Engineering Mathematics Test 9

P(A) = P(B) = 2 P(C)

P(A) + P(B) + P(C) = 1

5 P(C) = 1

P(C) = 1/5 = 0.2

P(A) = P(B) = 2P(C) = 0.4

Probability that B or C wins:

Probability of latter box contains 2 red and 6 green balls \(= \frac{{{5_{{C_2}}} \times {{10}_{{C_6}}}}}{{{{15}_{{C_8}}}}} = \frac{{10 \times 210}}{{6435}} = \frac{{140}}{{429}}\)

Box – 7 Apple pieces, 5 orange pieces

Imagine that 12 pieces of fruits are numbered as 1, 2, ….12.

There are 12! Possible orders at which the 12 pieces of fruit can be taken out of the box.

There are 7 × 11! Orders which the last element is an apple.

Let P be the premium and let X denote the earning of the insurance company per policy.

X = P; if there is no eventuality during period of the policy

= P – 1,00,000; if there is eventuality during period of the policy

Expectation, E(x) = 0.95 P + 0.05 (P – 1,00,000) = 1000

Let A denote the event that the person drinks regularly, and let B denote the event that the death is due to liver disease.

Given P(A) = 0.25

P(B) = 0.005

P(B/A) = 6 P(B/A^C)

By using the total probability theorem

P(B) = P(B/A) P(A) + P(B/A^C) P(A^C)

\( \Rightarrow 0.005 = P\left( {\frac{B}{A}} \right) \times 0.25 + P\left( {\frac{B}{A}} \right) \times \frac{1}{6} \times 0.75\) 

⇒ P(B/A) = 0.0133

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

\(P\left( {X \le \frac{2}{3}{\rm{|}}X > \frac{1}{3}} \right) = \frac{{P\left( {X \le \frac{2}{3} \cap X > \frac{1}{3}} \right)}}{{P\left( {X > \frac{1}{3}} \right)}}\)

\( = \frac{{P\left( {\frac{1}{3} < X \le \frac{2}{3}} \right)}}{{P\left( {X > \frac{1}{3}} \right)}}\)

\(P\left( {\frac{1}{3} < X \le \frac{2}{3}} \right) = \mathop \smallint \limits_{1/3}^{2/3} 4{x^3}dx = \left[ {{x^4}} \right]_{\frac{1}{3}}^{\frac{2}{3}} = \frac{{15}}{{81}} = \frac{5}{{27}}\)

\(P\left( {X > \frac{1}{3}} \right) = \mathop \smallint \limits_{1/3}^1 4{x^3}dx\;\)

\( = \left[ {{x^4}} \right]_{\frac{1}{3}}^1\)

\( = 1 - \frac{1}{{81}} = \frac{{80}}{{81}}\)

\(P\left( {X \le \frac{2}{3}{\rm{|}}X > \frac{1}{3}} \right) = \frac{{\frac{5}{{27}}}}{{\frac{{80}}{{81}}}} = \frac{5}{{27}} \times \frac{{81}}{{80}} = \frac{3}{{16}}\)

\(Var\left( {\frac{1}{X}} \right) = E\left( {\frac{1}{{{X^2}}}} \right) - {\left[ {E\left( {\frac{1}{X}} \right)} \right]^2}\) 

\(E\left( {\frac{1}{X}} \right) = \mathop \smallint \limits_0^1 \frac{1}{x} \times {x^2}\left( {2x + \frac{3}{2}} \right)dx\) 

\( = \mathop \smallint \limits_0^1 \left( {2{x^2} + \frac{3}{2}x} \right)dx\) 

\( = \left[ {\frac{2}{3}{x^3} + \frac{3}{4}{x^2}} \right]_0^1 = \frac{2}{3} + \frac{3}{4} = \frac{{17}}{{12}}\) 

\(E\left[ {\frac{1}{{{X^2}}}} \right] = \mathop \smallint \limits_0^1 \frac{1}{{{x^2}}} \times {x^2}\left( {2x + \frac{3}{2}} \right)dx\) 

\(\mathop \smallint \limits_0^1 \left( {2x + \frac{3}{2}} \right)dx\) 

