IGNOU MMT-008 Solved Assignment | 1st January, 2025 to 31st December, 2025 | PROBABILITY AND STATISTICS | MSc MACS

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IGNOU MMT-008 Solved Assignment | 1st January, 2025 to 31st December, 2025 | PROBABILITY AND STATISTICS | MSc MACS Sample Solution

MMT-008 Solved Assignment 2025

  1. a) A study about a population showed that the mobility of population of a state to a village, town and city is in the following percentages:
From To
Village Town City
Village 60 25 15
Town 10 70 20
City 10 30 60
From To Village Town City Village 60 25 15 Town 10 70 20 City 10 30 60| From | To | | | | :—: | :—: | :—: | :—: | | | Village | Town | City | | Village | 60 | 25 | 15 | | Town | 10 | 70 | 20 | | City | 10 | 30 | 60 |
What will be the proportion of population village, town and city after one year and two years, given that the present proportion of the population in the village, town and city are respectively 0.50 , 0.40 0.50 , 0.40 0.50,0.400.50,0.400.50,0.40 and 0.10 ?
b) In a certain state, 58 landfills are classified according to their concentration of three hazardous chemicals: arsenic, barium and mercury. Suppose that the concentration of each one of the three chemicals is characterized as either high or low. If a landfill is chosen at random from among 58 landfills, given the following configuration:
Barium
High Mercury Low Mercury
Arsenic High Low High Low
High 1 3 5 9
Low 4 8 10 18
Barium High Mercury Low Mercury Arsenic High Low High Low High 1 3 5 9 Low 4 8 10 18| | Barium | | | | | :—: | :—: | :—: | :—: | :—: | | | High Mercury | | Low Mercury | | | Arsenic | High | Low | High | Low | | High | 1 | 3 | 5 | 9 | | Low | 4 | 8 | 10 | 18 |
Find the probability that it has:
i) high concentration of Barium,
ii) high concentration of mercury and low concentration of both Arsenic and Barium and
iii) high concentratin of any one of the chemicals and low concentration of the other two.
  1. a) Consider the following system considering of two switches I and II between two points A A AAA and B B BBB :
original image
A signal is sent from the point A A AAA to point B B BBB and is received at B B BBB if both the switches I I III and II are closed. It is assumed that the probabilities of I and II being closed are 0.8 and 0.6 respectively and that P [ I I P [ I I P[II\mathrm{P}[I IP[II is closed ∣ I ∣ I ∣I\mid \mathrm{I}∣I is closed ] = P [ I I ] = P [ I I ]=P[II]=\mathrm{P}[I I]=P[II is closed ] ] ]]].
Find:
i) The probability that signal is received at B B BBB.
ii) The conditional probability that switch I was open, given that the signal was not received at B B BBB,
iii) The conditional probability that switch II was open, given that the signal was not received at B.
b) For the (M/M/K) : ( ∞ / ∞ / oo//\infty /∞/ FIF 0 ) queuing model with arrival rate λ λ lambda\lambdaλ and service rate per service channel μ μ mu\muμ, obtain the steady-state probability of n n nnn customers in the system, P n P n P_(n)\mathrm{P}_{\mathrm{n}}Pn. Hence or otherwise, find the probability that an arrival has to wait.
3. a) The certain item is manufactured by three factories, say 1,2 and 3 . It is known that 1 turns out twice as many items as 2 and that 2 and 3 turn out the same number of items (during a specified production period). It is also known that 2 percent of the items produced by 1 and 2 are defective while 4 percent of those manufactured by 3 are defective. All the items produced are put into one stock pile and then one item is chosen at random. The chosen item was found defective. What is the probability that it was produced in factory 1 ?
b) The joint distribution of the random variables X and Y is given by:
x x x\mathbf{x}x -1 0 1
-1 α α alpha\alphaα β β beta\betaβ α α alpha\alphaα
0 β β beta\betaβ 0 β β beta\betaβ
1 α α alpha\alphaα β β beta\betaβ α α alpha\alphaα
x -1 0 1 -1 alpha beta alpha 0 beta 0 beta 1 alpha beta alpha| $\mathbf{x}$ | -1 | 0 | 1 | | :—: | :—: | :—: | :—: | | -1 | $\alpha$ | $\beta$ | $\alpha$ | | 0 | $\beta$ | 0 | $\beta$ | | 1 | $\alpha$ | $\beta$ | $\alpha$ |
