Engineering Mathematics Test 10

Urn contains 5 red balls, 5 black balls.

One ball is picked at random.

Case (i): The first ball is red ball

Probability to get a red ball in the second draw is

\({P_1} = \frac{5}{{10}} \times \frac{4}{9} = \frac{2}{9}\) 

Case (ii): The first ball is black ball

Probability to get a red ball in the second draw is

\({P_2} = \frac{5}{{10}} \times \frac{5}{9} = \frac{5}{{18}}\) 

Required probability (P) \(= {P_1} + {P_2} = \frac{2}{9} + \frac{5}{{18}} = \frac{1}{2}\)

Explanation:

Total number of cases for selecting 5 cards out of 52 = ⁵²C₅

 One ace can be drawn out of 4 aces is ⁴C₁ ways. The remaining 4 can be drawn out of the remaining 48 cards is ⁴⁸C₄ ways.

∴ Favorable number of cases when the five cards drawn contain just one ace = ⁴⁸C₄ × ⁴C₁

Now,

\(Required\;probability\; = \frac{{\;Favourable\;cases}}{{Total\;cases}}\)

∴ Required probability \( = \frac{{\;{\;^4}{C_1}\; \times \;{^{48}}{C_1}}}{{\;{\;^{52}}{C_5}}}\)

Explanation:

P(at least one of A, B and C occurs)

= P(A ∪ B ∪ C)

P(A ∪ B ∪ C) = P(A) + P(B) + P(C) – P(A ∩ B) – P(B ∩ C) – P(C ∩ A) + P(A ∩ B ∩ C)       ---(1)

Since P(A ∩ B) = P(B ∩ C) = 0;

P(A ∩ B ∩ C) = 0 equation (1) Becomes

\(P\left( {A \cup B \cup C} \right) = \frac{3}{4} - 0 - 0 - \frac{1}{8}\)

\(\therefore P\left( {A \cup B \cup C} \right) = \frac{5}{8}\)

Explanation:

P (A⋃B) = P (A) + P (B) – P (A⋂B)

\(\begin{array}{l} \therefore \frac{5}{6} = P\left( A \right) + \frac{1}{2}-\frac{1}{3}\\ \Rightarrow P\left( A \right) = \frac{5}{6}-\frac{1}{2} + \frac{1}{3}\\ = \frac{{5-3 + 2}}{6} \end{array}\)

= 4/6

= 2/3

We have, P(A).P(B) = P(A⋂B), for independent events.

P(A).P(B) = (2/3) × (1/2) = 1/3

This is equal to P(A⋂B).

Thus events A and B are independent events.

Explanation:

Let, T: A speaks truth ⇒ P(T) = ¾

T̅ : A lies ⇒ P(T̅) = 1 - P(T) = ¼

Let, B: Shyam won the match.

There are three cases for matches. It can be won, drawn or lost.

The probability of winning a match, P (B/T) = 1/3

The probability of not winning a match, P(B,T̅) = 2/3

Using Baye’s theorem:

\(P\left( {\frac{T}{B}} \right) = \frac{{P\left( T \right).P\left( {\frac{B}{T}} \right)}}{{P\left( T \right).P\left( {\frac{B}{T}} \right) + P\left( {\bar T} \right).P\left( {\frac{B}{{\bar T}}} \right)}} \)

\(P(\frac{T}{B}) = \frac{{\frac{3}{4} \times \frac{1}{3}}}{{\frac{3}{4} \times \frac{1}{3} + \frac{1}{4} \times \frac{2}{3}}} = \frac{3}{5} = 0.6\)

Explanation:

\(Mean = \mu = \left( {\mathop \sum \limits_{i = 1}^n {x_i}{p_i}} \right) = \frac{1}{2} + 1 + \frac{5}{6} = \frac{7}{3}\)

\(\left( {\mathop \sum \limits_{i = 1}^n x_i^2{p_i}} \right) = \frac{1}{2} + 3 + \frac{{25}}{6} = \frac{{23}}{3}\)

\(var\left( X \right) = {\sigma ^2} = \;\mathop \sum \limits_{i = 1}^n x_i^2{p_i} - {\left( {\mathop \sum \limits_{i = 1}^n {x_i}{p_i}} \right)^2}\)

\(var\left( X \right) = {\sigma ^2} = \frac{{23}}{3}\; - {\left( {\frac{7}{3}} \right)^2}\)

\(var\left( X \right) = \frac{{20}}{9}\)

Explanation:

Let A = the event in which the item has been produced by machine A.

