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Let T denotes the event that person travels by train and A denotes that event that person travels by plane 

Given, P(T) = \(\frac{3}{5}\)  and P(A) = \(\frac{1}{4}\)  

We need to find the probability that person travels by plane or train i.e. P(T or A) = P(T∪A) 

Note: By definition of P(A or B) under axiomatic approach(also called addition theorem) we know that:

P(A∪B) = P(A) + P(B) – P(A ∩ B) 

∴ P(T∪A) = P(T) + P(A) – P(A ∩ T) 

∵ A person can never travel with both plane and train simultaneously. 

∴ P(A∩T) = 0 

Hence, 

P(T∪A) = P(T) + P(A) =   \(\frac{3}{5}\)  + \(\frac{1}{4}\) = \(\frac{17}{20}\)  

let A denote the event that a company executive will travel by plane and B denote the event of him travelling by train 

Given : P(A) = \(\frac{2}{5}\), P(B) = \(\frac{1}{3}\)

To find : Probability of a company executive will be travelling by plane or train = P(A or B)

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

Probability of a company executive will be travelling in both plane and train =P(A and B)= 0

(as he cannot be travelling by plane and train at the same time) 

P(A or B) = \(\frac{2}{5}\) + \(\frac{1}{3}\) – 0

P(A or B) = \(\frac{6+5}{15}\) = \(\frac{11}{15}\)

P(A or B) =  \(\frac{11}{15}\) 

Probability of a company executive will be travelling by plane or train= P(A or B) =  \(\frac{11}{15}\) 

let A denote the event that the card drawn is king and B denote the event that card drawn is queen. 

In a pack of 52 cards, there are 4 king cards and 4 queen cards 

Given : P(A) = \(\frac{4}{52}\), P(B) = \(\frac{4}{52}\)

To find : Probability that card drawn is king or queen = P(A or B) 

The formula used : Probability = 

\(\frac{favourable\,number\,of\,outcomes}{total\,number\,of\,outcomes}\)

P(A or B) = P(A) + P(B) - P(A and B) 

P(A) = \(\frac{4}{52}\) (as favourable number of outcomes = 4 and total number of outcomes = 52) 

P(B) = \(\frac{4}{52}\) (as favourable number of outcomes = 4 and total number of outcomes = 52) 

Probability that card drawn is king or queen = P(A and B)= 0

(as a card cannot be both king and queen in the same time)

P(A or B) =   \(\frac{4}{52}+\frac{4}{52}\) – 0 

P(A or B) = \(\frac{4+4}{52}\) = \(\frac{8}{52}\) =   \(\frac{2}{13}\)

P(A or B) =  \(\frac{2}{13}\)

Probability of a card drawn is king or queen = P(A or B) =  \(\frac{2}{13}\)

In a pack of 52 cards, there are 4 king cards and 4 queen cards

Let A denote the event that the card drawn is queen and B denote the event that card drawn is king.

Then,

P(A) = 4/52 and P(B) = 4/52

As a card cannot be both king and queen in the same time, so P(A and B) = 0

Using formula, P(A or B) = P(A) + P(B) – P(A and B)

P(A or B) = 4/52 + 4/52 – 0

= 2/13

Probability of a card drawn is king or queen = P(A or B) = 2/13

In a pack of 52 cards, there are 4 queen cards and 13 heart cards.

Let A denote the event that the card drawn is queen and B denote the event that card drawn is heart.

Then,

P(A) = 4/52 and P(B) = 13/52

As there is one card which is both queen and heart (queen of hearts), so P(A and B)= 1/52

Using formula, P(A or B) = P(A) + P(B) – P(A and B)

P(A or B) = 4/52 + 13/52 – 1/52

= 16/52

= 4/13

Probability of a card drawn is either a queen or heart = P(A or B) = 4/13

(d) 2²⁶ × ⁵²C₂₆ | ¹⁰⁴C₂₆

Since there are 52 distinct cards in a deck and each distinct card is 2 in number.

∴2 decks will also contain only 52 distinct cards, two each.

∴ Probability that the player gets all distinct cards = \(\frac{^{52}C_{26}\times2^{26}}{^{104}C_{26}}\).

Total numbers of elementary events are: 6 

Let E be the event of getting a number greater than 2 

Favorable outcomes are: 3, 4, 5, 6 

Numbers of favorable outcomes are: 4 

P (number>2) = P (E) = \(\frac{4}6\) = \(\frac{2}3\)

Given, 3 white (3 W), 3 red (3 R), 4 green (4 G), 4 black (4 B) balls 

Total no. of balls = 3 + 3 + 4 + 4 = 14 

Two balls are to be drawn, one by one without replacement. 

