Engineering Mathematics Test 8

A and B are mutually exhaustive events

Therefore, P(A ∪ B) = 1

 P(A ∪ B) = P(A) + P(B) – P(A ∩ B)

\(1=\frac{3}{4}+\frac{1}{2}-P\left( A\cap B \right)\)

\(\therefore P\left( A\cap B \right)=\frac{1}{4}\)

P(A’ ∪ B’) = P(A ∩ B)’

P(A ∩ B) + P(A ∩ B)’ = 1

\(\frac{1}{4}~+P{{\left( A\cap B \right)}^{'}}=1\)

\(P\left( {A}'\cup {B}' \right)=P{{\left( A\cap B \right)}^{'}}=\frac{3}{4}=0.75\)

Concept:

Let A and B be any two events associated with a random experiment. Then, the probability of occurrence of an event A under the condition that B has already occurred such that P(B) ≠ 0, is called the conditional probability and denoted by P(A | B)

i.e \(P\;\left( {A|\;B} \right) = \frac{{P\;\left( {A\; ∩ B} \right)}}{{P\left( B \right)}}\)

Similarly, \(P\left( {B\;|\;A} \right) = \frac{{P\left( {A\; ∩ B} \right)}}{{P\left( A \right)}},\;where\;P\left( A \right) \ne 0\)

Calculation:

Method I:

Original sample space = [HHH, HHT, HTH, TTT, TTH, THT, HTT, THH] = 8 events

Reduced sample space = [HHH, HHT, HTH, HTT]

After assuming first head = 4 events

Favourable cases = [HHT or HTH] = 2 events

∴ Required probability \(= \frac{{Favourable\;cases}}{{Reduced\;sample\;space}} = \frac{2}{4} = \frac{1}{2}\) 

Tips and Tricks:

The first toss outcome is head and it is given as a condition and hence the probability of first head becomes 1 as it is a sure event.

Since the first outcome is head, the probability that exactly two heads occur is P[HHT] or P[HTH] \(= 1 \times \frac{1}{2} \times \frac{1}{2} + 1 \times \frac{1}{2} \times \frac{1}{2} = \frac{1}{2}\) 

Concept:

Variance (X): It measures the spread of a distribution about its central value i.e smaller the variance and hence individual values lies closer to the average value.

Standard Deviation: It has same physical interpretation as that of variance. It is the square root of variance.

• Variance and Standard Deviation are always  ≥ 0 {0 in case of constant sequence}
• Variance is the fundamental property as compared to standard deviation
• Variance \(\propto \frac{1}{{Consistency}}\)  

Covariance: It measures the simultaneous variation of two random variables X and Y. It is formulated as, CoV (X, Y) = E (X, Y) – E (X) ⋅ E (Y)

• If X and Y are Independent Random variables CoV (X, Y) = 0

Var (aX ± bY) = a² Var (X) ± b² Var (Y) ± 2ab CoV (X, Y)

Calculation:

Given: Var (A) = 0.5, Var (B) = 0.04

∵ A and B are independent random variables,

∴ CoV (A, B) = 0

∴ Var (2A + 5B) = 2² × Var (A) + 5² × Var (B) + 2 × 2 × 5 CoV (A, B) = 4 × 0.25 + 25 × 0.04 + 0 = 2

∴ Standard Deviation \(= \sqrt {Variance} = \sqrt 2\)

Important Point:

E(ax ± by ± c) = a E(X) ± b E(y) ± E (c) = a E(x) ± b E(y) ± c

Where a, b and c are constants.

Concept:

Independent events: If occurrence of one event do not depend on the occurrence of other event, then both the events are called independent events.

In case of independent events,

P(A/B) = P(B)

∴ P(A ∩ B) = P(A) ⋅ P(B)

Addition theorem of probability:

P(A ∪ B) = P(A) + P(B) – P(A ∩ B)

Where, A ∪ B = Atleast one of A or B

Multiplication theorem of probability:

P(A ∩ B) = P(A/B) ⋅ P(B)

Where, A ∩ B = Simultaneous occurrence of A and B

Calculation:

Given: \(P\left( A \right) = \frac{1}{4},\;\;P\left( B \right) = \frac{1}{3}\) 

∵ A and B are independent events,

\(\therefore P\left( {\frac{A}{B}} \right) = P\left( A \right) = \frac{1}{4}\) 

By applying multiplication theorem,

Probability of simultaneous occurrence of A and B = P(A ∩ B) = P(A) ⋅ P(B) \(= \frac{1}{4} \times \frac{1}{3} = \frac{1}{{12}}\)

Comparing with the given form \(\frac{\alpha }{\beta }\), α = 1 and β = 2

∴ α + β = 1 + 12 = 13

Let the events be defined as:

A: Obtaining a sum of 8

B: Getting an even number on both dice

P(A U B) = P(A) + P(B) – P (A ∩ B)

Now cases favourable to A are (3, 5) (5, 3) (2, 6) (6, 2) (4, 4)

So, P(A) = 5/36

Cases favourable to B: (2, 2), (2, 4), (2, 6), (4, 2), (4, 4), (4, 6), (6, 2), (6, 2), (6, 4), (6, 6).

