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,

$Requiredprobability=\frac{Favourablecases}{Totalcases}$

∴ 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:

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

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

Explanation:

$Mean=\mu =\left(\sum _{i=1}^n{x}_i{p}_i\right)=\frac{1}{2}+1+\frac{5}{6}=\frac{7}{3}$

$\left(\sum _{i=1}^n{x}_i^2{p}_i\right)=\frac{1}{2}+3+\frac{25}{6}=\frac{23}{3}$

$var\left(X\right)={\sigma }^2=\sum _{i=1}^n{x}_i^2{p}_i-{\left(\sum _{i=1}^n{x}_i{p}_i\right)}^2$

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

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

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)

$\mathrm{P}\left(\frac{\mathrm{D}}{\mathrm{A}}\right)=\frac{2}{100}=P\left(\frac{D}{B}\right)$

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

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

Similarly, $P\left(B|A\right)=\frac{P\left(A\cap B\right)}{P\left(A\right)},whereP\left(A\right)\ne 0$

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 $=\frac{Favourablecases}{Reducedsamplespace}=\frac{2}{4}=\frac{1}{2}$ 

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,

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

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

∴ 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)

$\mathrm{i}.\mathrm{e}\mathrm{E}\left(\mathrm{x}\right)=\sum _{\mathrm{i}=1}^{\mathrm{n}}{\mathrm{X}}_{\mathrm{i}}\mathrm{f}\left({\mathrm{X}}_{\mathrm{i}}\right)$

$\mathrm{E}\left({\mathrm{x}}^2\right)=\sum _{\mathrm{i}=1}^{\mathrm{n}}{X}_i^2\mathrm{f}\left({\mathrm{X}}_{\mathrm{i}}\right)$

Variance:

V(X) = E(X²) – E(X)²

Calculation:

Given q = 0.4

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

$\begin{array}{l}E\left(X\right)=\sum _i{X}_i{p}_i=0\times 0.4+1\times 0.6=0.6\\ E\left(X^2\right)=\sum _i{X}_i^2{p}_i=0^2\times 0.4+1^2\times 0.6=0.6\\ \therefore V\left(X\right)=E\left(X^2\right)-{\left[E\left(X\right)\right]}^2=0.6-0.36=0.24\end{array}$