There are 4 possibilities.

Since all cases are mutually exclusive, therefore

Reqd. Probability = \(\frac{3}{14}\) x \(\frac{11}{13}\) + \(\frac{3}{14}\) x \(\frac{11}{13}\) + \(\frac{4}{14}\) x \(\frac{10}{13}\) + \(\frac{4}{14}\) x \(\frac{10}{13}\) = \(\frac{33+33+40+40}{14\times13}\) = \(\frac{146}{182}\) = \(\frac{73}{91}.\)

given: 2 white, 3 red, 5 green, 4 black

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

two balls are drawn one by one, we have to find the probability that they are of different colours

total possible outcomes are ¹⁴C₂ 

therefore n(S)= ¹⁴C₂ = 91

 let E be the event that all balls are of same colour 

E= {WW, RR, GG, BB} 

n(E)= ²C₂ + ³C₂ + ⁵C₂ + ⁴C₂ = 20 

probability of occurrence is

P(E) = \(\frac{n(E)}{n(S)}\)

P(E) = \(\frac{20}{91}\) 

Therefore, the probability of non-occurrence of the event (all balls are different) is

P(E') = 1- P(E) 

P(E') = \(1-\frac{20}{91}=\frac{71}{91}\) 

Given: A coin is tossed and a die is thrown. 

To Find: Write the sample space for the given experiment.

Explanation: Here, The coin is tossed and die is thrown. 

We know, when coin is tossed there will be 2 events either Head or Tail. 

And, when die is thrown then there will be 6 faces (1, 2, 3, 4, 5, 6)

So, The total number of Sample space together is 2×6 = 12 

S = {(H, 1), (H, 2), (H, 3), (H, 4), (H, 5), (H, 6), (T, 1), (T, 2), (T, 3), (T, 4), (T, 5), (T, 6)} 

Hence, Sample space are {(H, 1), (H, 2), (H, 3), (H, 4), (H, 5), (H, 6), (T, 1), (T, 2), (T, 3), (T, 4), (T, 5), (T, 6)}

Given: A coin is tossed twice. If the second throw results I a tail, A die is thrown. 

To Find: Write the sample space for the given experiment. 

Explanation: When a coin tossed twice, Then sample spaces for only coin will be: {HH, TT, HT , TH} 

Now, According to question , when we get Tail in second throw, then a dice is thrown. 

So, The total number of elementary events are 2+(2×6)=14 

And sample space will be 

S = {HH, TH, (HT, 1), (HT, 2), (HT, 3), (HT, 4), (HT, 5), (HT, 6), (TT, 1), (TT, 2), (TT, 3), (TT, 4), (TT, 5), (TT, 6)} 

Hence, this is the sample space for given experiment.

Sample space, 

S = {HH, HT, TH, TT}.

When we toss a coin, sample space, S = {H, T} 

Let A = event of getting head = {H} 

And B = event of getting tail = {T} 

Then, A ∩ B and A ∪ B = s = ϕ

∴ A, B are mutually exclusive and exhaustive.

Given, 

P(not A) = 0.65 

⇒ 1 – P(A) = 0.65 

⇒ P(A) = 0.35 

P(A ∪ B) = 0.65, p(B) =p . 

Since A, B are mutually exclusive, so–

 P(A ∪ B) = P(A) + P(B) 

⇒ 0.65 = 0.35 + P 

⇒ P = 0.30

Given: E₁ and E₂ are two events such that P(E₁) = \(\frac{1}{3}\) and P(E₂) = \(\frac{3}{5}\)

To Find: 

(i) P(E₁ ∪ E₂) when E₁ and E₂ are mutually exclusive.

We know that,

When two events are mutually exclusive P(E₁ ∩ E₂) = 0

Hence, P(E₁ ∪ E₂) = P(E₁) + P(E₂)

= \(\frac{1}{3}+\frac{3}{5}\)

= \(\frac{14}{15}\)

Therefore , P(E₁ ∪ E₂) = \(\frac{14}{15}\) when E₁ and E₂ are mutually exclusive.

(ii) P(E₁ ∩ E₂) when E₁ and E₂ are independent.

