Mathematics Test - 25

Concept:

P(A ∪ B) = P(A) + P(B) - P(A ∩ B)

For mutually exclusive events A and B, P(A ∩ B) = 0

Calculation:

We know, for mutually exclusive events A and B,  P(A ∩ B) = 0

∴ P(A ∪ B) = P(A) + P(B) - 0

=  P(A) + P(B)

Hence, option (2) is correct.

Explanation:

Leap Year

A year that has 29 days in February is said to be a leap year.

Such a year has 366 days.

It has 52 weeks and 2 days.

1 week has 7 days, ∴ 52 weeks has 52 × 7 = 364 days.

Days left = 366 – 364 ⇒ 2 days.

Sample space = {(Mon,Tue) – (Tue,Wed) – (Wed,Thu) – (Thu,Fri) – (Fri, Sat) – (Sat,Sun) – (Sun,Mon)}

⇒ Sample space = 7 days

Favourable condition = {(Tue,Wed) – (Wed,Thu)} ⇒ 2 days

\(\therefore {\rm{Probability\;of\;having\;}}53{\rm{\;Wednesday\;is}} = \frac{{{\rm{Favourable\;condition}}}}{{{\rm{Sample\;space}}}}\)
\( \Rightarrow \frac{2}{7}\)

Non-Leap Year

A year that has 28 days in February is said to be a non-leap year.

Such a year has 365 days.

It has 52 weeks and 1 day.

1 week has 7 days, ∴ 52 weeks has 52 × 7 = 364 days.

Days left = 365 – 364 ⇒ 1 day.

Sample space = {(Mon, Tue, Wed, Thus, Fri, Sat, Sun)}

⇒ Sample space = 7 days

Favourable condition = {Wednesday} ⇒ 1 day.

\(\therefore {\rm{Probability\;of\;having\;}}53{\rm{\;Wednesday\;is}} = \frac{{{\rm{Favourable\;condition}}}}{{{\rm{Sample\;space}}}}\)

\(\Rightarrow \frac{1}{7}\)

Concept:

The sample space when a coin is tossed three times is S = 2³ ⇒ {TTT, TTH, THT, HTT, THH, HTH, HHT, HHH}

Calculation:

Given:

Case-1: Probability of getting atleast two tails.

Favourable cases = {TTH, THT, HTT, TTT}

Probability of getting at least two cases = \(\frac{Favourable\;cases}{Sample\;space}\Rightarrow\frac{4}{8}=\frac{1}{2}\)

Case-2: Probability of getting almost two tails.

Favourable cases = {TTH, THT, HTT, THH, HTH, HHT, HHH}

Probability of getting almost two tails = \(\frac{Favourable\;cases}{Sample\;space}\Rightarrow\frac{7}{8}\)

Concept:

Probability Mass Function: 

Probability Mass Function gives the probability of a discrete random variable being exactly equal to some value.

Total probability = Σ p(x) = 1.

Calculation:

Given that \(\rm p(x)=\dfrac{kx}{3}, x=1,2,3\).

Therefore, \(\rm p(1)=\dfrac{k}{3}\), \(\rm p(2)=\dfrac{2k}{3}\) and \(\rm p(3)=\dfrac{3k}{3}\).

We know that Σ p(x) = 1.

⇒ p(1) + p(2) + p(3) = 1

⇒ \(\rm \dfrac{k}{3}+\dfrac{2k}{3}+\dfrac{3k}{3}=1\)

⇒ \(\rm \dfrac{6k}{3}=1\)

⇒ \(\rm k =\dfrac{1}{2}\).

Given:

P(A) = 0.3 and P(A ∪ B) = 0.8

Formula used:

P(A ∪ B)  = P(A) + P(B) - P(A) × P(B)

Two events are independent if the incidence of one event does not affect the probability of the other event.

Calculation:

We know that,

P(A ∪ B) = P(A) + P(B) - P(A) × P(B)

⇒ 0.8 = 0.3 + P(B)  - 0.3 × P(B)

⇒ 0.5 = P(B)[1 - 0.3] = 0.7P(B)

⇒ P(B) = 0.5/0.7

∴ P(B) = 5/7

Concept:

E = the event a man reports that it is even

E1 = the event of an even number is picked

E2 = the event of an odd number is picked

To find : The probability that it is actually even i.e. P(E1|E)

P(E1|E) = \(\rm \dfrac {P(E|E_1).P(E_1)}{P(E|E_1).P(E_1)+P(E|E_2).P(E_2)}\)

Calculations:

Given, a man speaks truth 2 out of 3 times. He picks one of the natural numbers in the set S = {1, 2, 3, 4, 5, 6, 7} and reports that it is even.

E₁ = the event of an even number is picked

E₂ = the event of an odd number is picked

⇒P(E1) = \(\dfrac 3 7\)

⇒P(E2) = \(\dfrac 4 7\)

E = the event a man reports that it is even

 A man speaks truth 2 out of 3 times.

