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Total number of discs = 90

(i) Let E₁ be the event of having a two digit number.

Number of discs bearing two digit number = 90 - 9 = 81

Let E₁ be the event of getting a good pen

Therefore P(getting  a two digit number) = P(E₁) = \(\frac{number\,of\,outcomes\,favorable\,to\,E_1}{number\,of\,all\,possible\,outcomes}\) = \(\frac{81}{90}\) = \(\frac{9}{10}\)

Thus, the probability that the discs bears a two digit number is \(\frac{9}{10}\).

(ii) Let E₂ be the event of getting a perfect squaring number.

Disc bearing perfect square numbers are 1,4,9,16,25,36,49,64 and 81.

number of discs bearing a perfect square number = 9

Therefore P(getting  a perfect square number) = P(E₂) = \(\frac{number\,of\,outcomes\,favorable\,to\,E_2}{number\,of\,all\,possible\,outcomes}\) = \(\frac{9}{90}\) = \(\frac{1}{10}\) 

Thus, the probability that the discs bears a perfect square number is \(\frac{1}{10}\)

(iii) Let E₃ be the event of getting a number divisible by 5.

Discs bearing numbers divisible by 5 are 5,10,15,20,25,30,35,40,45,50,55,60,65,70,75,80,85 and 90.

number of discs bearing a number divisible by 5 = 18

Therefore P(getting  a number divisible by 5) = P(E₃) = \(\frac{number\,of\,outcomes\,favorable\,to\,E_3}{number\,of\,all\,possible\,outcomes}\) = \(\frac{18}{90}\) = \(\frac{1}{5}\)

Thus, the probability that the discs bears a number divisible by 5 is \(\frac{1}{5}\).

No. of possible outcomes  =  90

Prime numbers less than 23  = 2,3,5,7,11,,13,17,19

No. of favourable outcomes = 8

P(prime no.less than 23) = 8/90 = 4/45

(i) P(not A) = P(A’) = 1 – P(A) = 0. 58 

(ii) P(not B} = P(B’) = 1 – P(B) = 0-52 

(iii) P(A or B) = P(A∪B) = 014

Answer: (D) = \(\frac{9}{40}\)

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

E₁ = Construction chosen is a bridge. 

E₂ = Construction chosen is a hospital. 

E₃ = Construction chosen is a hotel. 

A = Construction gets damaged. 

Since there are (200 + 400 + 600) = 1200 constructions,

\(P(E_1) =\frac{200}{1200} =\frac{1}{6}\) , \(P(E_2) =\frac{400}{1200} =\frac{1}{3} , P(E_3) =\frac{600}{1200} =\frac{1}{2}\)

Given, Probability that the construction that gets damaged is a bridge 

= P(A/E₁) = 0.01 

Similarly, P(A/E₂) = 0.15 and P(A/E₃) = 0.03 

\(\therefore\)  Probability that a hotel gets damaged in an earthquake =\(P\big(\frac{E_3}{A}\big)\) 

 = \(\frac{P(E_3) \times P(A/E_3)}{P(E_1) \times P(A/E_1) + P(E_2) \times P(A/E_2)+ P(E_3) \times P(A/E_3)}\)  (Using Bayes Th.)

= \(\frac{\frac{1}{2} \times 0.03}{\frac{1}{6} \times 0.01 + \frac{1}{3} \times 0.15 +\frac{1}{2} \times 0.03}\) 

= \(\frac{\frac{1}{2} \times 0.03}{\frac{1}{6} \times (0.01+0.3\times 0.09)}\)

=\(\frac{6}{2} \times \frac{0.03}{0.4} = \frac{6\times 3\times 10}{2 \times 4 \times 100}\)

= \(\frac{9}{40}\)

Correct answer is A.

