19}{20}$$$$P(\text{Target hit})=\frac{4}{5}+\frac{3}{4}-\left(\frac{4}{5}\times \frac{3}{4}\right)=\frac{19}{20}$$P(“Target hit”)=(4)/(5)+(3)/(4)-((4)/(5)xx(3)/(4))=(19)/(20)P(\text{Target hit}) = \frac{4}{5} + \frac{3}{4} – \left( \frac{4}{5} \times \frac{3}{4} \right) = \frac{19}{20}$$P(\text{Target hit})=\frac{4}{5}+\frac{3}{4}-(\frac{4}{5}\times \frac{3}{4})=\frac{19}{20}$$
2. Probability of the target being hit by exactly one person:
   $$P(\text{Exactly one hits})=\frac{4}{5}\times (1-\frac{3}{4})+\frac{3}{4}\times (1-\frac{4}{5})=\frac{7}{20}$$$$P(\text{Exactly one hits})=\frac{4}{5}\times \left(1-\frac{3}{4}\right)+\frac{3}{4}\times \left(1-\frac{4}{5}\right)=\frac{7}{20}$$P(“Exactly one hits”)=(4)/(5)xx(1-(3)/(4))+(3)/(4)xx(1-(4)/(5))=(7)/(20)P(\text{Exactly one hits}) = \frac{4}{5} \times \left(1 – \frac{3}{4}\right) + \frac{3}{4} \times \left(1 – \frac{4}{5}\right) = \frac{7}{20}$$P(\text{Exactly one hits})=\frac{4}{5}\times (1-\frac{3}{4})+\frac{3}{4}\times (1-\frac{4}{5})=\frac{7}{20}$$

Finding the pdf $f(x)$$f(x)$f(x)f(x)$f(x)$

1. For $0\le x\le \frac{1}{2}$$0\le x\le \frac{1}{2}$0 <= x <= (1)/(2)0 \leq x \leq \frac{1}{2}$0\le x\le \frac{1}{2}$:
   $$F(x)=x^2⟹f(x)=\frac{d}{dx}(x^2)=2x$$$$F(x)=x^2⟹f(x)=\frac{d}{dx}(x^2)=2x$$F(x)=x^(2)Longrightarrowf(x)=(d)/(dx)(x^(2))=2xF(x) = x^2 \implies f(x) = \frac{d}{dx}(x^2) = 2x$$F(x)=x^2⟹f(x)=\frac{d}{dx}(x^2)=2x$$
2. For $\frac{1}{2}\le x<3$$\frac{1}{2}\le x<3$(1)/(2) <= x < 3\frac{1}{2} \leq x < 3$\frac{1}{2}\le x<3$:
   $$F(x)=1-\frac{3(3-x{)}^2}{25}⟹f(x)=\frac{d}{dx}(1-\frac{3(3-x{)}^2}{25})$$$$F(x)=1-\frac{3(3-x{)}^2}{25}⟹f(x)=\frac{d}{dx}\left(1-\frac{3(3-x{)}^2}{25}\right)$$F(x)=1-(3(3-x)^(2))/(25)Longrightarrowf(x)=(d)/(dx)(1-(3(3-x)^(2))/(25))F(x) = 1 – \frac{3(3 – x)^2}{25} \implies f(x) = \frac{d}{dx}\left(1 – \frac{3(3 – x)^2}{25}\right)$$F(x)=1-\frac{3(3-x{)}^2}{25}⟹f(x)=\frac{d}{dx}(1-\frac{3(3-x{)}^2}{25})$$
   We will calculate this derivative.
3. For $x<0$$x<0$x < 0x < 0$x<0$ and $x\ge 3$$x\ge 3$x >= 3x \geq 3$x\ge 3$:
   The pdf is 0 since the CDF is constant in these ranges.

Evaluating $P(|X|\le 1)$$P(|X|\le 1)$P(|X| <= 1)P(|X| \leq 1)$P(|X|\le 1)$ and $P(\frac{1}{3}\le X<4)$$P\left(\frac{1}{3}\le X<4\right)$P((1)/(3) <= X < 4)P\left(\frac{1}{3} \leq X < 4\right)$P(\frac{1}{3}\le X<4)$