⇒P(E|E₁) = \(\dfrac 2 3\)

⇒P(E|E2) = \(\dfrac 1 3\)

To find : The probability that it is actually even i.e. P(E₁|E)

P(E1|E) = \(\rm \dfrac {P(E|E_1).P(E_1)}{P(E|E_1).P(E_1)+P(E|E_2).P(E_2)}\)

⇒P(E1|E) = \(\rm \dfrac {(\dfrac 2 3).(\dfrac 3 7)}{(\dfrac 2 3).(\dfrac 3 7)+(\dfrac 1 3).(\dfrac 4 7)}\)

⇒P(E1|E) = \(\rm \dfrac 6 {10}\)

⇒⇒P(E1|E) = \(\rm \dfrac 3 5\)

Hence, a man speaks truth 2 out of 3 times. He picks one of the natural numbers in the set S = {1, 2, 3, 4, 5, 6, 7} and reports that it is even. The probability that it is actually even is \(\rm \dfrac 3 5\)

Explanation

A and B are independent events 

⇒ P(A ∩ B) = P(A) × P(B)

According to question

⇒ P(A) × P(B) = 0.16   -----(A)

⇒ P(A') × P(B') = 0.36      ----(I)

⇒ [(1 - P(A)} × {1 - P(B)} = 0.36

⇒ [1 - P(B) - P(A) + P(A)P(B)] = 0.36

⇒ [1 - 0.36 + 0.16] = P(B) + P(A)

⇒ [1 - 0.36 + 0.16] = P(B) + P(A)

P(A) + P(B) = 0.8

By option elimination, we can conclude

P(A) = 0.4 and P(B) = 0.4

Concept:

The probability density function is always positive f(x) ≥ 0, and it follows the below condition.

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

Calculation:

Given:

\(f\left( x \right) = \left\{ {\begin{array}{*{20}{c}} {2k{x^4},\;\;0 \le x \le 1,}\\ {0,\;\;otherwise;} \end{array}} \right.\)

We know that \(\mathop \smallint \limits_{ - \infty }^\infty f\left( x \right)dx = 1\)

\(∴\mathop \smallint \limits_{ - \infty }^0 f\left( x \right)dx\;+\;\mathop \smallint \limits_{ 0 }^1 f\left( x \right)dx\;+\;\mathop \smallint \limits_{1 }^\infty f\left( x \right)dx =1\)

\(∴\mathop \smallint \limits_{ 0 }^1 2kx^4\;dx=1\)

\(∴\;2k\left [ \frac{x^5}{5} \right ]_{0}^{1}=1\)

∴ 2k = 5

\(\therefore\;k=\frac{5}{2}\)

Concept:

Conditional probability:

It gives the probability of happening of any event if the other has already occurred.

\({\rm{P}}\left( {\frac{{{{\rm{E}}_1}}}{{{{\rm{E}}_2}}}} \right) = {\rm{Probability\;of\;getting\;the\;event\;}}{{\rm{E}}_1}{\rm{\;when\;}}{{\rm{E}}_2}{\rm{is\;already\;occured}}.\)

\(P\left( {\frac{{{E_1}}}{{{E_2}}}} \right) = \frac{{P\left( {{E_1} \cap {E_2}} \right)}}{{P\left( {{E_2}} \right)}}\)

Calculation:

Given:

P (E₁) = Probability of stopping at the gas station and ask for tyre checked = 0.12

P (E₂) = Probability of stopping at the gas station and ask for oil checked = 0.29

P (E₁∩ E₂) = Probability of both checked = 0.07

\({\rm{P}}\left( {\frac{{{{\rm{E}}_2}}}{{{{\rm{E}}_1}}}} \right) = {\rm{Probability\;of\;person\;who\;has\;his\;tyre\;checked\;will\;also\;have\;oil\;checked}}\)   

∵ \(P\left( {\frac{{{E_2}}}{{{E_1}}}} \right) = \frac{{P\left( {{E_1} \cap {E_2}} \right)}}{{P\left( {{E_1}} \right)}}\)

∴ \(P\left( {\frac{{{E_2}}}{{{E_1}}}} \right) = \frac{{0.07}}{{0.12}} = 0.58\)

Concept:

For a random variable X = xi with probabilities P(X = x) = pi:

• Mean/Expected Value: μ = ∑pixi.
• Variance: Var(X) = σ² = ∑pi(xi)2 - (∑pixi)2 = ∑pi(xi)2 - μ2.
• Standard Deviation: σ = \(\rm \sqrt{Var(X)}\).

Calculation:

We have x1 = 1, x2 = 2, x3 = 3 and p1 = 0.3, p2 = 0.4, p3 = 0.3.

Now, ∑pixi = (0.3 × 1) + (0.4 × 2) +(0.3 × 3)

= 0.3 + 0.8 + 0.9

= 2

And, ∑pi(xi)2 = (0.3 × 1²) + (0.4 × 2²) +(0.3 × 3²)

= 0.3 + 1.6 + 2.7

= 4.6

∴ Var(X) = σ2 = ∑pi(xi)2 - (∑pixi)2 = 4.6 - 22 = 4.6 - 4 = 0.6.

The variance of the distribution is Var(X) = 0.6.