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False, because the outcome 3 is more likely, than the other numbers.

Total number of elementary events are : 6 

Let E be the event of getting a number less than 3 

Favorable events are: 1, 2 

Total numbers of favorable events are = 2 

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

Total numbers of elementary events are 2 + 2 = 4 

Let E be the event of getting one head 

Favorable outcome are: TH , HT 

Numbers of favorable outcomes are = 2 

P (exactly one head) = P (E) = \(\frac{2}4\) = \(\frac{1}2\)

There are total six colours on the disc. 

Sample space, 

S = {Red, Orange, Yellow, Blue, Green, Purple} 

∴ n(S) = 6 

∴ Arrow may stop on any one of the six colours.

There are total six colours on the disc.

Sample space,

S = {Red, Orange, Yellow, Blue, Green, Purple}

∴ n(S) = 6

∴ Arrow may stop on any one of the six colours.

Dates which are multiple of 5:

5, 10, 15, 20, 25, 30

∴ S = {Tuesday, Sunday, Friday, Wednesday, Monday, Saturday}

∴ n(S) = 6

∴ The days on which the date will be a multiple of 5 are Tuesday, Sunday, Friday, Wednesday, Monday and Saturday.

Given: Sample space is the set of first 500 natural numbers.

n (S) = 500

Let ‘A’ be the event of choosing the number such that it is divisible by 3

n (A) = [500/3]

= [166.67]

= 166 {where [.] represents Greatest integer function}

P (A) = n (A) / n (S)

= 166/500

= 83/250

Let ‘B’ be the event of choosing the number such that it is divisible by 5

n (B) = [500/5]

= [100]

= 100 {where [.] represents Greatest integer function}

P (B) = n (B) / n (S)

= 100/500

= 1/5

Now, we need to find the P (such that number chosen is divisible by 3 or 5)

P (A or B) = P (A ∪ B)

By using the 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)

[Since, we don’t have value of P(A ∩ B) which represents event of choosing a number such that it is divisible by both 3 and 5 or we can say that it is divisible by 15.]

n(A ∩ B) = [500/15]

= [33.34]

= 33

P (A ∩ B) = n(A ∩ B) / n (S)

= 33/500

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

= 83/250 + 1/5 – 33/500

= [166 + 100 – 33]/500

= 233/500

(b) \(\frac{55}{221}\)

S : Drawing 2 cards out of 52 cards 

⇒ n(S) = ⁵²C₂ = \(\frac{|\underline{52}}{|\underline{52}|\underline2}\) = \(\frac{52\times51}{2}\) = 1326

A : Event of drawing 2 black cards out of 26 black cards

⇒ n(A) = ²⁶C₂ = \(\frac{26\times25}{2}\) = 325

B : Event of drawing 2 kings out of 4 kings

⇒ n(B) = ⁴C₂ = \(\frac{|\underline{4}}{|\underline{2}|\underline2}\) = 6

⇒ A ∩ B : Event of drawing 2 black kings 

⇒ n(A ∩ B) = ²C₂ = 1

∴ P(A) = \(\frac{325}{1326}\), P(B) = \(\frac{6}{1326}\), P(A ∩ B) = \(\frac{1}{1326}\)

∴ P(Both cards are black or both kings) 

= P(A ∪ B) = P(A) + P(B) – P(A ∩ B)

= \(\frac{325}{1326}\) + \(\frac{6}{1326}\) - \(\frac{1}{1326}\) = \(\frac{330}{1326}\) = \(\frac{55}{221}\).

let A denote the event that the number is divisible by 4 and B denote the event that the number is divisible by 4. 

To find : Probability that the number is both divisible by 4 or 6 = 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) 

Numbers from 1 to 100 divisible by 4 are 4, 8, 12, 16, 20, 24, 28, 32, 36, 40, 44, 48, 52, 56, 60, 64, 68, 72, 76, 80, 84, 88, 92, 96, 100. 

