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Theory of probability was developed by Russian mathematician A.N. Kolmogorv in 1933.

Given that A and B are two independent events. C is the event in which exactly of A or B occurs. 

Let P (A) = x, P (B) = y 

Then P(C) = p(A ∩ B) + P(A ∩ B)

P(A)P(B) + P(A)P(B)

[∵ If A and B are independent so are 'A and B' and 'A and B']

P(C) = x(1 - y) + y(1 - x) ........(1)

Now consider, P(A ∪ ∩ B) p(bar A ∩ bar B)

= [P(A) + P(B) - P(A)P(B)][P(bar A)P(bar B)]

= (x + y - xy)(1 - x)(1 - y)

= (x + y) (1 – x) (1 – y) – xy (1 – x) (1 – y) ≤ (x + y) (1 – x) (1 – x) [∵ x, y(0, 1)] 

= x (1 – x) (1 – y) + y (1 – x) (1 – y) 

= x(1 – y) + y(1 – x) – x²(1 – y) – y²(1 – x) ≤ x(1 – y) + y(1 – x)³ 

= P (C) [Using eqⁿ (1)] 

Thus P(C) ≥ P(A ∪ B) P(A ∩ B) is proved. 

Let A denote the event that a person will get electrification contract and B denote the event that the person will get a plumbing contract 

Given : P(A) = \(\frac{2}{5}\) , P(not B) = P( \(\overline{B}\)) = \(\frac{4}{7}\), P(A or B) = \(\frac{2}{3}\) 

To find: Probability that he will get both electrification and plumbing contract = P(A and B) 

Formula used : P(B) = 1 – P( \(\overline{B}\))

P(A or B) = P(A) + P(B) - P(A and B)

P(B) = 1 - \(\frac{4}{7}\) =  \(\frac{3}{7}\)

P(B) = \(\frac{3}{7}\)

Probability of getting at least one contract = \(\frac{2}{3}\)

  \(\frac{2}{3}\)  = \(\frac{2}{5}+\frac{3}{7}\) - P(A and B) 

  \(\frac{2}{3}\)  = \(\frac{14+15}{35}\) - P(A and B)

P(A and B) = \(\frac{29}{35}-\frac{2}{3}\) =  \(\frac{87-70}{105}\) = \(\frac{17}{105}\)

P(A and B) = \(\frac{17}{105}\)

The probability that he will get both electrification and plumbing contract = \(\frac{17}{105}\)

Let C be the event that the person will have his teeth cleaned and F and E be the event of getting cavity filled or tooth extracted, respectively. We are given

P(C) = 0.48, P (F) = 0.25, P (E) = .20, P (C ∩ F) = .09,

P (C ∩ E) = 0.12, P (E ∩ F) = 0.07 and P (C ∩ F ∩ Ε) = 0.03

Now, P ( C ∪ F ∪ E) = P (C) + P (F) + P (E) – P (C ∩ F) – P (C ∩ E) – P (F ∩ E) + P (C ∩ F ∩ E)

= 0.48 + 0.25 + 0.20 – 0.09 – 0.12 – 0.07 + 0.03 = 0.68

Let A denote the event that a patient visiting a denist will have a tooth extracted and B denote the event that a patient will have a cavity filled 

Given : P(A) = 0.06 , P(B) = 0.2 , P(A or B) = 0.23 

To find: Probability that he will have a tooth extracted and a cavity filled= P(A and B)

Formula used : P(A or B) = P(A) + P(B) - P(A and B)

Probability that he will have a tooth extracted or a cavity filled = 0.23

0.23 = 0.06 + 0.2 - P(A and B)

0.23 = 0.26 P(A and B)

P(A and B) = 0.26 – 0.23 = 0.03

P(A and B) = 0.03

Probability that he will have a tooth extracted and a cavity filled=P(A and B) = 0.03

let A denote the event that the chosen individual is female and B denote the event that the chosen individual is over 50 years old.