where α , β > 0 α , β > 0 alpha,beta > 0\alpha, \beta>0α,β>0 with α + β = 1 / 4 α + β = 1 / 4 alpha+beta=1//4\alpha+\beta=1 / 4α+β=1/4.
i) Derive the marginal distribution of X and Y .
ii) Calculate the E ( X ) , E ( Y ) E ( X ) , E ( Y ) E(X),E(Y)\mathrm{E}(\mathrm{X}), \mathrm{E}(\mathrm{Y})E(X),E(Y) and E ( XY ) E ( XY ) E(XY)\mathrm{E}(\mathrm{XY})E(XY).
iii) Show that Cov ( X , Y ) = 0 Cov ( X , Y ) = 0 Cov(X,Y)=0\operatorname{Cov}(\mathrm{X}, \mathrm{Y})=0Cov(X,Y)=0.
iv) Show that the variables X and Y are dependent.
4. a) Let { X n ; n ≥ 0 } X n ; n ≥ 0 {X_(n);n >= 0}\left\{\mathrm{X}_{\mathrm{n}} ; \mathrm{n} \geq 0\right\}{Xn;n≥0} be a Markov chain with four states; 1, 2, 3, 4 and the following transition probability matrix:
1 2 3 4
1 0 0 1 0
2 1 0 0 0
1 2 3 4 1 0 0 1 0 2 1 0 0 0| | 1 | 2 | 3 | 4 | | :— | :— | :— | :— | :— | | 1 | 0 | 0 | 1 | 0 | | 2 | 1 | 0 | 0 | 0 |
3 1 / 2 1 / 2 1//21 / 21/2 1 / 2 1 / 2 1//21 / 21/2 0 0
4 1 / 3 1 / 3 1//31 / 31/3 1 / 3 1 / 3 1//31 / 31/3 1 / 3 1 / 3 1//31 / 31/3 0
3 1//2 1//2 0 0 4 1//3 1//3 1//3 0| 3 | $1 / 2$ | $1 / 2$ | 0 | 0 | | :— | :— | :— | :— | :— | | 4 | $1 / 3$ | $1 / 3$ | $1 / 3$ | 0 |
i) Find the probability P [ X 3 = 3 , X 2 = 1 ∣ X 1 = 2 ] P X 3 = 3 , X 2 = 1 ∣ X 1 = 2 P[X_(3)=3,X_(2)=1∣X_(1)=2]\mathrm{P}\left[\mathrm{X}_3=3, \mathrm{X}_2=1 \mid \mathrm{X}_1=2\right]P[X3=3,X2=1∣X1=2].
ii) Classify the states of the given Markov chain.
b) It is claimed that the function F X , Y = 1 16 xy ( x + y ) , 0 ≤ x ≤ 2 , 0 ≤ y ≤ 2 F X , Y = 1 16 xy ( x + y ) , 0 ≤ x ≤ 2 , 0 ≤ y ≤ 2 F_(X,Y)=(1)/(16)xy(x+y),0 <= x <= 2,0 <= y <= 2\mathrm{F}_{\mathrm{X}, \mathrm{Y}}=\frac{1}{16} \mathrm{xy}(\mathrm{x}+\mathrm{y}), 0 \leq \mathrm{x} \leq 2,0 \leq \mathrm{y} \leq 2FX,Y=116xy(x+y),0≤x≤2,0≤y≤2 is the joint distribution function of the random variables X X XXX and Y Y YYY. Then:
i) determine the corresponding joint probability density function f X , Y f X , Y f_(X,Y)f_{X, Y}fX,Y and
ii) calculate the probability P ( 0 ≤ X ≤ 1 , 1 ≤ Y ≤ 2 ) P ( 0 ≤ X ≤ 1 , 1 ≤ Y ≤ 2 ) P(0 <= X <= 1,1 <= Y <= 2)\mathrm{P}(0 \leq \mathrm{X} \leq 1,1 \leq \mathrm{Y} \leq 2)P(0≤X≤1,1≤Y≤2).
5. a) In an investigation related to a specific type of scores of men and women aged 65 to 70 , the mean verbal and performance scores for 101 subjects were found to be:
X ― = [ 55.24 34.97 ] X ¯ = 55.24 34.97 bar(X)=[[55.24],[34.97]]\overline{\mathrm{X}}=\left[\begin{array}{l} 55.24 \\ 34.97 \end{array}\right]X―=[55.2434.97]
The sample covariance matrix of the scores was
S = [ 210.54 126.99 126.99 119.68 ] S = 210.54 126.99 126.99 119.68 S=[[210.54,126.99],[126.99,119.68]]S=\left[\begin{array}{ll} 210.54 & 126.99 \\ 126.99 & 119.68 \end{array}\right]S=[210.54126.99126.99119.68]
In order to test the null hypothesis that observations came from a population with mean vector μ 0 = [ 60 50 ] μ 0 = 60 50 mu_(0)=[[60],[50]]\mu_0=\left[\begin{array}{l}60 \\ 50\end{array}\right]μ0=[6050], apply a suitable test statistic. You may consider α = 0.01 α = 0.01 alpha=0.01\alpha=0.01α=0.01 for the test.
b) What is the purpose of principal component analysis? Given the covariance matrix of order 2 × 2 2 × 2 2xx22 \times 22×2, explain how would you extract the principal components. Also, explain how would you find the proportion of total population variance for all the principal components.
6. a) Distinguish between ‘Age Replacement’ and ‘Block Replacement’ policies.
Let the lifetimes Y 1 , Y 2 , … Y 1 , Y 2 , … Y_(1),Y_(2),dotsY_1, Y_2, \ldotsY1,Y2,…. , are independently and identically distributed random variables and follows negative exponential distribution with parameter 5. If lifetimes T > 0 T > 0 T > 0\mathrm{T}>0T>0 and age replacement policy is to be employed, then:
i) find the mean renewal time and
ii) find the long-run average cost per unit time, given the costs C 1 = 4 C 1 = 4 C_(1)=4\mathrm{C}_1=4C1=4 and C 2 = 6 C 2 = 6 C_(2)=6\mathrm{C}_2=6C2=6 units of money.
