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We know that,

Probability of occurring = 1 - the probability of not occurring 

Let’s calculate for the probability of not occurring, i.e. probability such that both of them don’t have a birthday on the same day. For suppose the first person has a birthday on a particular day then the other person can have a birthday in the remaining 364 days 

Probability of not having the same birthday = \(\frac{364}{365}\)

Probability of having same birthday = 1 – probability of not having the same Birthday  =  \(1-\frac{364}{365}\) 

=   \(\frac{1}{365}\)

Conclusion: Probability of two persons having the same  birthday is  \(\frac{1}{365}\)

Correct option is (D) 0 and 1

Probability of an event always lie between 0 and 1.

\(\because0\leq P(E)\leq1\) for any event E.

Correct option is  D) 0 and 1

Correct option is (B) 1

The probability of a certain event is 1.

Correct option is  B) 1

Answer: (b) _{\(\frac{10}{31}\)}

Let E₁, E₂, E₃, E₄ and A be the events defined as follows: 

E₁ = Event of selecting school B₁ 

E₂ = Event of selecting school B₂ 

E₃ = Event of selecting school B₃ 

E₄ = Event of selecting school B₄ 

A = Event of selecting a girl. 

Since there are four schools and each school has an equal chance of being chosen,

P(E₁) = P(E₂) = P(E₃) = P(E₄) = \(\frac{1}{4}\) 

Now, P(Girl chosen is from school B₁) = \(P\big(\frac{A}{E_1}\big)\) = \(\frac{12}{100}\)

Similarly \(P\big(\frac{A}{E_2}\big)\) =\(\frac{20}{100}\)               

 \(P\big(\frac{A}{E_3}\big)\) =\(\frac{13}{100} \)                        

 \(P\big(\frac{A}{E_4}\big)\)= \(\frac{17}{100}\)

∴ P(Girl chosen is from school B₂

_{\( = \frac{P(E_2).P(A/E_2)}{P(E_1).P(A/E_1)+P(E_2).P(A/E_2)+P(E_3).P(A/E_3)+P(E_4).P(A/E_4)} \)  Using Baye's theorem}

_{\(\frac{\frac{1}{4}\times\frac{20}{100}}{\frac{1}{4}\times \frac{12}{100}+\frac{1}{4}\times \frac{20}{100}+\frac{1}{4}\times \frac{13}{100}+\frac{1}{4}\times\frac{17}{100}}\)}

_{= \(\frac{\frac{1}{4}\times\frac{20}{100}}{\frac{1}{4}\times\frac{62}{100}}\) = \(\frac{20}{62}\) = \(\frac{10}{31}\)}

Two coins are tossed:

Sample space = {HH, HT, TH, TT}

Total number of outcomes = 4

P(getting a tail) = 1/2.

Now,

(i) P(exactly 1 head) = 2/4 = 1/2

(ii) P(at most 1 head) = 3/4

(iii) P(at least 1 head) = 3/4

In a throw of a dice,

Possible outcomes = {1, 2, 3, 4, 5, 6}

Total number of possible outcomes = 6

Now,

(i)

Favorable outcomes = Even numbers = {2, 4, 6}

Number of favorable outcomes = 3

P(an even number) = 3/6 = 1/2.

(ii)

Favorable outcomes = number less than 5 = { 1, 2, 3, 4}

Number of favorable outcomes = 4

P(a number less than 5) = 4/6 = 2/3.

(iii)

Favorable outcomes = number greater than 2 ={ 3, 4, 5, 6}

Number of favorable outcomes = 4

P(a number greater than 2) = 4/6 = 2/3.

(iv)

Favorable outcomes = a number between 3 and 6 = { 4, 5}

Number of favorable outcomes = 2

P(a number between 3 and 6) = 2/6 = 1/3.

(v)

Favorable outcomes = a number other than 3 = {1, 2, 4, 5, 6}

Number of favorable outcomes = 5

P(a number other than 3) = 5/6.

(vi)

Favorable outcome = a number 5 = { 5}

Number of favorable outcomes = 1

P(a number 5) = 1/6.