Given : Town consists of 6000 people, 1200 are over 50 years old, and 2000 are females

To find : Probability that a randomly chosen individual from the town is either female or over 50 years = P(A or B) 

The formula used : Probability = 

  = \(\frac{favourable\,number\,of\,outcomes}{Total\,no.of\,outcomes}\)

P(A or B) = P(A) + P(B) - P(A and B)

For the event A , 

There are 2000 females present in a town of 6000 people

Favourable number of outcomes = 2000 

Total number of outcomes = 6000

P(A) = \(\frac{2000}{6000}\) = \(\frac{1}{3}\)

For the event B, 

There are 1200 are over 50 years of age in a town of 6000 people

Favourable number of outcomes = 1200 

Total number of outcomes = 6000 

P(A) = \(\frac{2000}{6000}\) = \(\frac{1}{5}\)

30% of the females are over 50 years

For the event A and B, 

\(\frac{30}{100}\times2000 = 600\) females are over 50

Favourable number of outcomes = 600

P(A and B) = \(\frac{600}{6000}\) = \(\frac{1}{10}\)

= P(A or B) = P(A) + P(B) - P(A and B)

P(A or B) = \(\frac{1}{3}+\frac{1}{5}+\frac{1}{10}\)

P(A or B) = \(\frac{10+6-3}{30}\) = \(\frac{13}{30}\)

P(A or B) =  \(\frac{13}{30}\) 

The probability that a randomly chosen individual from the town is either female or over 50 years = P(A or B) =  \(\frac{13}{30}\) 

B) Deductive

Correct option is  C) Theorem

D) Rama is immortal

P[Missing the target = 1 - \(\frac{7}{10}\)

\(= \frac{10 - 7}{10} = \frac{3}{10}\)

Correct option is (D) 1

Possible outcomes when a die is rolled are S = {1, 2, 3, 4, 5, 6}.

\(\therefore\) Total outcomes is n(S) = 6

E = Event of getting an even number = {2, 4, 6}

0 = Event of getting an odd number = {1, 3, 5}

\(\therefore\) n(E) = 3 & n(0) = 3

\(\therefore\) P(E) \(=\frac{n(E)}{n(S)}=\frac36=\frac12\) and

P(0) \(=\frac{n(0)}{n(S)}=\frac36=\frac12\)

\(\because\) P(E) + P(0) \(=\frac12+\frac12\) = 1

Hence, the sum of probabilities of getting an even number and an odd number when a die is rolled is 1.

Correct option is   D) 1

Correct option is (C) 1

The sum of the probabilities of all outcomes of a random experiment is always 1.

Correct option is  C) 1

When m = 6, n = 0 ⇒ |m-n| = 6

m = 5,n = 1 |m-n| = 4 

m = 4, n = 2 |m-nl=2 

m = 3,n = 3 |m-n| = 0 

m = 2, n = 4 |m-n|=2 

m = 1,n = 5 |m-n| = 4 

m = 0, n = 6 |m-n|=6 

∴ hence X can take values 0, 2, 4, 6.

Area of innermost ‘C’ circle = πr² 

= π x 1² = π sq. units. 

Area of the middle ‘B’ circle 

= π (2² – 1²) = π (4 – 1) = 3π sq. units. 

Area of the outermost ’A’ circle 

= π (3² – 2²) = π (9 – 4) = 5π sq. units.

Probability of hitting the circle B

= \(\frac {favourables\, area}{Total\,area}\)

= \(\frac {3\pi}{\pi + 3\pi + 5\pi} = \frac 39 = \frac 13\)

Total number of ladies: 200 

Number of ladies who like coffee: 82 

Number of ladies who dislike coffee: 118

Probability (P) = \(\frac{Number\,of\,a\,favrorable\,outcomes}{Total\,number\,of\,outcomes}\)

Let P ( No Coffee) be probability of ladies who dislike coffee

P (No Coffee) = \(\frac{Number\,of\,ladies\,who\,like\,coffe}{Total\,number\,ladies}\)

P (No Coffee) = \(\frac{118}{200}=\frac{59}{100}\)

Let the number of red card in a sample of 3 cards drawn be random variable X. Obviously X may have values 0,1,2,3.