Total trials = 600

No. Of getting prime number = 30 + 120 + 50 = 200

Let E be the event of getting a prime number so,

\(P(E)=\frac{200}{600}=\frac{2}{6}=\frac{1}{3}\)

(i) Given statement is incorrect. If 2 coins are tossed at the same time,

Total no. of possible outcomes = 4 {HH, HT, TH, TT}

P(HH) = P(HT) = P(TH) = P(TT) = 1/4 {∵Probability = (No.of favorable outcomes)/(Total no.of possible outcomes)}

i.e. for each outcome, probability of occurrence is 1/4

Outcomes can be classified as (2H, 2T, 1H & 1T) P(2H) = 1/4, P(2T) = 1/4, P(1H & 1T) = 2/4

Events are not equally likely because the event ‘one head & 1 tail’ is twice as likely to occur as remaining two.

(ii) This statement is true

When a die is thrown; total no. of possible outcomes = 6 {1, 2, 3, 4, 5, 6}

These outcomes can be taken as even no. & odd no.

P(even no.) = P(2, 4, 6) = 3/6 = 1/2

P(odd no.) = P(1,3,5) = 3/6 = 1/2

∴ Two outcomes are equally likely

Total no. of possible outcomes = 100 + 200 + 50 = 350 {100 red, 200 yellow & 50 blue}

(i) E ⟶ event of getting blue card.

No. of favourable outcomes = 50 {50 blue cards}

P(E) = 50/350 = 1/7

(ii) E ⟶ event of getting yellow card

No. of favourable outcomes = 200 {200 yellow}

P(E) = 200/350 = 4/7

Bar E⟶ event of not getting yellow card

P(Bar E)=1−P(E)

= 1−4/7 = 3/7

(iii) E ⟶ getting neither yellow nor a blue card

No. of favourable outcomes = 350 – 200 – 50 = 100 {removing 200 yellow & 50 blue cards}

P(E) = 100/350 = 2/7

1 year = 365 days

And a year has 52 weeks

So 52 × 7= 364 days

So now remaining 1 day can be any day i.e. Sunday, Monday, Tuesday, Wednesday, Thursday, Friday, Saturday

So P(probability of a non-leap year having 53 Sundays) = 1/7

Outcomes = (H,H), (H,T),(T,H),(T,T)

Atleast 1 head means that number of heads are one or more than 1.

Possible Outcomes = 3

P( that he get at least one head) = 3/4

Length of side of square JKLM = 6 cm

Area of square JKLM = 6² = 36 cm²

Since A & B are the mid points of KL & LM

KA = AL = LB = LM = 3 cm

Area of Δ AJB = area of square – area of Δ AKJ – area of Δ ALB – area of Δ BMJ

= 36−(1/2)x6x3 − (1/2)x6x3

= 36 − 9 − 4.5 − 9

= 13.5 sq. units

Probability that the point will be chosen from the interior of ΔAJB = (Area of ΔAJB)/(Area of square)

(i) Possible Outcomes: 2,3,5,7,11,13,17,19

P(a prime number) = 8/19

(ii) Possible Outcomes: 3,5,6,9,12,15,10,18

P(divisible by 3 or 5) = 8/19

(iii) Possible Outcomes: 1,2,3,4,6,7,8,9,11,12,13,14,16,17,18,19

P(neither divisible by 5 nor by 10) = 16/19

(iv) Possible Outcomes: 2,4,6,8,10,12,14,16,18

P(an even number) = 9/19

The correct option is: (C) 4/7

Explanation:

Probability that he will use the bicycle + Probability the he will use the car = 1 

.'. Probability that he will use the car = 3/7 

.'. Probability that he will use the bicycle = 1 - 3/7 = 4/7

The correct option is: (D) P → (iii), Q → (i), R → (iv), S → (ii)

The correct option is: (C)

Explanation:

Total number of possible outcomes = 19

(i) Prime numbers are 2, 3, 5,7 , 11, 13, 17 , 19. 

.'. Probability of numbers being prime = 8/19 

(ii) Numbers divisible by 3 or 5 are 3, 5, 6, 9, 10, 12, 15, 18. 

.'. Required probability = 8/19 

(iii) Numbers neither divisible by 5 nor by 10 are 1 , 2, 3, 4,6, 7, 8, 9, 11 , 12, 13, 14, ,16, 17 , 18, 19. 