\( = \left[ {{x^2} + \frac{3}{2}x} \right]_0^1 = 1 + \frac{3}{2} = \frac{5}{2}\) 

\(var\left( {\frac{1}{X}} \right) = \frac{5}{2} - {\left( {\frac{{17}}{{12}}} \right)^2}\) 

\( = \frac{5}{2} = \frac{{289}}{{144}} = \frac{{71}}{{144}}\) 

\(\mathop \smallint \limits_{ - \infty }^\infty {f_x}\left( x \right) = 1\;\) 

\( \Rightarrow \;\mathop \smallint \limits_0^1 \left( {k{x^3} + \frac{3}{2}{x^2}} \right)dx = 1\) 

\( \Rightarrow \left[ {\frac{k}{4}{x^4} + \frac{1}{2}{x^3}} \right]_0^1 = 1\) 

⇒ k = 2

The sample space is defined by two mutually-exclusive events – it rains or it does not rain.

Additionally, a third event occurs when the weatherman predicts rain.

Notation for these events are

Event A₁: It rains on Marie’s wedding.

Event A₂: It does not rain on Marie’s wedding.

Event B: The weatherman predicts rain.

In terms of probabilites, we know the following:

\(P\left( {{A_1}} \right) = \frac{5}{{365}} = 0.014\) [It rains 5 days out of the year]

\(P\left( {{A_2}} \right) = \frac{{360}}{{365}} = 0.986\) It does not rain 360 days ]

\(P\left( {B/{A_1}} \right) = 0.9\) [When it rains, the weatherman predicts rain 90% of the time]

\(P\left( {B/{A_2}} \right) = 0.1\) [When it does not rain, the weatherman predicts rain 10% of the time]

We want to know P(A₁/B), the probability it will rain on the day of Marie’s wedding. Given a forecast for rain by the weatherman.

Using Bayes’ theorem :

\(P\left( {{A_1}|B} \right) = \frac{{P\left( {{A_1}} \right)P\left( {B/{A_1}} \right)}}{{P\left( {{A_1}} \right)P\left( {B/{A_1}} \right) + P\left( {{A_2}} \right)P\left( {B/{A_2}} \right)}}\)

\(P\left( {{A_1}|B} \right) = \frac{{0.014 \times 0.9}}{{0.014 \times 0.9 + 0.986 \times 0.1}}\)

P(A₁|B) = 0.113

Standard deviation = √v, where v is the variance given by:

Variance (v) = E(x²) – μ²     -----Equation-(1)

Mean (μ) = E(x)

μ = E(x) = ∑ x_i p(x_i)

\( = \left( { - 6} \right)\left( {\frac{1}{6}} \right) + 4\left( {\frac{1}{2}} \right) + 6\left( {\frac{1}{3}} \right)\)

= -1 + 2 + 2 = 3

\({\rm{E}}{\left( {\rm{x}} \right)^2} = \sum x_i^2p\left( {{x_i}} \right)\)

\( = {\left( { - 6} \right)^2}\left( {\frac{1}{6}} \right) + {\left( 4 \right)^2}\left( {\frac{1}{2}} \right) + {\left( 6 \right)^2}\left( {\frac{1}{3}} \right)\)

= 6 + 8 + 12 = 26

From Equation-1, the variance is given by:

V = 26 – 9 = 17

Standard deviation \( = \sqrt {17} = 4.12\)

Case I: He drops the key which opens the lock

probability of dropping key that opens locks = 2/6

probability to open lock after dropping correct key : 1/5

Probability to open the lock \( = \frac{2}{6} \times \frac{1}{5} = \frac{2}{{30}}\)

Case II: He drops the key which doesn’t opens the lock

probability of dropping key that does not opens locks = 4/6

probability to open lock after dropping correct key : 2/5

Probability to open the lock \( = \frac{4}{6} \times \frac{2}{5} = \frac{8}{{30}}\)

Now, the total probability to open the lock

\( = \frac{2}{{30}} + \frac{8}{{30}} = \frac{{10}}{{30}} = \frac{1}{3}\)