b) In order to fit the regression line y = β 1 + β 2 x y = β 1 + β 2 x y=beta_(1)+beta_(2)xy=\beta_1+\beta_2 xy=β1+β2x on a data set consisting of 34 pairs of values ( z , y ) ( z , y ) (z,y)(\mathrm{z}, \mathrm{y})(z,y), the least square estimates of β 1 β 1 beta_(1)\beta_1β1 and β 2 β 2 beta_(2)\beta_2β2 are to be computed. From the data set, the following values are obtained:
∑ i x i = 100.73 , ∑ i y i = 16 , 703 ∑ i x i 2 = 304.7885 , ∑ i x i y i = 50 , 006.47 ∑ i   x i = 100.73 , ∑ i   y i = 16 , 703 ∑ i   x i 2 = 304.7885 , ∑ i   x i y i = 50 , 006.47 {:[sum _(i)x_(i)=100.73″,”sum _(i)y_(i)=16″,”703],[sum _(i)x_(i)^(2)=304.7885″,”sum _(i)x_(i)y_(i)=50″,”006.47]:}\begin{aligned} & \sum_i x_i=100.73, \sum_i y_i=16,703 \\ & \sum_i x_i^2=304.7885, \sum_i x_i y_i=50,006.47 \end{aligned}∑ixi=100.73,∑iyi=16,703∑ixi2=304.7885,∑ixiyi=50,006.47
Obtain the fitted regression line.
7. a) What is branching process? Give two real examples of branching process.
If P ( s ) P ( s ) P(s)P(s)P(s) and P n ( s ) P n ( s ) P_(n)(s)P_n(s)Pn(s) respectively is the probability generating function (pgf) of the i.i.d. random variables { ξ r } ξ r {xi _(r)}\left\{\xi_r\right\}{ξr} and the random variables { X n } X n {X_(n)}\left\{X_n\right\}{Xn}, where X n + 1 = ∑ r = 1 X r ξ r X n + 1 = ∑ r = 1 X r   ξ r X_(n+1)=sum_(r=1)^(X_(r))xi _(r)X_{n+1}=\sum_{r=1}^{X_r} \xi_rXn+1=∑r=1Xrξr.
Then show that:
P n ( s ) = P n − 1 ( P ( s ) ) and P n ( s ) = P ( P n − 1 ( s ) ) . P n ( s ) = P n − 1 ( P ( s ) ) and P n ( s ) = P P n − 1 ( s ) . {:[,P_(n)(s)=P_(n-1)(P(s))],[” and “quad,P_(n)(s)=P(P_(n-1)(s)).]:}\begin{array}{ll} & P_n(s)=P_{n-1}(P(s)) \\ \text { and } \quad & P_n(s)=P\left(P_{n-1}(s)\right) . \end{array}Pn(s)=Pn−1(P(s)) and Pn(s)=P(Pn−1(s)).
b) A community has two police cars, which operate independently of one another. The probability that a specific car will be available when needed is 0.99 .
i) What is the probability that neither car is available when needed?
ii) What is the probability that a car is available when needed?
  1. State whether the following statements are true or false. Justify your answer with a short proof or a counter example:
    i) Although in one-step transition probability matrix, P , of a Markov chain the sum of each row must necessarily be unity but in higher order transition probability matrices, P ( j ) ; j = 2 , 3 , 4 , … … P ( j ) ; j = 2 , 3 , 4 , … … P^((j));j=2,3,4,dots dotsP^{(j)} ; j=2,3,4, \ldots \ldotsP(j);j=2,3,4,……. This rule is not necessary.
    ii) For the joint pdf of random variables ( X , Y ) ( X , Y ) (X,Y)(\mathrm{X}, \mathrm{Y})(X,Y) given by:
f ( x , y ) = x 2 + xy 3 , 0 ≤ x ≤ 1 , 0 ≤ y ≤ 2 P [ X + Y ≥ ] = 65 72 f ( x , y ) = x 2 + xy 3 , 0 ≤ x ≤ 1 , 0 ≤ y ≤ 2 P [ X + Y ≥ ] = 65 72 {:[f(x”,”y)=x^(2)+(xy)/(3)”,”0 <= x <= 1″,”0 <= y <= 2],[P[X+Y >= ]=(65)/(72)]:}\begin{aligned} & \mathrm{f}(\mathrm{x}, \mathrm{y})=\mathrm{x}^2+\frac{\mathrm{xy}}{3}, 0 \leq \mathrm{x} \leq 1,0 \leq \mathrm{y} \leq 2 \\ & \mathrm{P}[\mathrm{X}+\mathrm{Y} \geq]=\frac{65}{72} \end{aligned}f(x,y)=x2+xy3,0≤x≤1,0≤y≤2P[X+Y≥]=6572
iii) Let the r.v. X follows the negative exponential distribution with parameter λ = 1 3 λ = 1 3 lambda=(1)/(3)\lambda=\frac{1}{3}λ=13. Then P [ X > 4 ∣ X > 3 ] = P [ X > 1 ] = e − 1 3 P [ X > 4 ∣ X > 3 ] = P [ X > 1 ] = e − 1 3 P[X > 4∣X > 3]=P[X > 1]=e^(-(1)/(3))\mathrm{P}[\mathrm{X}>4 \mid \mathrm{X}>3]=\mathrm{P}[\mathrm{X}>1]=\mathrm{e}^{-\frac{1}{3}}P[X>4∣X>3]=P[X>1]=e−13.
iv) One of the examples of non-Poisson queuing system is the (M/G/1):( ∞ ∞ oo\infty∞ (FlF0) queuing model.
v) If X 1 , X 2 , … , X n X 1 , X 2 , … , X n X_(1),X_(2),dots,X_(n)X_1, X_2, \ldots, X_nX1,X2,…,Xn be a random sample from N p ( μ , Σ ) N p ( μ , Σ ) N_(p)(mu,Sigma)N_p(\mu, \Sigma)Np(μ,Σ), then maximum likelihood estimators of μ μ mu\muμ and ∑ ∑ sum\sum∑ are:
μ ^ = X ¯ and Σ ^ = ∑ i = 1 n ( X i − X ¯ ) ′ ( X i − X ¯ ) μ ^ = X ¯ and Σ ^ = ∑ i = 1 n   X i − X ¯ ′ X i − X ¯ hat(mu)= bar(X)” and ” hat(Sigma)=sum_(i=1)^(n)(X_(i)-( bar(X)))^(‘)(X_(i)-( bar(X)))\hat{\mu}=\bar{X} \text { and } \hat{\Sigma}=\sum_{\mathrm{i}=1}^{\mathrm{n}}\left(X_{\mathrm{i}}-\bar{X}\right)^{\prime}\left(X_i-\bar{X}\right)μ^=X¯ and Σ^=∑i=1n(Xi−X¯)′(Xi−X¯)