.'. Required probability = 16/19 

(iv) Even numbers are 2, 4, 6, 8, 10, 12, 14, 16, 18. 

.'. Required probability = 9/19

Let the number of defective bulbs be represented by a random variable X. X may have value 0, 1, 2, 3, 4. 

If p is the probability of getting defective bulb in a single draw then p = 5/15 = 1/3

∴ q = Probability of getting non defective bulb = 1 - p = 1 - 1/3 = 2/3

Since each trial in this problem is Bernoulli trials, therefore we can apply binomial distribution as

P(X = x) = ⁿC_xq^{n - x}p^x, x = 0, 1, 2, ...n

P(X = r) = ⁴C_r(1/3)^r(2/3)^{4 - r}

Now, P(X = 1) = ⁴C₁(1/3)¹(2/3)³ = 4 x 1/3 x 8/27 = 32/81

P(X = 2) = ⁴C₂(1/3)²(2/3)² = 6 x 1/9 x 4/9 = 24/81

P(X = 3) = ⁴C₃(1/3)³(2/3)¹ = 4 x 1/27 x 2/3 = 8/81

P(X = 4) = ⁴C₄(1/3)⁴ = 1/81

Now probability distribution table is

Now mean E(X) = ∑p_ix_i = 0 x 16/81 + 1 x 32/81 + 2 x 24/81 + 3 x 8/81 + 4 x 1/81

= (32 + 48 + 24 + 4)/81 = 106/81 = 4/3

Given: A pair of dice is rolled , a coin is tossed. 

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

Explanation: A pair of dice is rolled, 

Then, No. of elementary events are 6² = 36 

Now, If outcomes is doublet means (1, 1)(2, 2)(3, 3)(4, 4)(5, 5)(6, 6), then a coin is tossed. 

If coin is tossed then no. of sample spaces is 2 

So, The total no. of elementary events including doublet = 6×2=12 

Thus, The Total number of elementary events are 30+12 = 42 

Hence, 42 events will occur for this experiments.

(i) We know that the bag containing the total balls = 5white+6red+4green = 15ballsGreen balls = 4

By using the formula,

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

∴ Probability of getting a green ball p (G) = number of green balls/total number of balls =4/15

(ii) We know that the bag containing the total balls = 5white +6red+4green = 15ballsWhite balls = 5

By using the formula,

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

∴ Probability of getting a white ball p (W) = number of white balls/total number of balls =5/15=1/3

(iii) We know that the bag containing the total balls = 4white +6red+4green = 15ballsNumber of outcomes (excluding red) = 5white+4green=9balls

By using the formula,

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

∴ Probability of getting a green ball p (G) = number of white balls/total number of balls =9/15=3/5

Total numbers of elementary events are 6 

Let E be the event of getting number greater than 4 

The favorable outcomes are: 5, 6 

Number of favorable outcomes are = 2 

P (number greater than 4) = P (E) = \(\frac{2}6\) = \(\frac{1}3\)

(i) We know that the bag containing the total balls = 4white + 5blue = 9ballsWhite balls = 4

By using the formula,

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

∴ Probability of getting a white ball p (W) = number of white balls/total number of balls =4/9

(ii) We know that the bag containing the total balls = 4white + 5blue = 9ballsBlue balls = 5

By using the formula,

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

∴ Probability of getting a blue ball p (B) = number of blue balls/total number of balls =5/9

We know that the coin has two sides head (H) and tail (T)

So the possible outcomes are X^m. (where x is the number of outcomes when a coin is tossed and m is number of coins)

(i) When there are 2 coins∴ 2²= 4 i.e. head and tail

∴ The possible outcomes are HH, HT, TH, TT.

Total possible outcomes =4

∴ Chances of getting a 2 tails = 1, i.e. TT

By using the formula,

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

∴ Probability of getting a tail p (bothT) = number of two tails occurred/total number of outcomes = ¼

(ii) When there are 2 coins∴ 2²= 4 i.e. head and tail

∴ The possible outcomes are HH, HT, TH, TT.