Total numbers of elementary events are: 6 

Let E be the event of getting an even number 

Favorable events are : 2, 4, 6 

Numbers of favorable events are: 3 

P (even number on dice) = P (E) = \(\frac{3}6\) = \(\frac{1}2\)

Correct option is (D) 0 ≤ P ≤ 1

\(\because\) Probability of any event is always greater or equal to 0 and less than or equal to 1.

i.e., 0 ≤ P ≤ 1

Correct option is  D) 0 ≤ P ≤ 1

n(S) = ⁶C₂ = \(\frac{6\times5}{2\times1}\) = 15 

(a) Let A: sum is at most 8 i.e max B 

A = {(O, 1) (0, 2) (0,3) (0,4) (0,5) (1,2) (1,3) (1.4) (1,5) (2,3) (2, 4) (2, 5) (3, 4) (3, 5)) 

n(A)=14

P(A) = \(\frac{n(A)}{n(S)}\) = \(\frac{14}{15}\)

(b) Let B = sum is at least 8 = {(5, 3) (4, 5)) ⇒ n(B) = 2 

⇒ n(B) = \(\frac{n(B)}{n(S)}\) = \(\frac{2}{15}\)

(c) Let C : Sum is prime number 

C = { (0. 1) (0, 2) (0, 3) (0, 5) (1, 2) (1. 4) (2, 3) (3, 4)) ⇒ n(C) = 8 

P(C) = \(\frac{n(B)}{n(S)}\) = \(\frac{8}{15}\)

When two coins are tossed, total outcomes are

{HH, HT, TH, TT} = 4

A = No tail occurs = {HH} = 1

Probability f A = P(A) = \(\cfrac14\)

B = No head occurs={TT} = 1

Probability of B = P(B) = \(\cfrac14\)

(A ∩ B) = No tail and no head occurs = not possible = 0

P (A ∩ B) = \(\cfrac04\) = 0

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

\(=\cfrac{0}{1/4} = 0\)

(c) 3/8

We know that the,

Total number of cards kept in the box = 8

∴ number of cards having a number less than 4 on it = 3

By using the formula,

Probability p () = number of favorable outcomes/ total number of outcomes

∴ Probability of selecting a card with a number less than 4 on it p(no.of cards less than 4) = number of cards having a number less than 4/total number of cards = 3/8

Total numbers of white shocks = 6 pairs 

Total numbers of black shocks = 3 pairs 

Total number pairs of shocks = 6 + 3 = 9

Probability of getting a white shock is = \(\frac{Total\,number\,of\,white\,shocks}{Total\,number\,of\,shocks}\) = \(\frac{6}{9}=\frac{2}{3}\)

Probability of getting a black shock is = \(\frac{Total\,number\,of\,black\,shocks}{Total\,number\,of\,shocks}\) = \(\frac{3}{9}=\frac{1}{3}\)

(b) 5/8

We know that the,

Total number of sectors= 3white+5green =8sectors

By using the formula,

Probability p () = number of favorable outcomes/ total number of outcomes

∴ Probability of getting a green sector p (G) = number of green sectors/total number of sectors = 5/8

Total numbers of green sectors = 3

Total numbers of blue sector = 1

Total numbers of red sector = 1

Total number of sectors = 3 + 1 + 1 = 5

Probability of getting a green sector is = Total number of green sectors/Total number of sectors

= 3/5

Probability of getting a blue sector is = Total number of blue sectors/Total number of sectors

= 1/5

Probability of getting a red sector is = Total number of red sectors/Total number of sectors

= 1/5

Yes, the probability of getting a green sector is maximum.

Total numbers of elementary event are: 50 

Let E be the event of getting a multiple of 3 or 5 

Favorable outcomes are: 3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, 36, 39, 42, 45, 48, 5, 10, 20, 25, 35, 40, 50 

Numbers of favorable outcomes are: 23 

P (multiple of 3 or 5) = P (E) = \(\frac{23}{50}\)

We have a set of natural numbers from 1 to 10 where a and b are two variables which can take values from 1 to 10.

So, total number of possible combination of a and b so that (a/2) is a fraction without replacement are: 

(1/2, 1/3, 1/4, .... 1/10)

Similarly we have 9 such sets of 10 elements each. So total number of possible combination,

= (9)(10)

= 90

Now the possible combination which makes (a/b) an integer without replacement are-

= (2/1, 3/1, 4/1, 5/1, 6/1, 7/1, 8/1, 9/1, 10/1, 4/2, 6/2, 6/3, 8/2, 8/4, 9/3, 10/2, 10/5)

= 17

Therefore the probability that (a/b) is an integer,

= [Possible combination which (a/b) an integer]/[Total Possible combination of (a/b)]

= 17/90.