There are 25 numbers from 1 to 100 divisible by 4 

Favourable number of outcomes = 25 

Total number of outcomes = 100 as there are 100 numbers from 1 to 100 

P(A) = \(\frac{25}{100}\)

Numbers from 1 to 100 divisible by 6 are 6,12,18,24,30,36,42,48,54,60,66,72,78,84,90,96 

There are 16 numbers from 1 to 100 divisible by 6 

Favourable number of outcomes = 16 

Total number of outcomes = 100 as there are 100 numbers from 1 to 100 

P(B) = \(\frac{16}{100}\) 

Numbers from 1 to 100 divisible by both 4 and 6 are 12,24,36,48,60,72,84,96 

There are 8 numbers from 1 to 100 divisible by both 4 and 6 

Favourable number of outcomes = 8 

P(A and B) = \(\frac{8}{100}\)

P(A or B) = P(A) + P(B) - P(A and B) 

P(A or B) = \(\frac{25}{100}+\frac{16}{100}-\frac{8}{100}\)

P(A or B) = \(\frac{25+16-8}{100}\) = \(\frac{33}{100}\)

P(A or B) = \(\frac{33}{100}\) 

The probability that the number is both divisible by 4 or 6 = P(A or B) =  \(\frac{33}{100}\) 

Given : A die is thrown twice 

To find : Probability that at least one of the two throws comes up with the number 4 

The formula used : Probability 

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

A die is numbered from 1 to 6 

When a die is thrown twice, total number of outcomes = 6² = 36 

Favourable outcomes = 

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

Favourable number of outcomes = 11 

Probability that at least one of the two throws comes up with the number 4 = \(\frac{11}{36}\) 

The probability that at least one of the two throws comes up with the number 4 =  \(\frac{11}{36}\) 

(b) \(\frac{11}{36}\)

Let S = total ways in which two dice can be rolled 

⇒ n(S) = 6 × 6 = 36 

Let A : Event of throwing 3 with 1st dice, 

B : Event of throwing 3 with 2nd dice. 

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

⇒ n(A) = 6 

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

A ∩ B = {(3, 3)} ⇒ n(A ∩ B) = 1 

∴ P(A) = \(\frac{n(A)}{n(S)}\) = \(\frac{6}{36}\), P(B) = \(\frac{n(B)}{n(S)}\) = \(\frac{6}{36}\), 

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

∴ Required probability = P(Throwing a 3 with at least one of the dice)

= P(A) + P(B) – P(A ∩ B)

= \(\frac{6}{36}\) + \(\frac{6}{36}\) - \(\frac{1}{36}\) = \(\frac{11}{36}\).

If a dice is thrown twice, it has a total of (6 × 6) = 36 possible outcomes.

If S represents the sample space then,

n (S) = 36

Let ‘A’ represent events the event such that 3 comes in the first throw.

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

P (A) = n (A) / n (S)

= 6 /36

= 1/6

Let ‘B’ represent events the event such that 3 comes in the second throw.

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

P (B) = n (B) / n (S)

= 6 /36

= 1/6

It is clear that (3,3) is common in both events so,

P (A ∩ B) = n (A ∩ B) / n (S)

= 1/ 36

Now we need to find the probability of event such that at least one of the 2 throws give 3 i.e. P (A or B) = P (A ∪ B)

By using the 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)

So, P (A ∪ B) = P (A) + P (B) – P (A ∩ B)

= 1/6 + 1/6 – 1/36

= 1/3 – 1/36

= 11/36

∴ P (at least one of the two throws comes to be 3) is 11/36

Given: 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 ace.

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 ace cards one from each suit.

