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The total number of elementary events are: 4,5,6 …..20 = 20 – 3= 17

Let E be the event of getting an even number on the random draw 

Then total favorable outcomes are: 4, 6, 8, 10, 12, 14, 16, 18, 20 

Total number of favorable outcomes = 9 

P (even number) = P (E) = \(\frac{9}{17}\)

Total numbers of elementary events are: 52 

Let E be the event of getting a non –face card 

The total number of favorable outcomes are ; 52 – 12 = 40: being four cards each of king , queen, jack being a face card , deleted from the favorable outcomes number 

P (non-face card) = P (E) = \(\frac{40}{52}\) = \(\frac{20}{26}\) = \(\frac{10}{13}\)

Total numbers of elementary events are: 5+8 + 7 = 20 

Let E be the event of getting a green ball or a white ball 

Total numbers of favorable outcomes are: 8+ 7 = 15 

P (white or green ball) = P (E) = \(\frac{15}{20}\) = \(\frac{3}{4}\)

Total numbers of elementary events are: 6 

Let E be the event of getting a prime number 

Favorable events are: 2, 3, 5 

Numbers of favorable events are = 3 

P (prime number) = P (E) = \(\frac{3}6\) = \(\frac{1}2\)

Correct option is: D) \(\frac{31}{366}\)

Since, \(\frac {2012}4 = 503\) 

Hence, 2012 is completely divisible by 4.

\(\therefore\) 2012 is a leap year in which total No of days are 366.

Total No of days in the month of July in 2012 = 31.

\(\therefore\) Probability that on individual person has a birthday in July in 2012 is

P = \(\frac {Total \,No\, of\, days \,in \,July\, in \, 2012}{Total\, No \,of\, days \,in \,year \,2012}\)= \(\frac {31}{366}\)

Correct option is: D) \(\frac{31}{366}\)

It is given that

Total number of students = 40

(i) We know that

Number of students having blood group O = 14

So we get

Required probability = number of students having blood group O/ Total number of students

By substituting the values

Required probability = 14/40 = 0.35

(ii) We know that

Number of students having blood group AB = 6

So we get

Required probability = number of students having blood group AB/ Total number of students

By substituting the values

Required probability = 6/40 = 0.15

It is given that

Total number of phone numbers = 200

(i) We know that

Number of phone numbers with 5 = 24

So we get

Required probability = number of phone numbers with 5/ Total number of phone numbers

By substituting the values

Required probability = 24/100 = 0.12

(ii) We know that

Number of phone numbers with 8 = 16

So we get

Required probability = number of phone numbers with 8/ Total number of phone numbers

By substituting the values

Required probability = 16/200 = 0.08

Correct option is: A) \(\frac{4}{5}\)

Total number of outcomes = 100 - 51 + 1 = 50.

Non - composite number from 51 to 100 are (53, 59, 61, 67, 71, 73, 79, 83, 89, 97).

\(\therefore\) Number of non-composite number from 51 to 100 is 10.

\(\therefore\) Number of composite number from 51 to 100 = 50 -10 = 40.

\(\therefore\) Probability of getting a composite number from 51 to 100 

= \(\frac {Number\, of \,composite\, number \,from \,51 \,to \,100}{Total \, number \, from \, 51 \,to \,100}\) = \(\frac {40}{50} = \frac 45\)

Correct option is: A) \(\frac{4}{5}\)

Correct option is: D) 6

\(\because\) \(\frac{5}{x}\) is a probability.

\(\therefore\) 0 \(\leq\) \(\frac{5}{x}\)  \(\leq\) 1

\(\Rightarrow\) \(\frac{5}{x}\)  \(\leq\) 1

\(\Rightarrow\) x \(\geq\) 5

\(\because\) 6 > 5

\(\therefore\) Possible value of x be 6.

Hence, for x = 6, \(\frac{5}{x}\) may be probability of an event.