Correct option is: A) No. of favourable outcomes

Correct option is: D) 0 ≤ P ≤ 1

Total numbers of elementary events are: 10 x 9 = 90 

Being numbers are of non-replacing nature and order matters 

Let E be the event of a/b as an integer 

Favorable events are: 

\(\frac{2}1\), \(\frac{3}1\) , \(\frac{4}1\), \(\frac{4}2\), \(\frac{5}1\) , \(\frac{6}1\), \(\frac{6}2\), \(\frac{6}3\) , \(\frac{7}1\), \(\frac{8}4\), \(\frac{8}2\) , \(\frac{8}1\), \(\frac{9}3\) , \(\frac{9}1\) , \(\frac{10}5\), \(\frac{10}2\) , \(\frac{10}1\) 

Numbers of favorable outcomes are: \(\frac{17}{90}\)

P (a/b is an integer) = P (E) = \(\frac{17}{90}\)

Correct option is: C) 3/5

Total numbers of elementary events are: 6 x 6 = 36 

Let E be the event of getting of showing different faces on both the dice

Numbers of favorable events on first dice are: 6 

Numbers of favorable events on second dice are: 5 

Total numbers of favorable events are = 6 × 5 = 30 

P (each dice showing different face) = \(\frac{30}{36}\) = \(\frac{10}{12}\) = \(\frac{5}6\)

Correct option is (C) \(\frac5{11}\)

Total number of S's in "ASSESSMENTS" = 5,

Total number of E's in "ASSESSMENTS" = 2,

A, M, N & T are 1 time occur in word "ASSESSMENTS".

\(\therefore\) Total letters in word "ASSESSMENTS" = 11

\(\because\) One letter is randomly selected from the word "ASSESSMENTS".

\(\therefore\) Probability that selected letter is 'S' \(=\frac{\text{Total S's in "ASSESSMENTS"}}{\text{Total letter in "ASSESSMENTS"}}\)

\(=\frac5{11}\)

Correct option is  C) S

Total numbers of elementary events are: 7 

Let E be the event of having exactly 52 Mondays 

Note- a leap year has 366 days or 52 weeks and 2 day. Two days could be any days amongst Sunday, Monday, Tuesday, Wednesday, Thursday, Friday or Saturday. 

Favorable outcomes are: Tuesday, Wednesday, Thursday, Friday, Saturday, 

Number of favorable outcome is = 5 

P (52 Mondays) = P (E) = \(\frac{5}7\)

Correct option is (C) 3/5

Red balls = n(R) = 2,

Blue balls = n(B) = 2 and

Green balls = n(G) = 1

N(S) = n(R) + n(B) + n(G)

= 2+2+1 = 5

Number of balls which are not red is

n(R') = 5 - n(R)

= 5 - 2 = 3

\(\therefore\) Probability that the selected ball is not a red ball is P(R')

\(=\frac{n(R')}{n(S)}=\frac35\)

Correct option is  C) 3/5

Correct option is (A) 0

A month have at most 31 days in a calendar year.

\(\therefore\) There are no month which have 32 days.

\(\therefore\) Favourable outcomes = 0

Total month = 12

\(\therefore\) Probability that random selected month to have 32 days \(=\frac0{12}=0.\)

Correct option is  A) 0

Total numbers of elementary events are: 12 

Let E be the event of having month as March or October 

Numbers of favorable events are = 2 

P (March or October) = P(E) = \(\frac{2}{12}\) = \(\frac{1}{6}\)

Correct option is: A) 0

Total numbers of green sectors = 3 

Total numbers of blue sector = 1 

Total numbers of red sector = 1 

Total number of sectors = 3 + 1 + 1 = 5

Probability of getting a green sector is = \(\frac{Total\,number\,of\,green\,sectors}{Total\,number\,of\,shocks}\) = \(\frac{3}{5}\)

Probability of getting a blue sector is = \(\frac{Total\,number\,of\,blue\,sectors}{Total\,number\,of\,shocks}\) = \(\frac{1}{5}\)

Probability of getting a red sector is = \(\frac{Total\,number\,of\,red\,sectors}{Total\,number\,of\,shocks}\) = \(\frac{1}{5}\)

Yes, probability of getting a green sector is maximum.

(i) List the outcomes for the event that the total is odd.

Possible outcomes of two dice are:

Outcomes for the event that the total is odd are: (2, 1), (4, 1), (6, 1), (1, 2), (3, 2), (5, 2), (2, 3), (4, 3), (6, 3), (1, 4), (3, 4), (5, 4), (2, 5), (4, 5), (6, 5), (1, 6), (3, 6), (5, 6)

(ii) Find probability of getting an odd total.