Now P(X = 0)= Probability of getting no red card = ²⁶C₃/⁵²C₃ = 2600/22100 = 2/17

P(X = 1)= Probability of getting one red card and two non-red cards =  (²⁶C₂ x ²⁶C₂)/⁵²C₃ = 8450/22100 = 13/34

P(X = 2)= Probability of getting two red card and one non-red card =  (²⁶C₂ x ²⁶C₁)⁵²C₃ = 8450/22100 = 13/34

P(X = 3)= Probability of getting 3 red cards = ²⁶C₃/⁵²C₃ = 2600/22100 = 2/17

Hence, the required probability distribution in table as

∴ Required mean = E(X) = ∑p_ix_i = 0 x 2/17 + 1 x 13/34 + 2 x 13/34 + 3 x 2/17

= 13/34 + 26/34 + 6/17 = (13 + 26 + 12)/34 = 51/34 = 3/2

(a) ↔ (iv) 

(b) ↔ (v) 

(c) ↔ (i)

(d) ↔ (iii)

(e) ↔ (ii)

(a) ↔ (iv) 

(b) ↔ (iii) 

(c) ↔ (ii) 

(d) ↔ (i)

Let B= boy G= girl

And let us consider, in a sample space, the first child is elder and second child is younger. 

Total possible outcome = {BB, BG, GB, GG} = 4

Let A = be the event that both the children are girls = 1

Therefore P(A) = \(\cfrac14\)

Case 1.

Let B = event that youngest is girl = {BG, GG} =2

{Since we have considered second is younger in a sample space}

Therefore P(B) = \(\cfrac24\)

And (A ∩ B) = both are girls and younger is also girl = (GG) = 1

Therefore , P (A ∩ B) = \(\cfrac14\)

We require P(\(\cfrac{A}{B}\))

\(P(\cfrac{A}{B})=\cfrac{P(A\cap B)}{P(B)}\)

= \(\cfrac{\frac14}{\frac24}=\cfrac12\)

Case 2.

Let B = event that at least one is girl = {BG,GB GG} =3

{Since we have considered second is younger in a sample space}

Therefore P(B) = \(\cfrac34\)

And (A ∩ B) = both are girls and atlas one is girl = (GG) = 1

Therefore , P (A ∩ B) = \(\cfrac14\)

We require P(\(\cfrac{A}B\))

\(P(\cfrac{A}{B})=\cfrac{P(A\cap B)}{P(B)}\)

= \(\cfrac{\frac14}{\frac34}\) = \(\cfrac13\)

Correct option is: C) 1/2

 Favourable Outcomes: If some outcomes out of all the elementary outcomes in the sample space of a random experiment indicate the occurrence of a certain event A, then these outcomes are called the favourable outcomes of the event A.

Equi-probable Events: If there is no apparent reason to believe that out of one or more events of a random experiment, any one event is more or less likely to occur than the other events, then those events are called as equi-probable events.

Sample Space: The set of all possible outcomes of a random experiment is called a sample space of that random experiment. It is denoted by U or S.

When P(A) + P(B) + P(C) = 1, we can say that three events A, B and C in a sample space are exhaustive.

The experiment which can be independently repeated under identical conditions and all its possible outcomes are known but it cannot be predicted with certainty which of the outcomes will appear is called a random experiment.

0, P(A ∩ B), P(A), P(A ∪ B), P(A) + P(B) are in the ascending order.

The sample space, S = {H₁, H₂, H₃, H₄, H₅, H₆, TH, TT}.

A coin has two sides, head (H) and tail (T). So, the number of elements in the sample space is 2ⁿ.

Total no. of possible outcomes = 8 + 6 + 4 = 18 {8 red, 6 white, 4 black}

(i) E ⟶ event of getting red or white ball

No. of favourable outcomes = 4 {4 black balls}

P(E) = 4/18 = 2/9

(Bar E) ⟶event of not getting black ball

P(Bar E)=1−P(E)=1−2/9 = 7/9

(ii) E ⟶ event of getting neither white nor black.

No. of favourable outcomes = 15 – 6 – 4 = 8 {Total balls – no. of white balls – no. of black balls}

P(E) = 8/18 = 4/9

Total no. of possible outcomes = 8 {3 red, 5 black}

(i) Let E ⟶ event of drawing red ball.

No. favourable outcomes = 1 {1 ace card}

P(E) = (No.of favorable outcomes)/(Total no.of possible outcomes) = 3/8

(ii) Let E ⟶ event of drawing black ball.

No. favourable outcomes = 5 {5 black balls}

P(E) = 5/8