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

By using the formula,

P (E) = favourable outcomes / total possible outcomes

= 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) = n (spade) / total number of cards

= 13 / 52

= 1/4

P (K) = 4/52

= 1/13

And P (S ⋂ K) = 1/52

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

P (Spade ‘or’ ace) = P(S ∪ K)

So, 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)

So, P (S ∪ K) = P (S) + P (K) – P (S ∩ K)

= 1/4 + 1/13 – 1/52

= 17/52 – 1/52

= 16/52

= 4/13

∴ P (S ∪ K) = 4/13

Since P (exactly one of A, B occurs) = q (given), we get

P (A∪B) – P ( A∩B) = q

⇒ p – P (A∩B) = q

⇒ P (A∩B) = p – q

⇒ 1 – P (A′∪B′) = p – q

⇒ P (A′∪B′) = 1 – p + q

⇒ P (A′) + P (B′) – P (A′∩B′) = 1 – p + q

⇒ P (A′) + P (B′) = (1 – p + q) + P (A′ ∩ B′)

= (1 – p + q) + (1 – P (A ∪ B))

= (1 – p + q) + (1 – p)

= 2 – 2p + q.

\(P(\bar A)=0.4\)

⇒ P(A) = 1 - 0.4 = 0.6

Now, \(P\left(\frac{A}{B}\right) = \frac{P(A\cap B)}{P(B)}\)

⇒ \(0.5=\frac{P(A\cap B)}{0.2}\)

⇒ P(A ∩ B) = 0.5 \(\times\) 0.2 = 0.1

Now P(A ∪ B) = P(A) + P(B) - P(A ∩ B)

= 0.6 + 0.2 - 0.1 = 0.8 - 0.1

= 0.7

Number divisible by 6 from 1 to 100 = 6, 12, 18, 24, 30, 36, 42, 48, 54, 60, 66, 72, 78, 84, 90, 96

Number divisible by 8 from 1 to 100 = 8, 16, 24, 32, 40, 48, 56, 64, 72, 80, 88, 96

∴ Number divisible by 6 or 8 but not by 24 from 1 to 100 = 6, 8, 12, 16, 18, 30, 32, 36, 40, 42, 54, 56, 60, 64, 66, 78, 80, 84, 88, 90.

∴ Required probability = \(\frac{20}{100}\) = \(\frac{1}{5}\)

P(A ∪ B) = P(A) + P(B) – P(A ∩ B)

⇒ 0.6 = 0.4 + p – P(A ∩ B)

⇒ P(A ∩ B) = 0.4 + p – 0.6 = p – 0.2

Since, A and B are independent events.

∴ P(A ∩ B) = P(A) × P(B)

⇒ p – 0.2 = 0.4 × p

⇒ p – 0.4 p = 0.2

⇒ 0.6 p = 0.2

⇒ p = \(\frac{0.2}{0.6}\) = \(\frac{1}{3}\)

Since, E and F are independent events.

⇒ P(E ∩ F) = P(E). P(F)

Now, P(E ∩ F') = P(E). P(E ∩ F)

= P(E) – P(E). P(F) = P(E)(1 – P(F))

⇒ P(E ∩ F′) = P(E). P(F′)

Hence, E and F′ are independent events.

Correct option is: D) All

When dice are thrown, sample space is– 

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)} 

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

B = {(1, 1), (1, 2), (2, 1)} 

C = {(5, 6), (6, 6)} 

Since, A ∩ B ≠ ϕ, B ∩ C ≠ ϕ, A ∩ C ≠ ϕ.

When two dice are rolled, sample space is– 

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)} 

A = {(1, 4}, (2, 3), (3, 2), (4, 1)} 

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

A ∩ B = {(2, 3), (3, 2)} and A ∪ B ≠ S. 

(i) A and B are not mutually exclusive. 

(ii) A and B are not exhaustive.

Let

E₁ = event that surgeries are rated as very complex

E₂ = event that surgeries are rated as complex

E₃ = event that surgeries are rated as routine

E₄ = event that surgeries are rated as simple

E₅ = event that surgeries are rated as very simple

Given: P (E₁) = 0.15, P (E₂) = 0.20, P (E₃) = 0.31, P (E₄) = 0.26, P (E₅) = 0.08

(a) P (complex or very complex) = P (E₁ or E₂) = P (E₁⋃ E₂)

By General Addition Rule:

P (A ∪ B) = P(A) + P(B) – P (A ∩ B)

⇒ P (E₁⋃ E₂) = P (E₁) + P (E₂) – P (E₁⋂ E₂)