Answer:

Question:-1

a) A study about a population showed that the mobility of population of a state to a village, town and city is in the following percentages:

From To
Village Town City
Village 60 25 15
Town 10 70 20
City 10 30 60
From To Village Town City Village 60 25 15 Town 10 70 20 City 10 30 60| From | To | | | | :—: | :—: | :—: | :—: | | | Village | Town | City | | Village | 60 | 25 | 15 | | Town | 10 | 70 | 20 | | City | 10 | 30 | 60 |
What will be the proportion of population village, town and city after one year and two years, given that the present proportion of the population in the village, town and city are respectively 0.50 , 0.40 0.50 , 0.40 0.50,0.400.50,0.400.50,0.40 and 0.10 ?

Answer:

To determine the population proportions in the village, town, and city after one year and two years, we can use the given transition matrix and initial population proportions. This is a Markov chain problem, where the population distribution evolves over time based on the transition probabilities.

Step 1: Define the Initial State and Transition Matrix

  • Initial population proportions (as a row vector):
    P ( 0 ) = [ 0.50 , 0.40 , 0.10 ] P ( 0 ) = [ 0.50 , 0.40 , 0.10 ] P(0)=[0.50,0.40,0.10]P(0) = [0.50, 0.40, 0.10]P(0)=[0.50,0.40,0.10]
    where 0.50 0.50 0.500.500.50 is the proportion in the village, 0.40 0.40 0.400.400.40 in the town, and 0.10 0.10 0.100.100.10 in the city.
  • Transition matrix (from the table):
    T = [ 0.60 0.25 0.15 0.10 0.70 0.20 0.10 0.30 0.60 ] T = 0.60 0.25 0.15 0.10 0.70 0.20 0.10 0.30 0.60 T=[[0.60,0.25,0.15],[0.10,0.70,0.20],[0.10,0.30,0.60]]T = \begin{bmatrix} 0.60 & 0.25 & 0.15 \\ 0.10 & 0.70 & 0.20 \\ 0.10 & 0.30 & 0.60 \end{bmatrix}T=[0.600.250.150.100.700.200.100.300.60]
    Here, each row represents the "from" location (village, town, city), and each column represents the "to" location (village, town, city). For example, T 12 = 0.25 T 12 = 0.25 T_(12)=0.25T_{12} = 0.25T12=0.25 is the probability of moving from village to town.

Step 2: Compute the Proportion After One Year

The population distribution after one year, P ( 1 ) P ( 1 ) P(1)P(1)P(1), is calculated by multiplying the initial state vector P ( 0 ) P ( 0 ) P(0)P(0)P(0) by the transition matrix T T TTT:
P ( 1 ) = P ( 0 ) ⋅ T P ( 1 ) = P ( 0 ) ⋅ T P(1)=P(0)*TP(1) = P(0) \cdot TP(1)=P(0)⋅T
Perform the matrix multiplication:
P ( 1 ) = [ 0.50 , 0.40 , 0.10 ] ⋅ [ 0.60 0.25 0.15 0.10 0.70 0.20 0.10 0.30 0.60 ] P ( 1 ) = [ 0.50 , 0.40 , 0.10 ] ⋅ 0.60 0.25 0.15 0.10 0.70 0.20 0.10 0.30 0.60 P(1)=[0.50,0.40,0.10]*[[0.60,0.25,0.15],[0.10,0.70,0.20],[0.10,0.30,0.60]]P(1) = [0.50, 0.40, 0.10] \cdot \begin{bmatrix} 0.60 & 0.25 & 0.15 \\ 0.10 & 0.70 & 0.20 \\ 0.10 & 0.30 & 0.60 \end{bmatrix}P(1)=[0.50,0.40,0.10]⋅[0.600.250.150.100.700.200.100.300.60]
Calculate each component:
  • Village (column 1): 0.50 ⋅ 0.60 + 0.40 ⋅ 0.10 + 0.10 ⋅ 0.10 = 0.30 + 0.04 + 0.01 = 0.35 0.50 ⋅ 0.60 + 0.40 ⋅ 0.10 + 0.10 ⋅ 0.10 = 0.30 + 0.04 + 0.01 = 0.35 0.50*0.60+0.40*0.10+0.10*0.10=0.30+0.04+0.01=0.350.50 \cdot 0.60 + 0.40 \cdot 0.10 + 0.10 \cdot 0.10 = 0.30 + 0.04 + 0.01 = 0.350.50⋅0.60+0.40⋅0.10+0.10⋅0.10=0.30+0.04+0.01=0.35
  • Town (column 2): 0.50 ⋅ 0.25 + 0.40 ⋅ 0.70 + 0.10 ⋅ 0.30 = 0.125 + 0.28 + 0.03 = 0.435 0.50 ⋅ 0.25 + 0.40 ⋅ 0.70 + 0.10 ⋅ 0.30 = 0.125 + 0.28 + 0.03 = 0.435 0.50*0.25+0.40*0.70+0.10*0.30=0.125+0.28+0.03=0.4350.50 \cdot 0.25 + 0.40 \cdot 0.70 + 0.10 \cdot 0.30 = 0.125 + 0.28 + 0.03 = 0.4350.50⋅0.25+0.40⋅0.70+0.10⋅0.30=0.125+0.28+0.03=0.435
  • City (column 3): 0.50 ⋅ 0.15 + 0.40 ⋅ 0.20 + 0.10 ⋅ 0.60 = 0.075 + 0.08 + 0.06 = 0.215 0.50 ⋅ 0.15 + 0.40 ⋅ 0.20 + 0.10 ⋅ 0.60 = 0.075 + 0.08 + 0.06 = 0.215 0.50*0.15+0.40*0.20+0.10*0.60=0.075+0.08+0.06=0.2150.50 \cdot 0.15 + 0.40 \cdot 0.20 + 0.10 \cdot 0.60 = 0.075 + 0.08 + 0.06 = 0.2150.50⋅0.15+0.40⋅0.20+0.10⋅0.60=0.075+0.08+0.06=0.215
So, after one year:
P ( 1 ) = [ 0.35 , 0.435 , 0.215 ] P ( 1 ) = [ 0.35 , 0.435 , 0.215 ] P(1)=[0.35,0.435,0.215]P(1) = [0.35, 0.435, 0.215]P(1)=[0.35,0.435,0.215]

