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When a fair die is tossed, the sample space is 

S = {1, 2, 3, 4, 5, 6} 

∴ n(S) = 6 

Let event A: 3 or 4 turns up. 

∴ A = {3, 4} 

∴ n(A) = 2

 ∴ P(A) = \(\frac {n(A)}{n(S)} = \frac {2} {6} = \frac {1}{3}\)

P(A’) = 1 - P(A) = 1 - 1/3 = 2/3

∴ Odds against the event A are P(A’) : P(A)

= 2/3 : 1/3

= 2:1

(a) 1 : 11

Total number of exhaustive cases = 6 × 6 = 36 A sum of 4 can be obtained as (1, 3) (2, 2) (3, 1) Therefore, there are 3 favourable outcomes and (36 – 3) = 33 unfavourable outcomes.

\(\therefore\) Odds in favour of sum of 4= \(\frac{3}{33}\) = \(\frac{1}{11}\)

Correct option is:(C) 1 : 2 

A fair dice is tossed twice. 

∴ n(S) = 36 

A: Getting 4, 5, or 6 on the first toss and Getting 1, 2, 3, or 4 on the second toss. 

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

 ∴ P(A) = \(\frac {n(A)}{n(S)} = \frac {12} {36} = \frac {1}{3}\)

∴ Required answer = P(A) : P(A’) = 1 : 2

given: two dices are thrown

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

total possible outcomes are ⁶C₁ 6C₁ 

therefore n(S)=6² = 36 

(i) let E be the event that total sum is 4 on dice 

E= {(1,3) (3,1) (2,2)} 

n(E)= ³C₁ = 3

P(E) = \(\frac{n(E)}{n(S)}\)

P(E) = \(\frac{3}{36}=\frac{1}{12}\) 

Therefore, probability of event E’ is 

P(E’) =1-P(E)

P(E’) = \(1-\frac{1}{12}=\frac{11}{12}\)

Odds in favour of getting sum as 4 is P(E):P(E’) =1:11 

(ii) let E be the event that total sum is 5 on dice 

E= {(1,4) (2,3) (3,2) (4,1)} 

n(E)= ⁴C₁ = 4

P(E) = \(\frac{n(E)}{n(S)}\)

P(E) = \(\frac{4}{36}=\frac{1}{9}\) 

Therefore, probability of event E’ is 

P(E’) = 1 - P(E)

P(E’) = \(1-\frac{1}{9}=\frac{8}{9}\)

Odds in favour of getting sum as 5 is P(E):P(E’) =1:8 

(iii) let E be the event that total sum is 6 on dice 

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

n(E)= ⁵C₁ = 5

P(E) = \(\frac{n(E)}{n(S)}\)

P(E) = \(\frac{5}{36}\) 

Therefore, probability of event E’ is 

P(E’) =1-P(E)

P(E') = \(1-\frac{5}{36}=\frac{31}{36}\) 

Odds against of getting sum as 6 is P(E’):P(E) =31:5

Given: class consisting of 10 boys and 8 girls

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

three students are selected at random, total possible outcomes are ¹⁸C₃ therefore n(S)= ¹⁸C₃ = 816 

(i) let E be the event that all are boys 

n(E)= ¹⁰C₃ = 120

P(E) = \(\frac{n(E)}{n(S)}\)

P(E) = \(\frac{120}{816}\) = \(\frac{5}{34}\) 

(ii) let E be the event that all are girls 

n(E) = ⁸C₃ = 56

P(E) = \(\frac{n(E)}{n(S)}\)

P(E) = \(\frac{56}{816}\) = \(\frac{7}{102}\) 

(iii) let E be the event that one boy and two girls are selected 

n(E) = ⁸C₁ ¹⁰C₂ = 360

P(E) = \(\frac{n(E)}{n(S)}\)

P(E) = \(\frac{360}{816}\) = \(\frac{35}{102}\) 

(iv) let E be the event that at least one girl is in the group 

E= {1,2,3} 

n(E)= ⁸C₁ ¹⁰C₂ + ⁸C₂ ¹⁰C₁+ ⁸C₃ ¹⁰C₀ = 696

P(E) = \(\frac{n(E)}{n(S)}\)

P(E) = \(\frac{693}{816}\) = \(\frac{29}{34}\) 

(v) let E be the event that at most one girl is in the group 

E= {0, 1}

n(E) = ⁸C₀ ¹⁰C₃ + ⁸C₁ ¹⁰C₂ = 480

P(E) = \(\frac{n(E)}{n(S)}\)