Correct option is: D) 6

Correct option is: C) \(\frac{1}{2}\)

When a die is rolled, possible outcomes can be 1, 2, 3, , 5 and 6.

\(\therefore\) Total number of outcomes = 6

Total prime numbers which can appear on die = 3 (2, 3 or 5)

\(\therefore\) P (Prime number) = \(\frac {Total \, prime\, numbers}{Total \, possible \, outcomes} = \frac 36 = \frac 12\)

Correct option is: C) \(\frac{1}{2}\)

Correct option is: A)  \(\frac{2}{6}\)

Probability of getting 2 or 3 when we throw a dice = \(\frac 26 = \frac 13\)

Correct option is: A) \(\frac{2}{6}\)

Correct option is: B)\(\frac{1}{8}\)

If there is 8 equally likely events. The probability of each event = \(\frac{1}{8}\)

Correct option is: B) \(\frac{1}{8}\)

We know– 

P(A ∪ B) = P(A) + P(B) – P(A ∩ B) 

⇒ P(A ∩ B) = 0 

[∵ P(A ∪ B) = P(A) + P(B)] 

∴ P(A ∩ B) = 0

[∵ P(A ∪ B) = P(A) + P(B)]

∴ The events A and B are mutually exclusive.

P(A) = P(B)

If P(A ∪ B) = P(A ∩ B), then the relation between P(A) and P(B) is P(A) = P(B).

No.

If A and B are exhaustive events, then P(A ∩ B) may not be 0.

Here, the sample space is– 

S = [HHH, HHT, HTH, HTT, THH, THT, TTH, TTT} 

A = {TTT} 

B = {HHT, HTH, THH} 

C = {THH, THT, TTH, TTT} 

A ∩ B = ϕ , B ∩ C = {THH} ≠ ϕ , A ∩ C = {TTT} ≠ ϕ .

(i) A, B are mutually exclusive. 

(ii) A is a simple event. 

(iii) B, C are compound events.

The total number of mapping from a set A into itself is  nⁿ 

[∵ there are h’ element in A] 

And the total number of one to one mapping is n 

∴ Required probability of getting one to one mapping = \(\frac{n!}{n^n}\).

Let ‘S’ be the sample space. 

∴ S = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}. 

⇒ n(S) = 36 

Let ‘A’ and ‘B’ be the events that the sum of the numbers on the two faces is divisible by 3 and 4 respectively. 

∴ A = {(1, 2), (1, 5), (2, 1), (2, 4), (3, 3), (3, 6), (4, 2), (4, 5), (5, 1), (5, 4), (6, 3), (6, 6)} 

⇒ n(A) = 12 

and, B = {(1, 3), (2, 2), (2, 6), (3, 5), (4, 4), (5, 3), (6, 2), (3, 1), (6, 6)} 

⇒ n(B) = 9

∴ P(A) = \(\frac{n(A)}{n(S)}=\frac{12}{36}\)

P(B) = \(\frac{n(A)}{n(S)}=\frac{12}{36}\)

P(B) = \(\frac{n(B)}{n(S)}=\frac{9}{36}\)

A ∩ B = {(6, 6)} ⇒ n(A ∩ B) = 1

∴ (A ∩ B) = \(\frac{n(A∩ B)}{n(S)}=\frac{1}{36}\)

We know– 

P(A ∪ B) = P(A) + P(B) – P(A ∩ B)

= \(\frac{13}{36}+\frac{9}{36}-\frac{1}{36}=\frac{20}{36}=\frac{5}{9}\)

∴ Probability of getting the sum multiple of 3 or 4 is \(\frac{5}{9}\)

⇒ Probability of getting the sum neither a multiple of 3 nor 4 = P(A ∪ B)

= \(p(\overline{A∪B})\)

= 1 – P(A ∪ B)

\(= 1 - \frac{5}{9}\)

= \(\frac{4}{9}\)

We know that,

Probability of occurrence of an event

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

Total outcomes are (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), 

(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), 

(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), 

(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6) , 

(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6) ,

(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6) 

Desired outcomes are all outcomes except (1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6) 

Total no. of outcomes are 36 and desired outcomes are 30 probability of not getting same number = \(\frac{30}{36}\) = \(\frac{5}{6}\) 

Conclusion: Probability of not getting the same number on the two dice is   \(\frac{5}{6}\) 

Let ‘S’ be the sample space associated with the given experiment. Then– 

n(S) = ²⁰C₃ = 1140. 