Step 3: Compute the Proportion After Two Years

The population distribution after two years, P ( 2 ) P ( 2 ) P(2)P(2)P(2), is obtained by multiplying P ( 1 ) P ( 1 ) P(1)P(1)P(1) by the transition matrix T T TTT again:
P ( 2 ) = P ( 1 ) ⋅ T P ( 2 ) = P ( 1 ) ⋅ T P(2)=P(1)*TP(2) = P(1) \cdot TP(2)=P(1)⋅T
Using P ( 1 ) = [ 0.35 , 0.435 , 0.215 ] P ( 1 ) = [ 0.35 , 0.435 , 0.215 ] P(1)=[0.35,0.435,0.215]P(1) = [0.35, 0.435, 0.215]P(1)=[0.35,0.435,0.215]:
P ( 2 ) = [ 0.35 , 0.435 , 0.215 ] ⋅ [ 0.60 0.25 0.15 0.10 0.70 0.20 0.10 0.30 0.60 ] P ( 2 ) = [ 0.35 , 0.435 , 0.215 ] ⋅ 0.60 0.25 0.15 0.10 0.70 0.20 0.10 0.30 0.60 P(2)=[0.35,0.435,0.215]*[[0.60,0.25,0.15],[0.10,0.70,0.20],[0.10,0.30,0.60]]P(2) = [0.35, 0.435, 0.215] \cdot \begin{bmatrix} 0.60 & 0.25 & 0.15 \\ 0.10 & 0.70 & 0.20 \\ 0.10 & 0.30 & 0.60 \end{bmatrix}P(2)=[0.35,0.435,0.215]⋅[0.600.250.150.100.700.200.100.300.60]
Calculate each component:
  • Village (column 1): 0.35 ⋅ 0.60 + 0.435 ⋅ 0.10 + 0.215 ⋅ 0.10 = 0.21 + 0.0435 + 0.0215 = 0.275 0.35 ⋅ 0.60 + 0.435 ⋅ 0.10 + 0.215 ⋅ 0.10 = 0.21 + 0.0435 + 0.0215 = 0.275 0.35*0.60+0.435*0.10+0.215*0.10=0.21+0.0435+0.0215=0.2750.35 \cdot 0.60 + 0.435 \cdot 0.10 + 0.215 \cdot 0.10 = 0.21 + 0.0435 + 0.0215 = 0.2750.35⋅0.60+0.435⋅0.10+0.215⋅0.10=0.21+0.0435+0.0215=0.275
  • Town (column 2): 0.35 ⋅ 0.25 + 0.435 ⋅ 0.70 + 0.215 ⋅ 0.30 = 0.0875 + 0.3045 + 0.0645 = 0.4565 0.35 ⋅ 0.25 + 0.435 ⋅ 0.70 + 0.215 ⋅ 0.30 = 0.0875 + 0.3045 + 0.0645 = 0.4565 0.35*0.25+0.435*0.70+0.215*0.30=0.0875+0.3045+0.0645=0.45650.35 \cdot 0.25 + 0.435 \cdot 0.70 + 0.215 \cdot 0.30 = 0.0875 + 0.3045 + 0.0645 = 0.45650.35⋅0.25+0.435⋅0.70+0.215⋅0.30=0.0875+0.3045+0.0645=0.4565
  • City (column 3): 0.35 ⋅ 0.15 + 0.435 ⋅ 0.20 + 0.215 ⋅ 0.60 = 0.0525 + 0.087 + 0.129 = 0.2685 0.35 ⋅ 0.15 + 0.435 ⋅ 0.20 + 0.215 ⋅ 0.60 = 0.0525 + 0.087 + 0.129 = 0.2685 0.35*0.15+0.435*0.20+0.215*0.60=0.0525+0.087+0.129=0.26850.35 \cdot 0.15 + 0.435 \cdot 0.20 + 0.215 \cdot 0.60 = 0.0525 + 0.087 + 0.129 = 0.26850.35⋅0.15+0.435⋅0.20+0.215⋅0.60=0.0525+0.087+0.129=0.2685
So, after two years:
P ( 2 ) = [ 0.275 , 0.4565 , 0.2685 ] P ( 2 ) = [ 0.275 , 0.4565 , 0.2685 ] P(2)=[0.275,0.4565,0.2685]P(2) = [0.275, 0.4565, 0.2685]P(2)=[0.275,0.4565,0.2685]

Verification

The proportions should sum to 1 (since they represent the total population):
  • After 1 year: 0.35 + 0.435 + 0.215 = 1 0.35 + 0.435 + 0.215 = 1 0.35+0.435+0.215=10.35 + 0.435 + 0.215 = 10.35+0.435+0.215=1
  • After 2 years: 0.275 + 0.4565 + 0.2685 = 1 0.275 + 0.4565 + 0.2685 = 1 0.275+0.4565+0.2685=10.275 + 0.4565 + 0.2685 = 10.275+0.4565+0.2685=1
Both sums check out, confirming the calculations are correct.

Final Answer

  • After one year, the proportions are:
    • Village: 0.35 0.35 0.350.350.35
    • Town: 0.435 0.435 0.4350.4350.435
    • City: 0.215 0.215 0.2150.2150.215
  • After two years, the proportions are:
    • Village: 0.275 0.275 0.2750.2750.275
    • Town: 0.4565 0.4565 0.45650.45650.4565
    • City: 0.2685 0.2685 0.26850.26850.2685

Question:-1(a)

b) In a certain state, 58 landfills are classified according to their concentration of three hazardous chemicals: arsenic, barium and mercury. Suppose that the concentration of each one of the three chemicals is characterized as either high or low. If a landfill is chosen at random from among 58 landfills, given the following configuration:

Barium
High Mercury Low Mercury
Arsenic High Low High Low
High 1 3 5 9
Low 4 8 10 18
Barium High Mercury Low Mercury Arsenic High Low High Low High 1 3 5 9 Low 4 8 10 18| | Barium | | | | | :—: | :—: | :—: | :—: | :—: | | | High Mercury | | Low Mercury | | | Arsenic | High | Low | High | Low | | High | 1 | 3 | 5 | 9 | | Low | 4 | 8 | 10 | 18 |
Find the probability that it has:
i) high concentration of Barium,
ii) high concentration of mercury and low concentration of both Arsenic and Barium and
iii) high concentration of any one of the chemicals and low concentration of the other two.