1. $P(|X|\le 1)$$P(|X|\le 1)$P(|X| <= 1)P(|X| \leq 1)$P(|X|\le 1)$:
   Since $X$$X$XX$X$ is a continuous variable and $F(x)$$F(x)$F(x)F(x)$F(x)$ is given for $X\ge 0$$X\ge 0$X >= 0X \geq 0$X\ge 0$, $P(|X|\le 1)=P(0\le X\le 1)$$P(|X|\le 1)=P(0\le X\le 1)$P(|X| <= 1)=P(0 <= X <= 1)P(|X| \leq 1) = P(0 \leq X \leq 1)$P(|X|\le 1)=P(0\le X\le 1)$.
   $$P(0\le X\le 1)=F(1)-F(0)$$$$P(0\le X\le 1)=F(1)-F(0)$$P(0 <= X <= 1)=F(1)-F(0)P(0 \leq X \leq 1) = F(1) – F(0)$$P(0\le X\le 1)=F(1)-F(0)$$
2. $P(\frac{1}{3}\le X<4)$$P\left(\frac{1}{3}\le X<4\right)$P((1)/(3) <= X < 4)P\left(\frac{1}{3} \leq X < 4\right)$P(\frac{1}{3}\le X<4)$:
   Since $F(x)=1$$F(x)=1$F(x)=1F(x) = 1$F(x)=1$ for $x\ge 3$$x\ge 3$x >= 3x \geq 3$x\ge 3$, $P(\frac{1}{3}\le X<4)=P(\frac{1}{3}\le X<3)$$P\left(\frac{1}{3}\le X<4\right)=P\left(\frac{1}{3}\le X<3\right)$P((1)/(3) <= X < 4)=P((1)/(3) <= X < 3)P\left(\frac{1}{3} \leq X < 4\right) = P\left(\frac{1}{3} \leq X < 3\right)$P(\frac{1}{3}\le X<4)=P(\frac{1}{3}\le X<3)$.
   $$P(\frac{1}{3}\le X<3)=F(3)-F\left(\frac{1}{3}\right)$$$$P\left(\frac{1}{3}\le X<3\right)=F(3)-F\left(\frac{1}{3}\right)$$P((1)/(3) <= X < 3)=F(3)-F((1)/(3))P\left(\frac{1}{3} \leq X < 3\right) = F(3) – F\left(\frac{1}{3}\right)$$P(\frac{1}{3}\le X<3)=F(3)-F\left(\frac{1}{3}\right)$$

1. For $0\le x\le \frac{1}{2}$$0\le x\le \frac{1}{2}$0 <= x <= (1)/(2)0 \leq x \leq \frac{1}{2}$0\le x\le \frac{1}{2}$:
   $$f(x)=2x$$$$f(x)=2x$$f(x)=2xf(x) = 2x$$f(x)=2x$$
2. For $\frac{1}{2}\le x<3$$\frac{1}{2}\le x<3$(1)/(2) <= x < 3\frac{1}{2} \leq x < 3$\frac{1}{2}\le x<3$:
   $$f(x)=\frac{6(3-x)}{25}$$$$f(x)=\frac{6(3-x)}{25}$$f(x)=(6(3-x))/(25)f(x) = \frac{6(3 – x)}{25}$$f(x)=\frac{6(3-x)}{25}$$
3. For $x<0$$x<0$x < 0x < 0$x<0$ and $x\ge 3$$x\ge 3$x >= 3x \geq 3$x\ge 3$:
   $$f(x)=0$$$$f(x)=0$$f(x)=0f(x) = 0$$f(x)=0$$

Evaluating $P(|X|\le 1)$$P(|X|\le 1)$P(|X| <= 1)P(|X| \leq 1)$P(|X|\le 1)$

• $F(0)=0$$F(0)=0$F(0)=0F(0) = 0$F(0)=0$
• For $1$$1$11$1$, since $1>\frac{1}{2}$$1>\frac{1}{2}$1 > (1)/(2)1 > \frac{1}{2}$1>\frac{1}{2}$ and $1<3$$1<3$1 < 31 < 3$1<3$, $F(1)=1-\frac{3(3-1{)}^2}{25}$$F(1)=1-\frac{3(3-1{)}^2}{25}$F(1)=1-(3(3-1)^(2))/(25)F(1) = 1 – \frac{3(3 – 1)^2}{25}$F(1)=1-\frac{3(3-1{)}^2}{25}$

Evaluating $P(\frac{1}{3}\le X<4)$$P\left(\frac{1}{3}\le X<4\right)$P((1)/(3) <= X < 4)P\left(\frac{1}{3} \leq X < 4\right)$P(\frac{1}{3}\le X<4)$