= 0.15 + 0.20 – 0 [given] [∵ All events are independent]

= 0.35

(b) P (neither very complex nor very simple) = P (E₁’ ⋂ E₅’)

= P (E₁⋃ E₅)’

= 1 – P (E₁⋃ E₅)

[∵By Complement Rule]

= 1 – [P (E₁) + P (E₅) – P (E₁⋂ E₅)] [∵ By General Addition Rule]

= 1 – [0.15 + 0.08 – 0]

= 1 – 0.23

= 0.77

(c) P (routine or complex) = P (E₃⋃ E₂)

= P (E₃) + P (E₂) – P (E₃⋂ E₂)

[∵ By General Addition Rule]

= 0.31 + 0.20 – 0 [given]

= 0.51

(d) P (routine or simple) = P (E₃⋃ E₄)

= P (E₃) + P (E₄) – P (E₃⋂ E₄)

[∵ By General Addition Rule]

= 0.31 + 0.26 – 0 [given]

= 0.57

Let E₁, E₂, E₃, E₄ denote the events that the person A, B C and D is selected respectively. 

Also, let the probability of selecting D be ‘k’. 

Then, P(E₄) = k 

∴ P(E₁) = 2P(E₂), P(E₂) P(E₃) = 2P(E₄) = 2k 

= 2 × 2k 

= 4k 

Since E₁, E₂, E₃, E₄ are mutually exclusive and exhaustive events so– 

P(E₁ ∪ E₂ ∪ E₃ ∪ E₄) = P(E₁) + P(E₂) + P(E₃) + P(E₄) 

⇒ 1 = 4k + 2k + 2k + k 

⇒ 9k = 1 

⇒ k = 1/9 

(i) P(E₃) = 2k = \(\frac{2}{9}\)

(ii) P(E₁’) = 1 – p(E₁)

= 1 – 4k = 1 – 4(10)

= 1 − \((\frac{1}{9})\) = 1 - \(\frac{4}{9}\)

= \(\frac{5}{9}\)

When a coin is tossed three times, the sample space is– 

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

E₁ = {TTT} 

E₂ = {HTT, THT, TTH} 

E₃ = {HHH, HHT, HTH, HTT, THH, THT, TTH} 

E₁ ∩ E₂ = ϕ , E₁ ∩ E₃ = ϕ , E₂ ∩ E₃ = {HTT, THT, TTH} 

≠ ϕ 

Also, E₁ ∩ E₂E₃ = S. 

∴ E₁, E₂, E₃ are not mutually eclusive but they are exhaustive events.

Total number of possible outcomes, n(S) = 36 

Number of favorable outcomes, 

n(E) = 6

∴ P(E) = \(\frac{n(E)}{n(S)}\) = \(\frac{6}{36}\) = \(\frac{1}{6}\)

Total number of possible outcomes = 10 

Average of the numbers = \(\frac{sum\,of\,the\,given\,n\,numbers}n\) = \(\frac{30}{10}\) = 3

∴ Number of favorable outcomes, 

n(E) = 3

∴ P(E) = \(\frac{n(E)}{n(S)}\) = \(\frac{3}{10}\)

Correct option is: C) \(\frac{1}{3}\)

After removing all heart cards from the deck of 52 cards, total No of cards remaining = 52 - 13 = 39.

Total diamond card in the remaining deck of cards = 13

\(\therefore\) Probability of picking a diamond = \(\frac {Total \,diamond\, cards }{Total \,remaining \,cards} = \frac {13}{39} = \frac 13\)

Correct option is: C) \(\frac{1}{3}\)

Correct option is: D) 0.6

Possible value for probability of an event is 0.6.

Correct option is: D) 0.6

We know that probability of any event varies between 0 to 1 and cannot be negative. Since 15% = 15/100 =0 .15

Therefore, only option is (ii)

(i) Equally likely outcome: When all the outcomes in sample space have the same probability, outcomes are called equally likely outcomes.