Answer:

To solve this problem, we’ll use the given table to calculate the probabilities based on the total number of landfills (58) and the counts provided for each combination of arsenic, barium, and mercury concentrations (high or low). Let’s break it down step-by-step for each part of the question.

Step 1: Understand the Table and Total Landfills

The table provides the number of landfills for each combination of high/low concentrations of arsenic, barium, and mercury. The total number of landfills is 58. Here’s the table with the data:
Arsenic Barium Mercury Count
High High High 1
High Low High 3
High High Low 5
High Low Low 9
Low High High 4
Low Low High 8
Low High Low 10
Low Low Low 18
First, verify the total:
1 + 3 + 5 + 9 + 4 + 8 + 10 + 18 = 58 1 + 3 + 5 + 9 + 4 + 8 + 10 + 18 = 58 1+3+5+9+4+8+10+18=581 + 3 + 5 + 9 + 4 + 8 + 10 + 18 = 581+3+5+9+4+8+10+18=58
The sum matches the given total, so the data is consistent.

Step 2: Solve Each Part

i) Probability of High Concentration of Barium

We need to find the probability that a randomly chosen landfill has a high concentration of barium, regardless of arsenic and mercury concentrations. From the table, we sum the counts where barium is high:
  • Arsenic High, Barium High, Mercury High: 1
  • Arsenic High, Barium High, Mercury Low: 5
  • Arsenic Low, Barium High, Mercury High: 4
  • Arsenic Low, Barium High, Mercury Low: 10
Total landfills with high barium:
1 + 5 + 4 + 10 = 20 1 + 5 + 4 + 10 = 20 1+5+4+10=201 + 5 + 4 + 10 = 201+5+4+10=20
Probability:
P ( High Barium ) = Number of landfills with high barium Total landfills = 20 58 = 10 29 P ( High Barium ) = Number of landfills with high barium Total landfills = 20 58 = 10 29 P(“High Barium”)=(“Number of landfills with high barium”)/(“Total landfills”)=(20)/(58)=(10)/(29)P(\text{High Barium}) = \frac{\text{Number of landfills with high barium}}{\text{Total landfills}} = \frac{20}{58} = \frac{10}{29}P(High Barium)=Number of landfills with high bariumTotal landfills=2058=1029

ii) Probability of High Concentration of Mercury and Low Concentration of Both Arsenic and Barium

We need the probability that a landfill has high mercury, low arsenic, and low barium. From the table, this corresponds to:
  • Arsenic Low, Barium Low, Mercury High: 8
Total landfills with this specific combination: 8.
Probability:
P ( High Mercury, Low Arsenic, Low Barium ) = Number of landfills with high mercury, low arsenic, low barium Total landfills = 8 58 = 4 29 P ( High Mercury, Low Arsenic, Low Barium ) = Number of landfills with high mercury, low arsenic, low barium Total landfills = 8 58 = 4 29 P(“High Mercury, Low Arsenic, Low Barium”)=(“Number of landfills with high mercury, low arsenic, low barium”)/(“Total landfills”)=(8)/(58)=(4)/(29)P(\text{High Mercury, Low Arsenic, Low Barium}) = \frac{\text{Number of landfills with high mercury, low arsenic, low barium}}{\text{Total landfills}} = \frac{8}{58} = \frac{4}{29}P(High Mercury, Low Arsenic, Low Barium)=Number of landfills with high mercury, low arsenic, low bariumTotal landfills=858=429

iii) Probability of High Concentration of Any One Chemical and Low Concentration of the Other Two

We need the probability that exactly one of the three chemicals (arsenic, barium, or mercury) is high, while the other two are low. We identify these cases from the table:
  1. High Arsenic, Low Barium, Low Mercury:
    • Arsenic High, Barium Low, Mercury Low: 9
  2. Low Arsenic, High Barium, Low Mercury:
    • Arsenic Low, Barium High, Mercury Low: 10
  3. Low Arsenic, Low Barium, High Mercury:
    • Arsenic Low, Barium Low, Mercury High: 8
Total landfills with exactly one high concentration:
9 + 10 + 8 = 27 9 + 10 + 8 = 27 9+10+8=279 + 10 + 8 = 279+10+8=27
Probability:
P ( Exactly one high, other two low ) = Number of landfills with exactly one high Total landfills = 27 58 P ( Exactly one high, other two low ) = Number of landfills with exactly one high Total landfills = 27 58 P(“Exactly one high, other two low”)=(“Number of landfills with exactly one high”)/(“Total landfills”)=(27)/(58)P(\text{Exactly one high, other two low}) = \frac{\text{Number of landfills with exactly one high}}{\text{Total landfills}} = \frac{27}{58}P(Exactly one high, other two low)=Number of landfills with exactly one highTotal landfills=2758

Final Answer

  • i) Probability of high concentration of barium: 10 29 10 29 (10)/(29)\frac{10}{29}1029
  • ii) Probability of high concentration of mercury and low concentration of both arsenic and barium: 4 29 4 29 (4)/(29)\frac{4}{29}429
  • iii) Probability of high concentration of any one of the chemicals and low concentration of the other two: 27 58 27 58 (27)/(58)\frac{27}{58}2758

Question:-2

a) Consider the following system consisting of two switches I and II between two points A A AAA and B B BBB:

original image
A signal is sent from the point A A AAA to point B B BBB and is received at B B BBB if both the switches I I III and II are closed. It is assumed that the probabilities of I and II being closed are 0.8 and 0.6 respectively and that P [ I I P [ I I P[II\mathrm{P}[I IP[II is closed ∣ I ∣ I ∣I\mid \mathrm{I}∣I is closed ] = P [ I I ] = P [ I I ]=P[II]=\mathrm{P}[I I]=P[II is closed ] ] ]]].
Find:
i) The probability that signal is received at B B BBB.
ii) The conditional probability that switch I was open, given that the signal was not received at B B BBB,
iii) The conditional probability that switch II was open, given that the signal was not received at B.