Now, E = {(1, 2, 3), (2, 3, 4), (3, 4, 5),….., (18, 19, 20)} 

⇒ n(E) = 18 

∴ P(E) = \(\frac{n(E)}{n(S)}=\frac{18}{1140}=\frac{3}{190}\)

∴ P(E’) = 1 − \(\frac{3}{190}=\frac{187}{190}\) 

∴ The required probability = \(\frac{187}{190}\).

It is given that

Total number of teachers = 75

(i) We know that

Probability that the selected teacher is 40 or more than 40 years old = (37 + 8)/ 75

So we get

Probability that the selected teacher is 40 or more than 40 years old = 45/75 = 3/5

(ii) We know that

Probability that the selected teacher is of any age lying between 30-39 years (including both) = 27/75

So we get

Probability that the selected teacher is of any age lying between 30-39 years (including both) = 9/25

(iii) We know that

Probability that the selected teacher is 18 years or more and 49 years or less = (3 + 27 + 37)/ 75

So we get

Probability that the selected teacher is 18 years or more and 49 years or less = 67/75

(iv) We know that

Probability that the selected teacher is 18 years or more old = (3 + 27 + 37 + 8)/75

So we get

Probability that the selected teacher is 18 years or more old = 75/75 = 1

(v) We know that

Probability that the selected teacher is above 60 years of age = 0/75 = 0

Given, 

P (exactly one of A or B occurs) = p 

P (Exactly one of B or C occurs) = p_'

P (exactly one of C or A occurs) = p_'

and p (All three occurs simultaneously) = p². 

i. e., P(A) + P(B) – 2P(A ∩ B) = p…(i) 

P(B) + P(A) – 2P(B ∩ C) = p…(ii) 

P(C) + P(A) – 2P(C ∩ A) = p …(iii) 

and, P(A ∩ B ∩ C) = p² …(iv) 

(i) + (ii) + (iii)

⇒ P(A) + P(B) + P(C) – P(A ∩ B) – P(B ∩ C) – P(C ∩ A) = \(\frac{3}{2}p\)

We know, 

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

\(=\frac{3}{2}p + p^2\)

\(= \frac{3p + 2p^2}{2}\)

∴ The probability that at least one of the three events A, B and C occurs is \(\frac{3p + 2p^2}{2}\).

It is given that

Total number of patients = 360

(i) We know that

Probability that his age is 30 years or more but less than 40 years = 60/360 = 1/6

(ii) We know that

Probability that his age is 50 years or more but less than 70 years = (50 + 30)/360

So we get

Probability that his age is 50 years or more but less than 70 years = 80/360 = 2/9

(iii) We know that

Probability that his age is 10 years or more but less than 40 years = (90 + 50 + 60)/360

So we get

Probability that his age is 10 years or more but less than 40 years = 200/ 360 = 5/9

(iv) We know that

Probability that his age is 10 years or more = 1

(v) We know that

Probability that his age is less than 10 years = 0

Let S be the sample space. Then, n(S) = 6² = 36.Let ‘A’ and ‘B’ be the events of getting a total of 7 and 9 respectively. 