Total possible outcomes =4

∴ Chances of getting atleast 1 tail = 3, i.e. HT, TH, TT

By using the formula,

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

∴ Probability of getting a tail p (atleast 1T) = number of times 1 tail occurred/total number of outcomes

=3/4

(iii) When there are 2 coins∴ 2²= 4 i.e. head and tail

∴ The possible outcomes are HH, HT, TH, TT.

Total possible outcomes =4

∴ Chances of getting atmost 1 tail = 3, i.e. HT, TH, TT

By using the formula,

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

∴ Probability of getting a tail p (atmost 1T) = number of times 1 tail occurred/total number of outcomes

=3/4

(i) A coin has two sides a head(H) and a tail(T), and there are two such coins.

∴ There are X^m possible outcomes.

That is 2² = 4 and they are HH, HT, TH, TT

∴  All possible outcomes are HH, HT, TH, TT.

Total number of outcomes = 4 

Chances of getting 2 tails = 1, that is TT

Probability (P) = \(\frac{Number\,of\,favorable\,outcomes}{Total\,number\,of\,outcomes}\)

∴ Probability of getting a tail P(both T)

= \(\frac{Number\,of\,times\,two\,tails\,occured}{Total\,number\,of\,outcomes\,when\,a\,coin\,is\,tossed}\) = \(\frac{1}{4}\)

(ii) A coin has two sides a head(H) and a tail(T), and there are two such coins.

∴ There are X^m possible outcomes.

That is 2² = 4 and they are HH, HT, TH, TT

∴ All possible outcomes are HH, HT, TH, TT.

Total number of outcomes = 4 

Chances of getting at least one tail = 3, that is HT, TH, TT.

Probability (P) = \(\frac{Number\,of\,favorable\,outcomes}{Total\,number\,of\,outcomes}\)

∴ Probability of getting a tail P(atleast 1 T)

= \(\frac{Number\,of\,times\,atleast\,1\,tail\,occured}{Total\,number\,of\,outcomes\,when\,a\,coin\,is\,tossed}\) = \(\frac{3}{4}\)

(iii) A coin has two sides a head(H) and a tail(T), and there are two such coins.

∴ There are X^m possible outcomes.

That is 2² = 4 and they are HH, HT, TH, TT

∴ All possible outcomes are HH, HT, TH, TT. 

Total number of outcomes = 4 

Chances of getting at most 1 tail = 3, that is HT, TH, TT.

Probability (P) = \(\frac{Number\,of\,favorable\,outcomes}{Total\,number\,of\,outcomes}\)

∴ Probability of getting a tail P(atmost 1 T)

= \(\frac{Number\,of\,times\,atleast\,1\,tail\,occured}{Total\,number\,of\,outcomes\,when\,a\,coin\,is\,tossed}\) = \(\frac{3}{4}\)

(b) \(\frac{23}{26}\)

Total number of ways in which 3 letters can be selected from 26 letters 

= ²⁶C₃. 

If A is not to be included in the choice, there are 25 letters left, so number of ways in which 3 letters can be selected without including A 

= ²⁵C₃. 

∴ Required probability = \(\frac{^{25}C_3}{^{26}C_3}\) = \(\frac{25\times24\times23}{26\times25\times24}\) = \(\frac{23}{26}\).

(b) \(\frac{2}{7}\)

Let S be the sample space having 9 balls (3R + 4B + 2G) 

Then n(S) = Total number of ways in which 3 balls can be drawn out of the 9 balls

= \(\frac{9\times8\times7}{3\times2}\) = 84

Let A : Event of drawing three balls of different colours from the urn.

⇒ A = Event of drawing 1 red ball out of 3 red balls, 1 blue ball out of 4 blue balls and 1 green ball out of 2 green balls 

⇒ n(A) = ³C₁ × ⁴C₁ × ²C₁ = 3 × 4 × 2 = 24.

∴ P(A) = \(\frac{n(A)}{n(S)}\) = \(\frac{24}{84}\) = \(\frac{2}{7}\).

(c) \(\frac{1}{4}\) 

Let A = Event that India wins the match. Then, 

\(\bar{A}\) = Event that India loses the match.