We know that when E₁ and E₂ are independent ,

P(E₁ ∩ E₂) = P(E₁) x P(E₂)

= \(\frac{1}{3}\times \frac{3}{5}\)

= \(\frac{1}{5}\)

Therefore, P(E₁ ∩ E₂) = \(\frac{1}{5}\) when E₁ and E₂ are independent.

Since E₁, E₂, E₃ are mutually exclusive and exhaustive events, so 

E₁ ∩ E₂ =ϕ , E₂ ∩ E₃ = ϕ, E₁ ∩ E₃ = ϕ, E₁ ∩ E₂ ∩ E₃ = ϕ and E₁ ∪ E₂ ∪ E₃ = S 

∴ p(E₁ ∪ E₂ ∪ E₃) = E₁ ∩ E₂ ∩ E₃ = ϕ p(E₁) + p(E₂) + p(E₃)

⇒ P(S) = P(E1) + \(\frac{2}{3}\)P(E₁) + 2P(F₁)

⇒ 1 = P(F₁) + \(\frac{8}{3}\)P(F₁)

⇒ \(\frac{11}{3}\)P(E₁) = 1

⇒ P(E₁) = \(\frac{3}{11}\)

(i) P (E₁ U E₂), when E₁ and E₂ are mutually exclusive

We know that when two events are mutually exclusive

P (E₁ ∩ E₂) = 0

So P (E₁ U E₂) = P (E₁) + P (E₂)

By substituting the values

= 1/3 + 3/5

= 14/15

Hence, P (E₁ U E₂) = 14/15, when E₁ and E₂ are mutually exclusive.

(ii) P (E₁ ∩ E₂), when E₁ and E₂ are independent

We know that when two events are independent

P (E₁ ∩ E₂) = P (E₁) × P (E₂)

By substituting the values

= 1/3 × 3/5

= 1/5

Hence, P (E₁ ∩ E₂) = 1/5 when E₁ and E₂ are independent.

When three fair coins are tossed, the sample space is 

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

∴ n(S) = 8 

Let event A: Getting three heads. 

∴ A = {HHH} 

Let event B: Getting at least two heads.

∴ B = {HHT, HTH, THH, HHH} 

∴ n(B) = 4

∴ P(B) = \(\frac {n(B)}{n(S)} = \frac {4} {8}\)

Now, A ∩ B = {HHH} 

∴ n(A ∩ B) = 1

 ∴ P(A ∩ B) = \(\frac {n(A∩B)}{n(S)} = \frac {1} {8}\)

∴ Probability of getting three heads, given that at least two coins show heads, is given by

P(A/B) = \(\frac {P(A∩B)}{P(B)} \)

= \(\frac {\frac{1}{8}}{\frac{4}{8}}\)

= 1/4.

Then

P (A) = P (B) = P (C) = \(\frac{2}{4}\) = \(\frac{1}{2}\) and

P (A ∩ B) = P ({HH}) = \(\frac{1}{4}\), P (A ∩ C) = P ({HT}) = \(\frac{1}{4}\)

P (B ∩ C) = P ({TH}) = \(\frac{1}{4}\), (A ∩ B ∩ C) = ϕ

∴ P (A ∩ B ∩ C) = P (ϕ) = 0 ≠ P (A). P (B). P (C)

Thus condition (i) is satisfied, i.e., the events are pairwise independent. But condition (ii) is not satisfied and so the three events are not independent

Correct option is B. dependent

since event A is two heads come and event B is last should be head, both are dependent.

Given, Sample space is the set of first 200 natural numbers. 

∴ n(S) = 200 

Let A be the event of choosing the number such that it is divisible by 6 

∴ n(A) = \([\frac{200}{6}] \)= [33.334] = 33 {where [.] represents Greatest integer function} 

∴ P(A) = \(\frac{n(A)}{n(S)} =\frac{33}{200}\)

Let B be the event of choosing the number such that it is divisible by 8 

∴ n(B) = \([\frac{200}{8}] \)= [25] = 25 {where [.] represents Greatest integer function} 

∴ P(B) = \(\frac{n(B)}{n(S)} = \frac{25}{200}\) 

We need to find the P(such that number chosen is divisible by 6 or 8) 

∵ P(A or B) = P(A∪B) 

Note: By definition of P(E or F) under axiomatic approach(also called addition theorem) we know that: 

P(E∪F) = P(E) + P(F) – P(E ∩ F) 

∴ P(A∪B) = P(A) + P(B) – P(A∩B) 

We don’t have value of P(A∩B) which represents event of choosing a number such that it is divisible by both 4 and 6 or we can say that it is divisible by 24. 

n(A∩B) = \([\frac{200}{24}]\) = [8.33] = 8 

∴ P(A∩B) = \(\frac{nP(A ∩ B)}{n(S)} =\frac{8}{200}\) 

∴ P(A∪B) = \(\frac{33}{200}+\frac{25}{200}-\frac{8}{200}\) 

= \(\frac{5}{200}= \frac{1}{4}\) 

Given, Sample space is the set of first 1000 natural numbers.