Answer:

Let’s solve this problem step by step. The system consists of two switches, I and II, in series between points A and B. A signal is received at B only if both switches are closed. We are given the following:
  • Probability that switch I is closed: P ( I ) = 0.8 P ( I ) = 0.8 P(I)=0.8P(I) = 0.8P(I)=0.8
  • Probability that switch II is closed: P ( I I ) = 0.6 P ( I I ) = 0.6 P(II)=0.6P(II) = 0.6P(II)=0.6
  • The conditional probability: P ( I I is closed ∣ I is closed ) = P ( I I is closed ) P ( I I is closed ∣ I is closed ) = P ( I I is closed ) P(II” is closed”∣I” is closed”)=P(II” is closed”)P(II \text{ is closed} \mid I \text{ is closed}) = P(II \text{ is closed})P(II is closed∣I is closed)=P(II is closed), which implies that the events of switch I being closed and switch II being closed are independent.
Define:
  • I I III: Event that switch I is closed, so P ( I ) = 0.8 P ( I ) = 0.8 P(I)=0.8P(I) = 0.8P(I)=0.8, and switch I is open with probability P ( I c ) = 1 − 0.8 = 0.2 P ( I c ) = 1 − 0.8 = 0.2 P(I^(c))=1-0.8=0.2P(I^c) = 1 – 0.8 = 0.2P(Ic)=1−0.8=0.2.
  • I I I I IIIIII: Event that switch II is closed, so P ( I I ) = 0.6 P ( I I ) = 0.6 P(II)=0.6P(II) = 0.6P(II)=0.6, and switch II is open with probability P ( I I c ) = 1 − 0.6 = 0.4 P ( I I c ) = 1 − 0.6 = 0.4 P(II^(c))=1-0.6=0.4P(II^c) = 1 – 0.6 = 0.4P(IIc)=1−0.6=0.4.
  • S S SSS: Event that the signal is received at B, which requires both switches to be closed, i.e., S = I ∩ I I S = I ∩ I I S=I nn IIS = I \cap IIS=I∩II.
Since P ( I I is closed ∣ I is closed ) = P ( I I is closed ) P ( I I is closed ∣ I is closed ) = P ( I I is closed ) P(II” is closed”∣I” is closed”)=P(II” is closed”)P(II \text{ is closed} \mid I \text{ is closed}) = P(II \text{ is closed})P(II is closed∣I is closed)=P(II is closed), the events I I III and I I I I IIIIII are independent. Thus, P ( I ∩ I I ) = P ( I ) ⋅ P ( I I ) P ( I ∩ I I ) = P ( I ) ⋅ P ( I I ) P(I nn II)=P(I)*P(II)P(I \cap II) = P(I) \cdot P(II)P(I∩II)=P(I)⋅P(II).

i) Probability that the Signal is Received at B B BBB

The signal is received at B B BBB if both switches are closed, i.e., event S = I ∩ I I S = I ∩ I I S=I nn IIS = I \cap IIS=I∩II. Since I I III and I I I I IIIIII are independent:
P ( S ) = P ( I ∩ I I ) = P ( I ) ⋅ P ( I I ) = 0.8 ⋅ 0.6 = 0.48 P ( S ) = P ( I ∩ I I ) = P ( I ) ⋅ P ( I I ) = 0.8 ⋅ 0.6 = 0.48 P(S)=P(I nn II)=P(I)*P(II)=0.8*0.6=0.48P(S) = P(I \cap II) = P(I) \cdot P(II) = 0.8 \cdot 0.6 = 0.48P(S)=P(I∩II)=P(I)⋅P(II)=0.8⋅0.6=0.48
So, the probability that the signal is received at B B BBB is:
P ( S ) = 0.48 P ( S ) = 0.48 P(S)=0.48P(S) = 0.48P(S)=0.48

ii) Conditional Probability that Switch I was Open, Given that the Signal was Not Received at B B BBB

