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Here, there are total 6 letters O, R, A, N, G, E in the word ORANGE.
And there are 3 vovels O, A, E

∴ The total number of ways of arranging these six letters,

n = ₆P⁶ = 6! = 6 × 5 × 4 × 3 × 2 × 1 = 720

A = Event of getting vowels in the first, third and sixth place.

∴ The favourable outcomes for the event A are obtained as follows:

∴ m = ₃P³ × ₃P³ = 3! × 3!

= 6 × 6 = 36

Hence, P(A) = \(\frac{m}{n} = \frac{36}{720} = \frac{1}{20}\)

Seven speakers A, B, C, D, E, F, G can deliver speech in,
n = ⁷P₇ = 7! = 7 × 6 × 5 × 4 × 3 × 2 × 1
= 5040 ways.

A = Event that speaker B delivers speech immediately after speaker A.

The favourable outcomes for the event A are obtained as follow:

All 7 speakers can deliver speech in random in the following manners:

• If A comes first, the remaining 6 speakers can deliver their speech in ₆P¹ ways.
• If A comes first and B at second, the remaining 5 speakers can deliver their speech in ₅P¹ ways.
• If C comes after A and B, the remaining 4 speakers can deliver their speech in ₄P¹ ways.
• If D comes after A, B and C, the remaining 3 speakers can deliver their speech in ₃P¹ ways.
• If E comes after A, B, C and D, the remaining 2 speakers can deliver their speech in ₂P¹ ways.
• If F comes after A, B, C, D and E, the speaker G delivers his speech in ₁P¹ ways.

∴ Favourable outcomes for the event A is

m = ₆P¹ × ₅P¹ × ₄P¹ × ₃P¹ × ₂P¹ × ₁P¹

= 6 × 5 × 4 × 3 × 2 × 1 = 720

Hence. P(A) = \(\frac{m}{n} = \frac{720}{5040} = \frac{1}{7}\)

Alternative Method:
Seven speakers A, B, C, D, E, F, G can deliver speech in n = ₇P⁷ = 7! = 5040 ways.

A = Event that speaker B delivers speech immediately after speaker A.

The favourable outcomes for A are obtained as follows:
Considering speakers A and B as one unit, total 6 speakers can deliver speech in ₆P⁶ ways and B can deliver speech immediately after speaker A in ₁P¹ way.

∴ Favourable outcomes for A

m = ₆P⁶ × ₁P¹

= 6! × 1 = 720

∴ P(A) = \(\frac{m}{n} = \frac{720}{5040} = \frac{1}{7}\)

Five members of a family can be arranged in,

n = ₅P⁵ = 5! = 5 × 4 × 3 × 2 × 1 = 120 ways.

A = Event that the husband and wife are seated next to each other.

∴ The favourable outcomes for the event A are obtained as follows:

∴ m = ₂P² × ₄P⁴

= 2! × 4!

= 2 × 24 = 48

Hence, P(A) = \(\frac{m}{n} = \frac{48}{120} = \frac{2}{5}\)

Given:

P(occurrence of event A) = 0.7

P(occurrence of event B) = 0.3

and P(occurrence of both) = 0.4

We know that,

P(A ⋂ B) = P(A) × P(B)

[when A and B are independent events]

⇒ P(A ⋂ B) = 0.7 × 0.3

⇒ P(A ⋂ B) = 0.21

But given that P(A ⋂ B) = 0.4

Hence, the given statement is False.

Probability of each student getting distinction in their final examination is less than or equal to 1

Since, the two given events are not related to the given Sample Space

∴, The sum of the probabilities of two may be 1.2.

Hence, the given statement is True.

Let E₁ and E₂ be two events such that 

E₁ = A coming to the school in time. 

E₂ = B coming to the school in time. 

Here P (E₁) = 3/7 and P (E₂) = 5/7

⇒ P(bar E₁) = 4/7, P(bar E₂) = 2/7

P (only one of them coming to the school in time) = P(E₁)P(bar E₂) + P(bar E₁)P(E₂)

= 3/7 x 2/7 + 5/7 x 4/7 = (6 + 20)/49 = 26/49

Coming to school in time i.e., punctuality is a part of discipline which is very essential for development of an individual. 