∴ A = {(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)} 

⇒ n(A) = 6 

B = {(3, 6) (4, 5), (5, 4), (6, 3)}, ⇒ n(B) = 4 

A ∩ B = ϕ ⇒ n(A ∩ B) = 0 

∴ P(A) = \(\frac{n(A)}{n(S)} = \frac{6}{36}, P(B) = \frac{n(B)}{n(S)} = \frac{4}{36}\)
and O(A ∩ B) = \(\frac{n(A∩B)}{n(S)}= 0\)

We know– 

P(A ∪ B) = P(A) + P(B) – P(A ∩ B)

= \(\frac{6}{36}+\frac{4}{36}-0\)

= \(\frac{10}{36}\)

= \(\frac{5}{18}\)

∴ The probability of getting a total of 7 or 9 is \(\frac{5}{15}\) .

Let ‘L’ and ‘H’ be the events that the young man has lungs problem and heart problem respectively.

 ∴ P(L) = 0.55, P(H) = 0.29, P(∪ H) 0 = 0.57 

P(L ∩ H) = ? 

(i) We know that, 

P(L ∪ H) = P(L) + P(H) – P(L ∩ H) 

⇒ P(L ∩ H) = P(L) + P(H) – P(L ∩ H) 

= 0.55 + 0.29 – 0.57 

= 0.27 

∴ The probability that he has both the problems is 0.27. 

(ii) p(L ∩ H’) = p(L) – p(L ∩ H) 

= 0.55 – 0.27 

= .28 

∴ The probability that he has lungs problem but not heart problem is 0.28 

∴ Out of 1000 persons, the number of persons having both the problems 

= 0.27 × 1000 = 270.32

Let E₁ : First ball drawn is white,

  E₂  =  First ball drawn G green,

A : Second ball drawn is white

The required probability by Bayes' Theorem, 

In the month of August there are 31 days.
4 weeks a month. So there are 4 × 7 = 28 days and 3 extra days in the month of August. In a week each day comes only once. So in 4 weeks each day comes 4 times.

∴ The sample space for 3 extra days is expressed as follows :

U = {(Sunday, Monday, Tuesday), (Monday, Tuesday, Wednesday), (Tuesday, Wednesday, Thursday), (Wednesday, Thursday, Friday), (Thursday, Friday, Saturday), (Friday, Saturday, Sunday), (Saturday, Sunday, Monday)}

∴ Total number of primary outcomes n = 7

A = Event that having 5 Tuesdays in the month of August of any year.
= {(Sunday, Monday, Tuesday), (Monday, Tuesday, Wednesday), (Tuesday, Wednesday, Thursday)}

∴ Favourable outcomes for the event A is m = 3

Hence, P(A) = \(\frac{m}{n} = \frac{3}{7}\)

Let length of side of smaller square = a

Then length of side of bigger square = 1.5a

Area of smaller square = a²

Area of bigger square = (1.5)²a² = 2.25a².

Probability that dart will land in the interior of the smaller square = (Area of smaller square)/(Area of bigger square) = a²/2.25a² = 1/2.25

∵ Geometrical probability,

P(E) = (measure of specified region part)/(measure of the whole region)

Area of circle with radius 0.5 m

A circle = (0.5)2 = 0.25πm²

Area of rectangle = 3 ×2=6m²

Probability (geometric) = (measure of specified region part)/(measure of whole region)

Probability that tie will land inside the circle with diameter 1m 

= (area of circle)/(area of rectangle) 

= 0.25πm²/6m²

= (1/4)x(π/6)

= π/24

Total number of possible outcomes = 12 {3 red balls, 5 black balls & 4 white balls}

(i) E ⟶ event of getting white ball

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

Probability, P(E) = 4/12 = 1/3

(ii) E ⟶ event of getting red ball

No. of favourable outcomes = 5 {3 red balls}

P (E) = 3/12 = 1/4

(iii) E ⟶ event of getting black ball

No. of favourable outcomes = 5 {5 black balls}

P (E) = 5/12

(iv) E ⟶ event of getting red

No. of favourable outcomes = 3 {3 black balls}

P(E) = 3/12 = 1/4

(Bar E) ⟶ event of not getting red.

P(Bar E) = 1 – P(E)

= 1 – 1/4

= 3/4