We need to find P ( I c ∣ S c ) P ( I c ∣ S c ) P(I^(c)∣S^(c))P(I^c \mid S^c)P(Ic∣Sc), where S c S c S^(c)S^cSc is the event that the signal is not received at B B BBB, and I c I c I^(c)I^cIc is the event that switch I is open.
First, compute the probability that the signal is not received:
P ( S c ) = 1 − P ( S ) = 1 − 0.48 = 0.52 P ( S c ) = 1 − P ( S ) = 1 − 0.48 = 0.52 P(S^(c))=1-P(S)=1-0.48=0.52P(S^c) = 1 – P(S) = 1 – 0.48 = 0.52P(Sc)=1−P(S)=1−0.48=0.52
Now, we need P ( I c ∣ S c ) P ( I c ∣ S c ) P(I^(c)∣S^(c))P(I^c \mid S^c)P(Ic∣Sc). Using the definition of conditional probability:
P ( I c ∣ S c ) = P ( I c ∩ S c ) P ( S c ) P ( I c ∣ S c ) = P ( I c ∩ S c ) P ( S c ) P(I^(c)∣S^(c))=(P(I^(c)nnS^(c)))/(P(S^(c)))P(I^c \mid S^c) = \frac{P(I^c \cap S^c)}{P(S^c)}P(Ic∣Sc)=P(Ic∩Sc)P(Sc)
  • Compute P ( I c ∩ S c ) P ( I c ∩ S c ) P(I^(c)nnS^(c))P(I^c \cap S^c)P(Ic∩Sc): The event I c ∩ S c I c ∩ S c I^(c)nnS^(c)I^c \cap S^cIc∩Sc means switch I is open and the signal is not received. If switch I is open ( I c I c I^(c)I^cIc), the signal cannot be received regardless of the state of switch II, because both switches must be closed for the signal to pass. Thus, whenever I c I c I^(c)I^cIc occurs, S c S c S^(c)S^cSc automatically occurs. Therefore:
    P ( I c ∩ S c ) = P ( I c ) = 0.2 P ( I c ∩ S c ) = P ( I c ) = 0.2 P(I^(c)nnS^(c))=P(I^(c))=0.2P(I^c \cap S^c) = P(I^c) = 0.2P(Ic∩Sc)=P(Ic)=0.2
  • Compute the conditional probability:
    P ( I c ∣ S c ) = P ( I c ∩ S c ) P ( S c ) = 0.2 0.52 = 0.2 0.52 = 0.2 ÷ 0.04 0.52 ÷ 0.04 = 5 13 P ( I c ∣ S c ) = P ( I c ∩ S c ) P ( S c ) = 0.2 0.52 = 0.2 0.52 = 0.2 ÷ 0.04 0.52 ÷ 0.04 = 5 13 P(I^(c)∣S^(c))=(P(I^(c)nnS^(c)))/(P(S^(c)))=(0.2)/(0.52)=(0.2)/(0.52)=(0.2-:0.04)/(0.52-:0.04)=(5)/(13)P(I^c \mid S^c) = \frac{P(I^c \cap S^c)}{P(S^c)} = \frac{0.2}{0.52} = \frac{0.2}{0.52} = \frac{0.2 \div 0.04}{0.52 \div 0.04} = \frac{5}{13}P(Ic∣Sc)=P(Ic∩Sc)P(Sc)=0.20.52=0.20.52=0.2÷0.040.52÷0.04=513
So, the conditional probability that switch I was open, given that the signal was not received, is:
P ( I c ∣ S c ) = 5 13 P ( I c ∣ S c ) = 5 13 P(I^(c)∣S^(c))=(5)/(13)P(I^c \mid S^c) = \frac{5}{13}P(Ic∣Sc)=513

iii) Conditional Probability that Switch II was Open, Given that the Signal was Not Received at B B BBB

Now we need P ( I I c ∣ S c ) P ( I I c ∣ S c ) P(II^(c)∣S^(c))P(II^c \mid S^c)P(IIc∣Sc), where I I c I I c II^(c)II^cIIc is the event that switch II is open.
Using conditional probability:
P ( I I c ∣ S c ) = P ( I I c ∩ S c ) P ( S c ) P ( I I c ∣ S c ) = P ( I I c ∩ S c ) P ( S c ) P(II^(c)∣S^(c))=(P(II^(c)nnS^(c)))/(P(S^(c)))P(II^c \mid S^c) = \frac{P(II^c \cap S^c)}{P(S^c)}P(IIc∣Sc)=P(IIc∩Sc)P(Sc)
  • Compute P ( I I c ∩ S c ) P ( I I c ∩ S c ) P(II^(c)nnS^(c))P(II^c \cap S^c)P(IIc∩Sc): The event I I c ∩ S c I I c ∩ S c II^(c)nnS^(c)II^c \cap S^cIIc∩Sc means switch II is open and the signal is not received. If switch II is open ( I I c I I c II^(c)II^cIIc), the signal cannot be received regardless of the state of switch I. Thus, whenever I I c I I c II^(c)II^cIIc occurs, S c S c S^(c)S^cSc automatically occurs. Therefore:
    P ( I I c ∩ S c ) = P ( I I c ) = 0.4 P ( I I c ∩ S c ) = P ( I I c ) = 0.4 P(II^(c)nnS^(c))=P(II^(c))=0.4P(II^c \cap S^c) = P(II^c) = 0.4P(IIc∩Sc)=P(IIc)=0.4
  • Compute the conditional probability:
    P ( I I c ∣ S c ) = P ( I I c ∩ S c ) P ( S c ) = 0.4 0.52 = 0.4 ÷ 0.04 0.52 ÷ 0.04 = 10 13 P ( I I c ∣ S c ) = P ( I I c ∩ S c ) P ( S c ) = 0.4 0.52 = 0.4 ÷ 0.04 0.52 ÷ 0.04 = 10 13 P(II^(c)∣S^(c))=(P(II^(c)nnS^(c)))/(P(S^(c)))=(0.4)/(0.52)=(0.4-:0.04)/(0.52-:0.04)=(10)/(13)P(II^c \mid S^c) = \frac{P(II^c \cap S^c)}{P(S^c)} = \frac{0.4}{0.52} = \frac{0.4 \div 0.04}{0.52 \div 0.04} = \frac{10}{13}P(IIc∣Sc)=P(IIc∩Sc)P(Sc)=0.40.52=0.4÷0.040.52÷0.04=1013
So, the conditional probability that switch II was open, given that the signal was not received, is:
P ( I I c ∣ S c ) = 10 13 P ( I I c ∣ S c ) = 10 13 P(II^(c)∣S^(c))=(10)/(13)P(II^c \mid S^c) = \frac{10}{13}P(IIc∣Sc)=1013

Final Answer

  • i) Probability that the signal is received at B B BBB: 0.48 0.48 0.480.480.48
  • ii) Conditional probability that switch I was open, given that the signal was not received at B B BBB: 5 13 5 13 (5)/(13)\frac{5}{13}513
  • iii) Conditional probability that switch II was open, given that the signal was not received at B B BBB: 10 13 10 13 (10)/(13)\frac{10}{13}1013
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