The sample space has 216 outcomes.

Now A = (1,1,4) (1,2,4) ... (1,6,4) (2,1,4) (2,2,4) ... (2,6,4) (3,1,4) (3,2,4) ... (3,6,4) (4,1,4) (4,2,4) ...(4,6,4) (5,1,4) (5,2,4) ... (5,6,4) (6,1,4) (6,2,4) ...(6,6,4) 

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

and A ∩ B = {(6,5,4)}.

Now, P(B) = 6/216 and P(A ∩ B) = 1/216

Then P(A/B) = (P(A ∩ B))/P(B) = (1/216)/(6/216) = 1/6

Let E be the event that ‘number 4 appears at least once’ and F be the event that ‘the sum of the numbers appearing is 6’.

Then, E = {(4,1), (4,2), (4,3), (4,4), (4,5), (4,6), (1,4), (2,4), (3,4), (5,4), (6,4)} 

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

We have P(E) = 11/36 and P(E) = 5/36

Also E ∩ F = {(2, 4), (4, 2)}

Therefore P(E ∩ F) = 2/36

Hence, the required probability P(E/F) = (P(E ∩ F))/P(F) = (2/36)/(5/36) = 2/5

Let the first observation be from the black die and second from the red die. 

When two dice (one black and another red) are rolled, the sample space 

S = 6 x 6 = 36 (equally likely sample events) 

(i) Let E : set of events in which sum greater than 9 and F : set of events in which black die resulted in a 5 

E = {(6,4), (4,6), (5, 5), (5,6), (6, 5), (6,6)} ⇒ n(E) = 6 

and F = {(5, 1), (5,2), (5, 3), (5, 4), (5, 5), (5,6)} ⇒ n(F) = 6 

⇒ E ∩ F = {(5, 5), (5,6)} ⇒ n(E ∩ F) = 2

The conditional probability of obtaining a sum greater than 9, given that the black die resulted in a 5, is given by P(E/F)

P(E) = 6/36 and P(F) = 6/36

Also, P(E ∩ F) = 2/36

∴ P(E/F) = (P(E ∩ F))/P(F) = (2/36)/(6/36) = 2/6 = 1/3

(ii) Let E : set of events having 8 as the sum of the observations, 

F : set of events in which red die resulted in a (in any one die) number less than 4 

⇒ E = {(2,6), (3, 5), (4,4), (5, 3), (6,2)} ⇒n(E) = 5 

and F = {(1, 1), (1, 2),….. (3, 1), (3, 2)…… (5, 1), (5, 2),……….} ⇒ n(F) = 18 

⇒ E ∩ F = {(5, 3), (6, 2)} ⇒ n(E ∩ F) = 2 

The conditional probability of obtaining a sum greater than 9, given that the black die resulted in a 5, is given by P(E/F)

P(E) = 5/36 and P(F) = 18/36

Also, P(E ∩ F) = 2/36

∴ P(E/F) = (P(E ∩ F))/P(F) = (2/36)/(18/36) = 2/18 = 1/9

Probability P(E) = 3/4. 

Probability of “not E” P (\(\overline{E}\)) = 1 – P(E)

\(=1-\frac{3}{4}=\frac{1}{4}\)

Total no. of possible outcomes when 2 dice are thrown = 6×6 = 36 which 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) }

(i) E ⟶ event of 5 not coming up on either of them

No. of favourable outcomes = 25 which 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) } Probability, P(E) = (No.of favorable outcomes)/(Total no.of possible outcomes)

P(E) = 25/36

(ii) E ⟶ event of 5 coming up at least once {(1, 5) (2, 5) (3, 5) (4, 5) (5, 5) (5, 1) (5, 2) (5, 3) (5, 4) (5, 6) (6, 5)}

P(E) = 11/36

(iii) E ⟶ event of getting 5 on both dice

No. of favourable outcomes = 1 { (5, 5) }

P(E) = 1/36

The events “ walk to work” and “ take a taxi to work” are mutually exclusive events, therefore are complementary events. 