B = the event in which the item has been produced by machine B.

C = the event in which the item has been produced by machine C.

Let D = the event of the item being defective

\(P\left( A \right) = \frac{1}{2},\;P\left( B \right) = P\left( C \right) = \frac{1}{4}\)

P(D/A) = (an item is defective, given that A has produced it)

\({\rm{P}}\left( {\frac{{\rm{D}}}{{\rm{A}}}} \right){\rm{\;}} = \frac{2}{{100}} = P\left( {\frac{D}{B}} \right)\)

\(P\left( {\frac{D}{C}} \right) = \frac{4}{{100}}\)

By theorem of total probability,

\(P\left( D \right) = P\left( A \right) \times P\left( {\frac{D}{A}} \right) + P\left( B \right) \times P\left( {\frac{D}{B}} \right) + P\left( C \right) \times P\left( {\frac{D}{C}} \right)\)

\(P\left( D \right) = \frac{1}{2} \times \frac{2}{{100}} + \frac{1}{4} \times \frac{2}{{100}} + \frac{1}{4} \times \frac{4}{{100}}\)

\(\therefore P\left( D \right) = \frac{1}{{40}}\)

∴ P (D) = 0.025

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:

\(S.D = \sqrt {E\left( {{X^2}} \right) - {{\left\{ {E\left( X \right)} \right\}}^2}} \)

Calculation:

\(E\left( X \right) = 2\left( {\frac{1}{3}} \right) + \left( { - 1} \right)\left( {\frac{2}{3}} \right)\)

∴ E(X) = 0

Now,

\(E\left( {{X^2}} \right) = \sum x_i^2P\left( {{x_i}} \right)\)

\(E\left( {{X^2}} \right) = {2^2}\left( {\frac{1}{3}} \right) + {\left( { - 1} \right)^2}\left( {\frac{2}{3}} \right)\)

∴ E(X²) = 2

We know that,

\(S.D\;\left( \sigma \right) = \sqrt {E\left( {{X^2}} \right) - {{\left\{ {E\left( X \right)} \right\}}^2}} \) 

\(\therefore S.D\;\left( \sigma \right) = \sqrt {{2^2} - 2} \)

∴ σ = √2 = 1.414

Concept:

Mean:

Let X is a discrete random variable having the possible values x₁, x₂, ……x_n

If P(X = x_i) = f(x_i), where i = 1, 2……n, then

E(X) = mean = x₁ f(x₁) + x₂ f(x₂)+ …….x_n f(x_n)

\({\rm{i}}.{\rm{e\;E}}\left( {\rm{x}} \right) = \mathop \sum \limits_{{\rm{i}} = 1}^{\rm{n}} {{\rm{X}}_{\rm{i}}}{\rm{\;f}}\left( {{{\rm{X}}_{\rm{i}}}} \right)\)

\({\rm{E}}\left( {{{\rm{x}}^2}} \right) = \mathop \sum \limits_{{\rm{i}} = 1}^{\rm{n}} X_i^2{\rm{\;f}}\left( {{{\rm{X}}_{\rm{i}}}} \right)\)

Variance:

V(X) = E(X²) – E(X)²

Calculation:

Given q = 0.4

Required value = V(X) = E(X²) – [E(X)]²

\(\begin{array}{l} E\left( X \right) = \mathop \sum \limits_i {X_i}{p_i} = 0 \times 0.4 + 1 \times 0.6 = 0.6\\ E\left( {{X^2}} \right) = \mathop \sum \limits_i X_i^2{p_i} = {0^2} \times 0.4 + {1^2} \times 0.6 = 0.6\\ \therefore V\left( X \right) = E\left( {{X^2}} \right) - {\left[ {E\left( X \right)} \right]^2} = 0.6 - 0.36 = 0.24\; \end{array}\)