P(E) = \(\frac{480}{816}\) = \(\frac{10}{17}\) 

given: pack of 52 playing cards

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

one card is drawn from the given pack of cards, total possible outcomes are ⁵²C₁ therefore 

n(S)=⁵²C₁ = 52 

(i) let E be the event of getting a spade n(E) = ¹³C₁ = 13

P(E) = \(\frac{n(E)}{n(S)}\)

P(E) = \(\frac{13}{52}=\frac{1}{4}\) 

Therefore, probability of event E’ is 

P(E’) =1 - P(E)

P(E') = \(1-\frac{1}{4}=\frac{3}{4}\) 

Odds in favour of getting a spade is P(E):P(E’) =1:3 

(ii) let E be the event of getting a king 

n(E)= ⁴C₁ = 4

P(E) = \(\frac{n(E)}{n(S)}\)

P(E) = \(\frac{4}{52}=\frac{1}{13}\) 

Therefore, probability of event E’ is 

P(E’) =1 - P(E)

P(E') = \(\frac{1}{13}=\frac{12}{13}\) 

Odds in favour of getting a king is P(E):P(E’) =1:12

Given: box containing 6 red marbles numbered 1-6, 4 white marbles numbered 12-15.

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

one marble is drawn from the given box, total possible outcomes are ¹⁰C₁ therefore n(S)= ¹⁰C₁ = 10 

(i) let E be the event of getting white marble

n(E)=  ⁴C₁ = 4

P(E) = \(\frac{n(E)}{n(S)}\)

P(E) = \(\frac{4}{10}\) =  \(\frac{2}{5}\) 

(ii) let E be the event of getting white marble with odd numbered 

E= {13,15} 

n(E)= 2

P(E) = \(\frac{n(E)}{n(S)}\)

P(E) = \(\frac{2}{10}\) =  \(\frac{1}{5}\)  

(iii) let E be the event of getting even numbered marble 

E= {2, 4, 6, 12, 24} 

n(E)= 5

P(E) = \(\frac{n(E)}{n(S)}\)

P(E) = \(\frac{5}{10}\) =  \(\frac{1}{2}\)  

(iv) let E1 be the event of getting red marble

P(E₁) = \(\frac{6}{10}\) (from(i))

Let E2 be the event of getting even numbered marble 

P(E₂) = \(\frac{5}{10}\)  (from (ii)) 

Therefore (E₁ꓵ E₂) = red coloured and even numbered 

n (E₁ ꓵ E₂) =3 

P(E₁ꓵ E₂) = \(\frac{3}{10}\) 

By law of addition P(E₁∪ E₂) = P(E₁) + P(E₂) - P(E₁ ꓵ E₂)

P(E₁ U E₂) = \(\frac{6}{10}+\frac{5}{10}-\frac{3}{10}=\frac{8}{10}=\frac{4}{5}\)

Given: box containing 10 red, 20 blue and 30 green marbles.

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

five marbles are drawn from the given box, total possible outcomes are ⁶⁰C₅ therefore 

n(S) =  ⁶⁰C₅  

(i) let E be the event of getting all blue balls

n(E)= ²⁰C₅

P(E) = \(\frac{n(E)}{n(S)}\)

P(E) = \(\frac{^{20}C_5}{^{60}C_5}\) = \(\frac{34}{11977}\) 

(ii) let E be the event of getting at least one green let E’ be the event of getting no green 

E’= {(5B) (1R 4B) (2R 3B) (3R 2B) (4R 1B) (5R)}

5B = ²⁰C₅ 

1R 4B = ¹⁰C₁ ²⁰C₄ 

2R 3B = ¹⁰C₂ ²⁰C₃ 

3R 2B = ¹⁰C₃ ²⁰C₂ 

4R 1B= ¹⁰C₄ ²⁰C₁ 

5R= ¹⁰C₅ 

P(E) =1 - P(E’)

P(E’) = \(\frac{^{20}C_5+^{10}C_1^{20}C_4+^{10}C_2^{20}C_3+^{10}C_3^{20}C_2+^{10}C_4^{20}C_1+^{10}C_5}{^{60}C_5}\) 

  P(E’) = \(\frac{117}{4484}\)  

P(E) = 1 - \(\frac{117}{4484}\)  = \(\frac{4367}{4484}\) 

Correct option is: C) +1

Sure event is an event that is surely to happen.