Hence, P (take a taxi to work) = 1 – P ( walk to work) =1- \(\frac{3}{5}\) = \(\frac{2}{5}\)

(i) Number of white balls = 27 – (10 + 2) = 15

P(white ball) = \(\frac{Number\,of\,white\,balls}{Total\,number\,of\,balls}\) = \(\frac{15}{27}\) = \(\frac{5}{9}\)

(ii) P (red ball or green ball) = P (red) + P (green)

= \(\frac{10}{27}+\frac{2}{27}\) = \(\frac{12}{27}\) = \(\frac{4}{9}\) .

Odds of an event

If there are p outcomes favourable to a certain event and q outcomes unfavourable to the event in a sample space

then, odds in favour of the event = \(\frac{Number\,of\,favourable\,outcomes}{Number\,of\,unfavourable\,outcomes}\) = \(\frac{p}{q}\)

Odds against the event = \(\frac{Number\,of\,unfavourable\,outcomes}{Number\,of\,favourable\,outcomes}\) = \(\frac{q}{p}\)

If odds in favour of an event A are m : n , then 

Probability of happening of event A= P(A) = \(\frac{m}{m+n}\)

Probability of not happening of event A = P(\(\bar A\)) = \(\frac{n}{m+n}\)

Combined events: There are many times when two or more events occur together. Some common examples are: 

(i) Tossing two coins together 

There are four equally likely outcomes: {HH, HT, TH, TT} 

(ii) Tossing three coins together 

There are eight equally likely outcomes: {HHH, HTT, THH, TTH, HHT, HTH, THT, TTT} 

Ex. P ( 3 heads or 3 tails) = \(\frac{1}{8}+\frac{1}{8}\) = \(\frac{2}{8}\)= \(\frac{1}{4}\)

(iii) Throwing two dice together 

There are 36 equally likely outcomes: 

{(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), (6,2), (6,3), (6,4), (6,5), (6,6)} 

Ex. P ( a sum of 5) = \(\frac{4}{36}\) = \(\frac{1}{9}\)( \(\therefore\) There are four combinations to get a sum of 5. (1,4 ), (2,3), (3,2) and (4,1)).

Independent events: If the outcome of event A does not affect the outcome of event B, then the events A and B are called independent events. 

P ( A and B) = P(A). P(B)

Ex. Two dice are thrown. Find the probability of getting a prime number on one dice and even number on the other dice.

Here both the events are independent, so 

P ( a prime number) = \(\frac{3}{6}\) = \(\frac{1}{2}\) (\(\therefore\) There are 3 prime numbers 2, 3, 5)

P ( an even number) =\(\frac{3}{6}\) = \(\frac{1}{2}\) (\(\therefore\) There are 3 even numbers 2, 4, 6)

\(\therefore\) Reqd. probability = \(\frac{1}{2}\) x \(\frac{1}{2}\) = \(\frac{1}{4}\) 

It is something that happens, whose probability we may want to measure, as drawing a spade from a pack of 52 cards.

Sample Space is the set of all possible outcomes of a random experiment generally denoted by S. For example, when a die is thrown, S = {1, 2, 3, 4, 5, 6,}. Each element of a sample space is called a sample point.

(C) The probability that sum of the two numbers will be a multiple of 4 is 1/4. 

An event containing more than one sample point is called a compound or composite event. It is a subset of the sample spaces. 

For example, 

(i) If a die is thrown, then the statement “getting an odd number on the top” is a compound event containing the sample points 1, 3 and 5. 

Let A denote the given event, then A = {1, 3, 5} 

Sample space S = {1, 2, 3, 4, 5, 6} 

Clearly A ⊂ S

(ii) If two coins are tossed simultaneously then the sample space S is given by : 

S = {(HH), (HT), (TH), (TT)} 

If A be the event that “one head and one tail” turn up, then 

A = {(HT), (TH)}

No. Since the probability of an event cannot be more than 1.

The correct answer is(C) 5/12.

Correct option is: C) Complementary

When a card is picked up from a deck of cards, it should be either a red or a black card because these events be complementary events.

Correct option is: C) Complementary

The two dice are thrown 

Possible outcomes = 6 x 6 = 36

The sum of both faces be 9 they are,

(3,6); (6,3); (4,5) ; (5,4) = 4

P(E) = 4/36 = 1/9

correct answer is (D) 1/2.

