Engineering Mathematics Test 8

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:

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

Concept:

The player will win majority of races if he wins either 3 races or all 4 races.

The probability of winning ‘r’ races from ‘n’ races is given by

P = ⁿC_r(p)^r(q)^{n - r}

Calculations:

Probability to win majority races

= probability to win races + probability to win 4 races

Probability to win 3 races = ⁴C₃(0.6)³ (0.4)¹

Probability to win 4 races = ⁴C₄(0.6)⁴ (0.4)⁰

= ⁴C₃(0.6)³ (0.4)¹ + ⁴C₄(0.6)⁴ (0.4)⁰

= 4 (0.6)³ (0.4) + (0.6)4 (1)

= 0.3456 + 0.1296 = 0.4752 

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

X and Y are distributed uniformly over the interval [0, 1].

The probability that the problem get solved in 20 minutes is 1/3 for both X and Y.

Case (i): Only Sachin solved in less than 20 minutes

\({P_1} = \frac{1}{3} \times \frac{2}{3} = \frac{2}{9}\) 

Case (ii): Only Saurabh solved in less than 20 minutes

\({P_2} = \frac{2}{3} \times \frac{1}{3} = \frac{2}{9}\)

Case (iii): Both Sachin and Saurabh solved in less than 20 minutes

\({P_3} = \frac{1}{3} \times \frac{1}{3} = \frac{1}{9}\) 

Total probability that the problem will be solved in less than 20 minutes is,

\(P = \frac{2}{9} + \frac{2}{9} + \frac{1}{9} = \frac{5}{9}\) 

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

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

Let X denote the number of questions asked to Girish.

The Last one he answers wrongly.

P(X = k) = P^k-1 q, k = 1, 2, 3, ….

The probability that he will quit after answering an odd number of questions = 0.9

\( \Rightarrow \;\mathop \sum \limits_{k = 0}^\infty P\left( {X = 2k + 1} \right) = 0.9\;\)

\( \Rightarrow \;\mathop \sum \limits_{k = 0}^\infty {P^{2k}}q = 0.9\) 

⇒ (q + P²q + P⁴ q + …) = 0.9

Concept:

Poisson distribution:

A Poisson random variable describes the total number of events that happen in a certain time period.

A discrete random variable X is said to have a Poisson distribution with parameter λ (λ > 0) if the pdf of X is

\(P\left( {X = x} \right) = \frac{{{\lambda ^x}{e^{ - \lambda }}}}{{x!}}\;;x = 0,\;1,\;2,\; \ldots \)

Calculation:

Twice the average incidence would be 4 cases.

Let the random variable X = number of cases in 1 million people.

Has Poisson distribution with parameter 2.

P (X ≥ 4) = 1 – P (X ≤ 3)

\( = 1 - \left( {{e^{ - 2}}\frac{{{2^0}}}{{0!}} + {e^{ - 2}}\frac{{{2^1}}}{{1!}} + {e^{ - 2}}\frac{{{2^2}}}{{2!}} + {e^{ - 3}}\frac{{{2^3}}}{{3!}}} \right) = 0.143\)