P(A) = \(\frac{1}{2}\) and P(\(\bar{A}\)) = 1 - \(\frac{1}{2}\) = \(\frac{1}{2}\)  (∵ P(A) + P(\(\bar{A}\)) = 1)

P(Third match is India’s second winning match) 

= P(India wins the 1^st match, loses 2^nd match, wins 3^rd match) + P(India loses 1^st match, wins 2^nd match, wins 3^rd match)

(or)

⇒ Required probability = P(A \(\bar{A}\) A) +  P( \(\bar{A}\) A A)

= P(A) x P(\(\bar{A}\)) x P(A) + P(\(\bar{A}\)) x P(A) x P(A)

= \(\frac{1}{2}\) x \(\frac{1}{2}\) x \(\frac{1}{2}\) + \(\frac{1}{2}\) x \(\frac{1}{2}\) x \(\frac{1}{2}\) = \(\frac{1}{8}\) + \(\frac{1}{8}\) = \(\frac{2}{8}\) = \(\frac{1}{4}\).

Let A: Anil qualify the examination B: Ashima qualify the examination 

Given: P(A) = 0.05, P(B) = 010 and P(A∩B) = 0. 02 

(a) A’ : Anil will not qualify the exam and B’ : Asima will not qualify the exam. 

To find: P(A’∩B’) = P((A∪B)’) = 1 -P(A∩B) 

We have, P(A∪B) = P(A) +P(B)~ P(A∩B) 

= 0 05 + 0.10-0 02 

P(A∪B) = 013 

P(A’ ∩ B’) = 1 – P(A ∪ B) = 1 – 013 = 0.87

(b) We have 

P(at least one of them will not qualify the examination) 

= 1 – P (both of them will qualify) 

= 1 – P(A∩B) = 1- 0.02 =0.98 

(c) Only one of them will qualify 

= (Anil will qualify and Ashima will not) 

or 

(Anil will not qualify and Ashima will) 

= (A∩B’) or (A’∩B) = (A∩B’)∪(A’∩B) 

Required probability 

= P((A∩B’)∪(A’∩B) 

= P(A∩B’) + P(A’∩B) 

= P(A)-P(A∩B) + P(B) – P(A∩B) 

= 0-05+ 010 – 2(0.02) 

= 0 15 – 0 04 = 0.11 

Correct option is: B) 1

Let S = Sample – space. 

 A = The event that the two cards drawn are red. 

B = The event that the two cards drawn are queen. 

A∩ B = The event that the two cards drawn are queen of red colour. |S|=C (52,2), |A| =C (26,2), |B| =C (4,2) 

n(A∩ B) =C (2,2) 

P(A) = |A| / |S|=C (26,2) /C (52.2) , P(B) = |B| / |S|=C (4,2) / C(52,2) 

P(A B) = |A∩ B| / |S|= C(2,2) /C (52,2) 

P(A⋃ B) = ? 

We have P(A⋃ B) = P(A) + P(B) – P(A∩ B) = C(26,2) /C (52.2)+C( 4,2) / C(52,2)–C (2,2) /C (52,2) = 55/221

S = { (1,1), (1,2)…, (1,6), (2,1), (2,2), … (2,6), (3,1), (3,2)… (3,6), ..,(5,1), (5,2), … (5,6), (6,1,), (6,2), -…- (6,6) } |S|= 6 x 6 = 36 

(i) Let E1 = Event of getting same number on both side: E1 = { (1,1), (2,2), (3,3), (4,4), (5,5), (6,6) }; 

|E1| = 6 

 P(E1) = |E1|/|S| = 6/36 = 1/6 

(ii) Let E2 = Event of getting an even number as the sum. 

 E2 = { (1,1), (1,3), (1,5), (2,2), (2,4), (2,6), (3,1), (3,3), (3,5), (4,2), (4,4), (4,6), (5,1), (5,5), (6,2), (6,4), (6,6) } 

 |E2|= 18 hence P(E2) = |E2|/|S| = 18/36 = 1/2 

(iii) Let E3 = Event of getting a prime number as the sum.. 