• $F(3)=1$$F(3)=1$F(3)=1F(3) = 1$F(3)=1$
• For $\frac{1}{3}$$\frac{1}{3}$(1)/(3)\frac{1}{3}$\frac{1}{3}$, since $\frac{1}{3}<\frac{1}{2}$$\frac{1}{3}<\frac{1}{2}$(1)/(3) < (1)/(2)\frac{1}{3} < \frac{1}{2}$\frac{1}{3}<\frac{1}{2}$, $F\left(\frac{1}{3}\right)={\left(\frac{1}{3}\right)}^2$$F\left(\frac{1}{3}\right)={\left(\frac{1}{3}\right)}^2$F((1)/(3))=((1)/(3))^(2)F\left(\frac{1}{3}\right) = \left(\frac{1}{3}\right)^2$F\left(\frac{1}{3}\right)={\left(\frac{1}{3}\right)}^2$

1. For $P(|X|\le 1)$$P(|X|\le 1)$P(|X| <= 1)P(|X| \leq 1)$P(|X|\le 1)$:
   $$P(0\le X\le 1)=F(1)-F(0)=\frac{13}{25}-0=\frac{13}{25}$$$$P(0\le X\le 1)=F(1)-F(0)=\frac{13}{25}-0=\frac{13}{25}$$P(0 <= X <= 1)=F(1)-F(0)=(13)/(25)-0=(13)/(25)P(0 \leq X \leq 1) = F(1) – F(0) = \frac{13}{25} – 0 = \frac{13}{25}$$P(0\le X\le 1)=F(1)-F(0)=\frac{13}{25}-0=\frac{13}{25}$$
2. For $P(\frac{1}{3}\le X<4)$$P\left(\frac{1}{3}\le X<4\right)$P((1)/(3) <= X < 4)P\left(\frac{1}{3} \leq X < 4\right)$P(\frac{1}{3}\le X<4)$:
   $$P(\frac{1}{3}\le X<3)=F(3)-F\left(\frac{1}{3}\right)=1-\frac{1}{9}=\frac{8}{9}$$$$P\left(\frac{1}{3}\le X<3\right)=F(3)-F\left(\frac{1}{3}\right)=1-\frac{1}{9}=\frac{8}{9}$$P((1)/(3) <= X < 3)=F(3)-F((1)/(3))=1-(1)/(9)=(8)/(9)P\left(\frac{1}{3} \leq X < 3\right) = F(3) – F\left(\frac{1}{3}\right) = 1 – \frac{1}{9} = \frac{8}{9}$$P(\frac{1}{3}\le X<3)=F(3)-F\left(\frac{1}{3}\right)=1-\frac{1}{9}=\frac{8}{9}$$

1. Lifespan of Electronic Components

2. Queueing Theory in Operations Research

3. Radioactive Decay in Physics

7. (a) A manufacturer, who produces bolts, find that $0.1\mathrm{\%}$$0.1\mathrm{\%}$0.1%0.1 \%$0.1\mathrm{\%}$ of the bolts are defective. The bolts are packed in boxes containing 500 bolts. A service centre buys 100 boxes from the producer. Find the expected number of boxes which will contain at least two defective bolts.

• The probability of a bolt being defective is $0.1\mathrm{\%}$$0.1\mathrm{\%}$0.1%0.1\%$0.1\mathrm{\%}$ or $0.001$$0.001$0.0010.001$0.001$.
• Each box contains 500 bolts.

1. The probability of a box having at least two defective bolts is approximately $0.0902$$0.0902$0.09020.0902$0.0902$.
2. Therefore, the expected number of boxes out of 100 that will contain at least two defective bolts is approximately $9.02$$9.02$9.029.02$9.02$.

(i) Probability that the Repair Time Exceeds 2 Hours

(ii) Conditional Probability that a Repair Takes at Least 10 Hours Given that its Duration Exceeds 9 Hours

1. Probability that the repair time exceeds 2 hours:
   $$P(X>2)=e^{-1}\approx 0.3679$$$$P(X>2)=e^{-1}\approx 0.3679$$P(X > 2)=e^(-1)~~0.3679P(X > 2) = e^{-1} \approx 0.3679$$P(X>2)=e^{-1}\approx 0.3679$$
   This means there is approximately a 36.79% chance that the repair time will exceed 2 hours.
2. Conditional probability that a repair takes at least 10 hours given that its duration exceeds 9 hours:
   $$P(X>10|X>9)=e^{-\frac{1}{2}}\approx 0.6065$$$$P(X>10|X>9)=e^{-\frac{1}{2}}\approx 0.6065$$P(X > 10|X > 9)=e^(-(1)/(2))~~0.6065P(X > 10 \,|\, X > 9) = e^{-\frac{1}{2}} \approx 0.6065$$P(X>10|X>9)=e^{-\frac{1}{2}}\approx 0.6065$$
   This means, given that the repair has already taken more than 9 hours, there is approximately a 60.65% chance that it will take at least 10 hours in total.

1. Bayesian Inference and Learning

2. Project Management and PERT