∴ n(S) = 1000 

Let A be the event of choosing the number such that it is multiple of 2 

∴ n(A) = [1000/2] = [500] = 500 {where [.] represents Greatest integer function}

∴P(A) = \(\frac{n(A)}{n(S)}\) = \(\frac{500}{1000}\) 

Let B be the event of choosing the number such that it is multiple of 9 

∴ n(B) = [1000/9] = [111.11] = 111 {where [.] represents Greatest integer function}

∴P(A) = \(\frac{n(A)}{n(S)}\) = \(\frac{111}{1000}\) 

We need to find the P(such that number chosen is multiple of 2 or 9) 

∵ P(A or B) = P(A∪B)

Note: By definition of P(E or F) under axiomatic approach(also called addition theorem) we know that: 

P(E∪F) = P(E) + P(F) – P(E ∩ F) 

∴ P(A ∪ B) = P(A) + P(B) – P(A ∩ B)

We don’t have value of P(A∩B) which represents event of choosing a number such that number is a multiple of both 2 and 9 or we can say that it is a multiple of 18.

n(A∩B) = [1000/18] = [55.55] = 55

∴ P(A ∩ B) = \(\frac{n(A∪B)}{n(s)}\) = \(\frac{55}{1000}\)

∴ P(A ∩ B) =   \(\frac{500}{1000}+\frac{111}{1000}-\frac{55}{1000}\) 

= \(\frac{556}{1000}=\frac{139}{250}​​ ​\)

When a coin is tossed 4 times. A total of 2⁴ = 16 outcomes are possible. 

Let S be the set consisting of all such outcomes. 

∴ n(S) = 16 

Let A be the event of getting 2 tails. 

∴ A = {TTHH,THTH,THHT,HTTH,HTHT,HHTT} 

∴ n(A) = 6 

∴ P(A) = \(\frac{6}{16} \)= \(\frac{3}{8}\) 

Let B be the event of getting 3 tails. 

∴ B = { TTTH ,TTHT, THTT,HTTT } 

⇒ n(B) = 4

∴ P(B) = \(\frac{4}{16}\)= \(\frac{1}{4}\)

We need to find the probability of getting 2 tails or 3 tails i.e. 

P(A∪B) = ? 

As we can’t get 2 and 3 tails at the same time. So A and B are mutually exclusive events. 

∴ P(A∪B) = P(A) + P(B) =  \(\frac{3}{8}\) + \(\frac{1}{4}\)= \(\frac{5}{8}\)

As a card is drawn from a deck of 52 cards 

Let S denotes the event of card being a spade and K denote the event of card being King. 

As we know that a deck of 52 cards contains 4 suits (Heart ,Diamond ,Spade and Club) each having 13 cards. The deck has 4 king cards one from each suit. 

We know that probability of an event E is given as-

P(E) = \(\frac{number\,of\,favourable\,outcomes}{total\,number\,of\,outcomes}\)   = \(\frac{n(E)}{n(s)}\)

Where n(E) = numbers of elements in event set E 

And n(S) = numbers of elements in sample space. 

Hence,

P(S) =  \(\frac{n(spade)}{total\,number\,of\,cards}\) =  \(\frac{13}{52}=\frac{1}{4}\) 

P(K) = \(\frac{4}{52}=\frac{1}{13}\)

And P(S∩K) = \(\frac{1}{52}\)

We need to find the probability of card being spade or king, i.e.

P(Spade ‘or’ King) = P(S∪K) 

Note: By definition of P(A or B) under axiomatic approach(also called addition theorem) we know that: 

P(A∪B) = P(A) + P(B) – P(A∩B) 

∴ P(S∪K) = P(S) + P(K) - P(S∩K) 

⇒ P(S∪K) = \(\frac{1}{4}+\frac{1}{13}-\frac{1}{52}\) 

= \(\frac{17}{52}-\frac{1}{52}\) = \(\frac{16}{52}=\frac{4}{13}\)

∴ P(S∪K) = 4/13 

∴ Option (c) is the correct choice.