P(B) = 9/36

Now, (2, 6) (6, 2) and (4, 4) are common to both events A and B

So, P (A ∩ B) = 3/36

Concept:

Bayes' Theorem:

Let E1, E2, ….., En be n mutually exclusive and exhaustive events associated with a random experiment and let S be the sample space. Let A be any event which occurs together with any one of E1 or E2 or … or En such that P(A) ≠ 0. Then

\(P\left( {{E_i}\;|\;A} \right) = \frac{{P\left( {{E_i}} \right)\; \times \;P\left( {A\;|\;{E_i}} \right)}}{{\mathop \sum \nolimits_{i\; = 1}^n P\left( {{E_i}} \right) \times P\left( {A\;|\;{E_i}} \right)}},\;i = 1,\;2,\; \ldots .\;,\;n\)

Law of Total probability: Let E₁, E₂, E₃ are mutually exclusive and exhaustive events and A is an event which can occur with all E₁, E₂, E₃, then

\(P\left( A \right) = P\left( {{E_1}} \right) \cdot P\left( {\frac{A}{{{E_1}}}} \right) + P\left( {{E_2}} \right) \cdot P\left( {\frac{A}{{{E_2}}}} \right) + P\left( {{E_3}} \right) \cdot P\left( {\frac{A}{{{E_3}}}} \right)\) 

Calculation:

Let A = Man reports that it is 6

E₁ = Six actually occurs                   ∴ \(P\left( {{E_1}} \right) = \frac{1}{6}\)

E₂ = Six does not occurs                 ∴ \(P\left( {{E_2}} \right) = \frac{5}{6}\)

∵ E₁ and E₂ are mutually exhaustive events,

∴ P(E₁) + P(E₂) = 1

Now,

∵ Man speaks truth 2 out of 3 times.

P [Man telling truth] \(= P\left( {\frac{A}{{{E_1}}}} \right) = \frac{2}{3}\) 

P [Man telling lie] \(= P\left( {\frac{A}{{{E_2}}}} \right) = \frac{1}{3}\) 

By law of total probability,

\(P\left( A \right) = P\left( {{E_1}} \right) \times P\left( {\frac{A}{{{E_1}}}} \right) + P\left( {{E_2}} \right) \times P\left( {\frac{A}{{{E_2}}}} \right) = \frac{1}{6} \times \frac{2}{3} + \frac{5}{6} \times \frac{1}{3} = \frac{1}{9} + \frac{5}{{18}} = \frac{7}{{18}}\) 

By Baye’s theorem;

∴ The probability that it is actually a \(6 = \frac{{P\left( {{E_1}} \right) \cdot P\left( {\frac{A}{{{E_1}}}} \right)}}{{P\left( A \right)}} = \frac{{\frac{1}{6} \times \frac{2}{3}}}{{\frac{7}{{18}}}} = \frac{{\left( {\frac{1}{9}} \right)}}{{\left( {\frac{7}{{18}}} \right)}} = \frac{1}{7}\) 

Case 1: Head, white from U₁, white from U₂

\(= \left( {\frac{1}{2}} \right)\left( {\frac{3}{5}} \right)\left( {\frac{2}{2}} \right) = \frac{3}{{10}}\)

Case 2: Head, red from U₁, white from U₂

\(= \left( {\frac{1}{2}} \right)\left( {\frac{2}{5}} \right)\left( {\frac{1}{2}} \right) = \frac{1}{{10}}\)

Case 3: Tail, 2 white from U₁, white from U₂

\(= \left( {\frac{1}{2}} \right)\left( {\frac{{3{c_2}}}{{5{c_2}}}} \right)\left( {\frac{3}{3}} \right) = \frac{3}{{20}}\)

Case 4: Tail, white and red from U₁, white from U₂

\(= \left( {\frac{1}{2}} \right)\left( {\frac{{3{c_2} \times 2{c_1}}}{{5{c_2}}}} \right)\left( {\frac{2}{3}} \right) = \frac{1}{5}\)

Case 5: Tail, 2 red from U₂, white from U₁

\(\begin{array}{l}= \left( {\frac{1}{2}} \right)\left( {\frac{{2{c_2}}}{{5{c_2}}}} \right)\left( {\frac{1}{3}} \right) = \frac{1}{{60}}\\{\rm{Total\;Probability\;}} = \frac{3}{{10}} + \frac{1}{{10}} + \frac{3}{{20}} + \frac{1}{5} + \frac{1}{{60}} = \frac{{23}}{{30}}\end{array}\)

Given lines are:

4x + 3y + 7 = 0

3x + 4y + 8 = 0

Let  4x + 3y + 7 = 0 be the line of x on y and 3x + 4y + 8 = 0 be the line of y on x.

\(4x + 3y + 7 = 0 \)

\(x = - \frac{{7}}{4} - \frac{3}{4}y\)

\(3x + 4y + 8 = 0 \)

\(y = - \frac{8}{4} - \frac{3}{4}x\)

The regression coefficient of x on y is:

\({b_{xy}} = - \frac{3}{4}\)

The regression coefficient of y on x is:

\({b_{yx}} = - \frac{3}{4}\)

The correlation coefficient will be:

\(r = \sqrt {{b_{xy}}.{b_{yx}}} = \sqrt {\left( { - \frac{3}{4} \times - \frac{3}{4}} \right)}\)

\(\Rightarrow r = \frac{3}{4} = 0.75\)

Since \({b_{xy}},\;{b_{yx}}\) are both negative, r is also negative. ⇒ r = -0.75

Final Cumulative frequency = 28 and \(\frac{{28}}{2} = 14\) 

The cumulative frequency just greater than 14 is 18 and the corresponding class is (15 - 20) and the value of the class interval (size of the median class) \(\rm C = 5.\)

\(\therefore {\rm{Median}} = {\rm{L}} + \frac{{\left( {\frac{{\rm{N}}}{2} - {\rm{m}}} \right)}}{{\rm{f}}}{\rm{C}}\)

\(= 15 + \frac{{\left( {14 - 13} \right)}}{5}5 = 15 + 1 = 16\)