E3 = { (1,1), (1,2), (1,4), (1,6), (2,1), (2,3), (2,5), (3,2), (3,4), (4,1), (4,3), (5,2), (5,6), (6,1), (6,5),} 

|E3|= 15 

P(E2) = |E3| / |S| = 15/36 = 5/12 

(iv) Let E4 = Event of getting a multiple of ‘3’ as the sum. 

E4 = { (1,2), (1,5), (2,1), (2,4), (3,3), (3,6), (4,2), (4,5), (5,1), (5,4),(6,3), (6,6),} 

|E4|= 12 P(E4) = 

|E4|/|S| = 12/36 = 1/3 

(v) Let E5 = Event of getting a total of at least 10. 

 E5 = { (4,6), (5,5), (5,6), (6,4), (6,5), (6,6), } 

|E5|= 6 P(E5) = |E5|/|S| = 6/36 = 1/6 

(vi) Let E6 = Event of getting a doublet of even numbers. 

 E6 = { (2,2), (4,4), (6,6), } |E6|= 3 

P(E6) = |E6|/ |S| = 3/36 = 1/12 

(vii) Let E7 = Even of getting a multiple of ‘2” on one dice and a multiple of ‘3’ on the other dice. E7 = { (2,3), (2,6), (4,3), (4,6), (6,3), (3,2), (3,4), (3,6), (6,2), (6,4) } 

 |E7|= 11

 P(E7) = |E7| / |S| = 11/36

Let ‘S’ be the sample – space. Then S = { HHH, HHT, HTH, THH, HTT, THT, TTH, TTT } 

(i) Let ‘E1’ = Event of getting all heads, Then E1 = { HHH } 

 |E1| = 1 

 P(E1) = |E1| / |S| = 1 / 8 

(ii) Let E2 = Event of getting ‘2’ heads. Then: 

 E2 = { HHT, HTH, THH } 

 |E2| = 3 P (E2) = 3 / 8 

(iii) Let E3 = Event of getting at least one head. Then: 

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

 |E3| = 7 

 P (E3) = 7 / 8 

(iv) Let E4 = Event of getting at least one head, Then: 

 E4 = { HHH, HHT, HTH, THH, } 

 |E4| = 4 P 

(E4) = 4/8 = 1/2

Let A be the event that student ‘A’ fails in exam and B be the event that student ‘B’ fails in exam. 

Given, P(A) = 0.2 and P(B) = 0.3 

We have to find P(A∪B) 

We know that P(A∪B) = P(A) + P(B) – P(A∩B) 

As P(A∩B) not given, but we know that – 0 ≤ P(A ∩ B) ≤ 1 

∴ P(A∪B) = 0.2 + 0.3 – P(A∩B) = 0.5 - P(A∩B) 

Clearly a +ve number will be subtracted form 0.5 to make it equal to P(A∪B). 

∴ P(A∪B) ≤ 0.5 

As our answer matches with option (c) 

∴ Option (c) is the only correct choice.

As 6 boys and 6 girls are sitting in a row. So these 12 persons can sit in 12! Ways 

Now group all 6 girls together and treat them as 1. 

Now, all girls together can sit in 7! Ways 

And girls can sit among self in 6! Ways. 

∴ total ways in which all 6 girls sit together = 7! × 6!

∴ P(E) = \(\frac{7!\times 6!}{12!}\) = \(\frac{720}{8\times9\times10\times11\times12}\) = \(\frac{1}{132}\)

Hence,  

P(E) = \(\frac{1}{132}\)

As our answer matches only with option (d) 

∴ Option (d) is the only correct choice.

3 numbers can be chosen out of 20 in ²⁰C₃ ways. 

As we have to choose 3 consecutive numbers. So we can’t choose 19 and 18. 

Choosing consecutive number is same as choosing a single number because other 2 numbers are automatically chosen. 

This can be chosen 18 ways. 