In a toss of a coin occurrence of head or tail have equal probabilities. Thus it has an equally likely outcome of head and tail

(ii) Equally likely outcome: When all the outcomes in sample space have the same probability, outcomes are called equally likely outcomes.

Birth of a boy or a girl has an equal probability. Now Mr Sharma has one child. The child is a boy, or a girl has an equal probability and hence it is an equally likely outcome.

(iii) Equally likely outcome: When all the outcomes in sample space have the same probability, outcomes are called equally likely outcomes.

For a true false question there are only two probabilities: True or False. Now the sample space of the attempt contains two possibilities True or False.

There can be only one correct answer for the event. Hence there is equal possibilities of answer being correct or incorrect and therefore it is an equally likely outcome.

(iv) Equally likely outcome: When all the outcomes in sample space have the same probability, outcomes are called equally likely outcomes.

A player plays a ball and misses it. Now there are many possibilities of the path took bby ball after being missed by batsman. Hence occurrence of ball hitting wicket is not an equally likely outcome with missing. As it may go to wicketkeeper or it may go pass wicket keeper. Thus there are many possibilities for the event.

We know that 0<P(Event)<1.

we know that the probability of any event varies between 0 to 1 and cannot be negative.

(i) greater than 1 so it cannot be possible.

(ii) Since 67% = 67/100 = 0.67 possible

(iii) it cannot be negative.

(iv) It is within the range, so possible

(v) It is within the range, so possible

So (i) and (iii)

(i) Equally Likely Event:- Events that have same theoretical probability.

it's not equally likely Event as the car may have a mechanical breakdown or fuel problems or may be can be loose. So it depends upon a various thing, so it is not an equally likely event.

(ii) it is not equally likely. If he shoots the basket the ball will drop inside the basket, or he will miss it, this depends upon the skill of the individual basket ball player.

(i) Usually car starts. But if it has any defect, car will not start. Hence this experiment do not have equally likely outcome. 

(ii) This experiment do not have equally likely outcome because result depends upon training and aim of the player. 

(iii) This experiment has equally likely outcome becasue in truefalse question, answer is one and no possiblity for two. 

(iv) A baby is born. There is a possibility of male or female. Because out of two, one is sure event. Hence this experiment has equally likely outcome.

(i) 1 (one) 

(ii) 0; impossible event. 

(iii) 1 (one): possible event. 

(iv) 1 (one). 

(v) 0 (zero): 1 (one).

1st circle ⟶ with radius 3

2nd circle ⟶ with radius 7

3rd circle ⟶ with radius 9

Area of 1st circle = (3)² = 9π

Area of 2nd circle= (7)² = 49π

Area of 3rd circle = (9)² = 81π

Area of shaded region = Area of 2nd circle – area of 1st circle

= 49π – 9π

= 40π

Probability that will land on the shaded region = (area of shaded region)/(area of 3rd circle) =40π/81π = 40/81

Radius of circle = 1cm

Length of side of square = 1 + 1 = 2cm

Area of square = 2 × 2 = 4cm²

Area of shaded region = area of square – 4 × area of quadrant

= 4 – 4(1/4)π(1)²

= (4 − π) cm²

Probability that the point will be chosen from the shaded region = (Area of shaded region)/(Area of square ABCD) = (4 - π)/4 = 1 - π/4

Since geometrical probability,

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

We know total possible outcomes when two dice are thrown = 36

A = sum is 7

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

P(A) = \(\cfrac6{36}\)

B =second die exhibit odd number

{(1,1),(1,3),(1,5),(2,1),(2,3),(2,5),(3,1),(3,3),(3,5),(4,1),(4,3),(4,5),(5,1),(5,3),(5,5),(6,1),(6,3),(6,5)} = 18

P(B) = \(\cfrac{18}{36}\)

(A ∩ B) = sum of two number is 7 and also second die exhibit odd number

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

P(A ∩ B) = \(\cfrac{3}{36}\)

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

= \(\cfrac{\frac3{36}}{\frac{18}{36}}\) = \(\cfrac{3}{18}=\cfrac16\)