Total numbers of outcomes from two dice are 36 

From above we get that the total number of outcomes for the event that the total is odd are 18

Probability of getting an event that the total is odd = \(\frac{Total\,number\,of\,events\,with\,of\,odd\,total}{Total\,number\,of\,events}\) = \(\frac{18}{36}=\frac{1}{2}\)

(iii) List the outcomes for the event that total is less than 5. 

Total numbers of outcomes from two dice are 36 

Total number of outcomes of the events that total is less than 5 are: (1, 1), (2, 1),(3, 1), (1, 2), (2, 2) and (1, 3)

(iv) Find the probability of getting a total less than 5? 

Total numbers of outcomes from two dice are 36 

Total number of events that total is less than 5 are: (1, 1), (2, 1),(3, 1), (1, 2), (2, 2) and (1, 3)

Probability of getting an event that total is less than 5 = \(\frac{Total\,number\,of\,events\,with\,total\,less\,than\,5}{Total\,number\,of\,events}\) = \(\frac{6}{36}=\frac{1}{6}\)

Therefore the probability of getting an event that total is less than 5 is \(\frac{1}{6}\) 

Total number of ways in which 10 person can sit around a circular table = 9! 

(∵ We shall keep one place fixed and the rest of the 9 places will be filled in (9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1) ways asthere is no repetition in such case. The first variable place can be filled by any of the remaining 9 person, second place by any of the remaining 8 persons and so on.)

If two particular persons (say A and B) sit together, then the number of ways in which 9 person (considering the pair of A and B as one unit) can sit round the table = 2! × 8!. (Here A and B can interchange places amongst themselves in 2! ways)

∴ Favourable number of cases = 2 × 8!

∴ Required probability = \(\frac{2\times8!}{9!}\) = \(\frac{2\times8!}{9\times8!}\) = \(\frac{2}{9}.\)

given: word “CLIFTON”

Formula: P(E) = \(\frac{favorable\ outcomes}{total\ possible\ outcomes}\) 

In the random arrangement of the alphabets of word “CLIFTON” we have to find the probability that vowels come together 

total possible outcomes of arranging the alphabets are 7! 

therefore n(S)=7! 

let E be the event that vowels come together 

number of vowels in SOCIAL is I, O 

therefore, number of ways to arrange them so (I, O) come together 

n(E)= 6! × 2!

P(E) = \(\frac{n(E)}{n(S)}\)

P(E) = \(\frac{6!2!}{7!}\) = \(\frac{2}{7}\) 

There are 7 letters in the word SOCIETY. 

∴ Total number of ways of arranging all the 7 letters = n(S) = 7!. When the case of three vowels being together is taken, then the three vowels are considered as one unit, so the number of ways in which 5 letters (SCTY–4, IEO–1) can be arranged = 5!

Also the 3 vowels can be arranged amongst themselves in 3! ways 

∴ Total number of favourable cases = 5! × 3! 

∴ Required probability = \(\frac{5!\times3!}{7!}\) = \(\frac{1}{7}\)

Trial is the performance of an experiment, such as throwing a dice or tossing a coin.

a) Head = 45 times; Tails = 55 times 

Total = 100 

P(H), 

Probability of getting Head = 45/100 

P(T), 

Probability of getting Tail = 55/100

[P = \(\frac {No.\, of\,favourable\, outcomes}{Total \,No.\ of\, outcomes,}\)]

b) P(H) + P(T) = 45/100 + 55/100 = 100/100 = 1

Correct option is: B) Trial

Correct option is: A) event

Total number of possible outcomes, n(S) = 100 + 200 + 50 = 350 

(i) Number of favorable outcomes, 

n(E) = 50

∴ P(E) = \(\frac{n(E)}{n(S)}\) = \(\frac{1}{7}\)

(ii) Number of favorable outcomes, n(E) = 100 + 50 = 150

∴ P(E) = \(\frac{n(E)}{n(S)}\) = \(\frac{3}{7}\)

(iii) Number of favorable outcomes, 

n(E) = 100

∴ P(E) = \(\frac{n(E)}{n(S)}\) = \(\frac{2}{7}\)

Let A: student passes first examination B: student passes second examination 

Given: P(A) = 0.8, P(B) = 0 .7 and 

P(A∪B) = 0.95 

To find: P(A∩B) 

∵ P(A∩B) = P(A) + P(B) – P(A∪B) = 0.55