∴ P(chosen numbers are consecutive) = \(\frac{18}{{20}^C_3}\) Our answer matches with option (d) 

∴ Option (d) is the only correct choice

Total numbers of 4digit that can be formed using 0,2,3,5 without repetition= 3× 3 × 2 × 1 = 18 

For number to be divisible by 5 unit digit must end with 5 or 0. 

∴ total such numbers(4 digit) ending with zero are = 3× 2× 1(0 is fixed at the unit digit) = 6 

∴ total such numbers(4 digit) ending with zero are = 2× 2× 1(5 is fixed at the unit digit) = 4 

Total 4 digit numbers using given digit divisible by 5 are 6 + 4 =10 

∴ P(a number divisible by 5) = \(\frac{10}{18}\) = \(\frac{5}{9}\)

As our answer matches only with option (d) 

∴ Option (d) is the only correct choice

As P(A) and P(B) are mutually exclusive event. 

∴ P(A ∪ B) = P(A) + P(B) 

As P(A ∪ B) ≤ 1 

∴ P(A) + P(B) ≤ 1 

⇒ P(A) ≤ 1 – P(B) 

⇒ P(A) ≤ P(B’) 

Our answer matches with option (a) 

∴ Option (a) is the only correct choice

Given: A and B are the events such that P(A) = \(\frac{1}{2}\) and P(B) = \(\frac{7}{12}\) and 

P(not A or not B) = \(\frac{1}{4}\)

To Find: 

(i) If A and B are mutually exclusive

Since P(not A or not B) = \(\frac{1}{4}\) i.e., P\((\overline A\cup\overline B)\) = \(\frac{1}{4}\)

we know that , P\((\overline A\cup\overline B)\)) = P(A ∩ B)' = 1 - P(A ∩ B) = 0

But here P(A ∩ B) \(\neq\) 0

Therefore , A and B are not mutually exclusive.

(ii) If A and B are independent

The condition for two events to be independent is given by

P(E₁ ∩ E₂) = P(E₁) x P(E₂)

= \(\frac{1}{2}\times \frac{7}{12}\)

= \(\frac{7}{24}\) Equation 2

Since Equation 1 \(\neq\) Equation 2

\(\Rightarrow\) A and B are not independent

Finite sample space: The sample space obtained for the random experiment throwing a balanced’ six face die U = {1, 2, 3, 4, 5, 6} is the illustration of finite sample space.

Infinite sample space: The sample space obtained for the random experiment of selecting a card from a pack of 52 cards till it is ace of heart is the illustration of infinite sample space.

Impossible event: Event to get the number greater than 6 on the upper side of a balanced die.

Certain event: Event to get head or tail in tossing a balanced coin.

Let ‘S’ be the sample space. Then, n(S) = 52. 

Let A, B and C be the events of getting a king, a heart and a red card respectively. 

Then, n(A) = 4, n(B) = 13, n(C) = 26

∴ P(A) = \(\frac{n(A)}{n(S)}= \frac{4}{52}\)

P(B) = \(\frac{n(B)}{n(S)}= \frac{13}{52}\)

P(C) = \(\frac{n(C)}{n(S)}=\frac{26}{52}\)

n(A ∩ B) = 1, n(B ∩ C) = 13, n(C ∩ A) = 2, n(A ∩ B ∩ C) = 1

⇒ P(A ∩ b) = \(\frac{1}{52}\) ⇒ P(B ∩ C) = \(\frac{13}{52}\)

⇒ P(A ∩ B) = \(\frac{1}{52}\) ⇒ n(A ∩ B ∩ C) = \(\frac{1}{52}\)

We know– 

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

= \(\frac{4}{52}+\frac{13}{52}+\frac{26}{52}-\frac{1}{52}-\frac{13}{52}-\frac{2}{52}+\frac{1}{52}\)

= \(\frac{28}{52}\)

= \(\frac{7}{13}\)

Thus, the probability of getting a king, a heart of a red card is 7/13.

Number of all possible outcomes = 52

Number of favourable outcomes = 26+2 = 28

P(a king or a red) = 28/52 = 7/13

P(E) = 0.20

P(not E) = 1 - P(E)

= 1 - 0.20

= 0.80