Random drawing of cards ensures equally likely outcomes

(i) Number of face cards (King, Queen and jack of each suits) = 4 × 3 = 12

Total number of cards in deck = 52

∴Total number of possible outcomes = 52

P (drawing a face card) = 13/525 = 3/13

(ii) Number of red face cards = 2 × 3 = 6

Number of favourable outcomes of drawing red face card = 6

P (drawing of red face red)  = 6/52 = 3/26

Odd numbers = 1,3,5

Numbers less than 4 = 1,2,3

.^. . No. of favour outcomes = 4

.^. . P (an odd no. or no. < 4) = 4/6 = 2/3

Total outcomes = 52

Red face card = 6

P(red face card ) = 6/52 = 3/26

a) The possible outcomes are 1, 2, 3, 4, 5 and 6.

b) Yes. All the outcomes are equally likely since every event has equal chance of occurrence or no event has priority to occur.

c) Even : Turning up of a composite number 

Possible outcomes = 4, 6 

No. of possible outcomes = 2 

Total outcomes = 1, 2, 3, 4, 5 and 6 

Number of total outcomes = 6

Probability = \(\frac {No.\, of\, favourable\, outcomes}{Total \, no. \, of\, outcomes}\) = 2/6 = 1/3

Total number of cards in the deck = 52. 

Total number of hearts in the deck of cards =13. 

When Hearts are removed, remaining cards = 52 – 13 = 39. 

i)Picking out an Ace: 

Number of outcomes favourable to Ace = 3 [∵ ♦ A, ♥ A, ♠ A, ♣ A] 

Total number of possible outcomes from the remaining cards = 39 

– after removing Hearts. 

Probability = P(A)

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

\(=\frac{3}{39}= \frac{1}{13}\)

ii) Picking out a diamond: 

Number of favourable outcomes to diamonds (♦) = 13 

Total number of possible outcomes = 39 

∴ p(♦) \(=\frac{3}{39}= \frac{1}{13}\)

iii) Picking out a card that is ‘not a heart’: 

As all hearts are removed, the remain-ing cards are all non-heart cards. So the picked card will be definitely a non-heart card. So this is a sure event. 

Hence its probability is one 

P(E) \(=\frac{39}{39}= 1\)

Let P(E) = The probability that two students not having the same birthday = 0.992 

Then P(\(\overline{E}\)) = The complementary event of E, i.e., two students having the same birthday 

Also, P(E) + p(\(\overline{E}\)) = 1 

∴ The probability that two students have the same birthday P(\(\overline{E}\)) = 1 – P(E) 

= 1 – 0.992 = 0.008

Number of favourable outcomes to red king (♥ K, ♦ K) = 2.

Number of total outcomes = 52 (∵ Number of cards in a deck of cards = 52) 

∴ Probability of getting a red king P (Red king)

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

\(=\frac{2}{52}=\frac{1}{26}\)

i) When a die is thrown for one time, total number of outcomes = 6 

No. of outcomes favourable to a prime number (2, 3, 5) = 3 

∴ Probability of getting a prime =

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

= 3/6 

=1/2

ii) No. of outcomes favourable to a number lying between 2 and 6 (3, 4, 5) = 3 

∴ Probability of getting a number between 2 and 6

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

= 3/6 

=1/2

iii) Number of outcomes favourable to an odd number (1, 3, 5) = 3 

∴ Probability of getting an odd number P(odd)

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

= 3/6 

=1/2

Let E₁, E₂ be two events such that 

E₁ = the captain of team ‘A’ gets a six. 

E₂ = the captain of team ‘B’ gets a six. 

Here P (E₁) = 1/6 , P (E₂) = 1/6

P(E₁) = 1 - 1/6 = 5/6, P(E₂) = 1 - 1/6 = 5/6

Now P (winning the match by team A) 

= 1/6 + 5/6 x 5/6 x 1/6 + 5/6 x 5/6 x 5/6 x 5/6 x 1/6 + ....

= 1/6 + (5/6)² x 1/6 + (5/6)⁴ x 1/6 + ...

= (1/6)/(1 - 25/36) = 6/11

∴ P (winning the match by team B) =  1 - 6/11 = 5/11

The decision of re free was not fair because the probability of winning match is more for that team who start to throw dice.