Also, the probability of sure event is 1.

Correct option is: C) +1

Correct option(d) 2.13!/14!

Explanation:

15 persons can be seated round a table in 14! ways. If 4 persons are to be included between A and B, they can be arranged in ¹³C₄  x 4!  ways, now total number of arrangements is ¹³C₄ x 4! × 9! × 2! = 2.13!

Hence the probaility is 2.13!/14! = 2/14 = 1/7

Let A : Event of A getting a head 

⇒ \(\bar{A}\) : Event of A not getting a head 

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

Similarly, B : Event of B getting a head 

\(\bar{B}\) : Event of B not getting a head.

So P(B) = \(\frac{1}{2}\) and P(\(\bar{B}\)) = 1 - \(\frac{1}{2}\) = \(\frac{1}{2}\).

Let A start the game. A's wins if he throws a head in the 1st throw or 3rd throw or 5th throw or throw and so on.

Probability of A's winning in 1st throw = P(A) = \(\frac{1}{2}\)

Probability of A's winning in 3rd throw = P(not A). P(not B). P(A) = P(\(\bar{A}\)) x P(\(\bar{B}\)) x P(A)

= \(\frac{1}{2}\)x \(\frac{1}{2}\)x \(\frac{1}{2}\) = \(\big(\frac{1}{2}\big)^3\)

Similarly, probability of A's winning in 5th throw =  P(\(\bar{A}\)) x P(\(\bar{B}\)) x  P(\(\bar{A}\)) x P(\(\bar{B}\)) x P(A)

= \(\frac{1}{2}\)x \(\frac{1}{2}\)x \(\frac{1}{2}\)x \(\frac{1}{2}\)x \(\frac{1}{2}\) = \(\big(\frac{1}{2}\big)^5\) and so on. Since all these events are mutually exclusive,

p(A winning the game first) = \(\frac{1}{2}\) + \(\big(\frac{1}{2}\big)^3\) + \(\big(\frac{1}{2}\big)^5\) + \(\big(\frac{1}{2}\big)^7\) ... 

= \(\frac{1}{2}\)\(\bigg[\) 1 + \(\big(\frac{1}{2}\big)^2\) + \(\big(\frac{1}{2}\big)^4\) + \(\big(\frac{1}{2}\big)^6\) + ...\(\bigg]\)

= \(\frac{1}{2}\)\(\bigg[\frac{1}{1-\big(\frac{1}{2}\big)^2}\bigg]\) = \(\frac{1}{2}\) x \(\frac{1}{\frac{3}{4}}\) = \(\frac{1}{2}\) x \(\frac{4}{3}\) = \(\frac{2}{3}\).

(∵ For an infinite GP, S_∞ = \(\frac{\text{First term}}{\text{1-common ratio}}\))

Since A and B are mutually exclusive events, as either of them will win, P(B winning the game first) = 1 - \(\frac{2}{3}\) = \(\frac{1}{3}\).

A: Event of drawing a red ball and placing a green ball in the urn 

B: Event of drawing a green ball and placing a red ball 

C: Event of drawing a red ball in the second draw

P(A) = 5/7

P(B) = 2/7

P(C/A) = 4/7

P(C/B) = 6/7

Required probability 

P(C) = P(A) P(C/A) + P(B) P(C/B)

\(\frac {5}{7} \times \frac {4}{7} +\frac{2}{7}\times\frac{6}{7}\)

= 32/49

n(S) = 6, A = {2, 4, 6} 

P(A) = \(\frac{n(A)}{n(S)}\) = \(\frac{3}{6}\)= \(\frac{1}{2}\)

Here, p (x) = \(\frac{x+2}{25}\)

Putting, x = 1, 2, 3, 4, 5

P(1) = \(\frac{1+2}{25} = \frac{3}{25}\)

P(2) = \(\frac{2+2}{25} = \frac{4}{25}\)

P(3) = \(\frac{3+2}{25} = \frac{5}{25}\)

P(4) = \(\frac{4+2}{25} = \frac{6}{25}\)

P(5) = \(\frac{5+2}{25} = \frac{7}{25}\)

Now, by the definition of discrete probability distribution, we must have

(1) p(x) > 0 and (2) Σp(x) = 1.

Now, p(x_i) > 0 for (i = 1, 2, 3, 4, 5) and

Σp(x_i) = p(1) + p(2) + p(3) + p(4) + p(5)

= \(\frac{3}{25}+\frac{4}{25}+\frac{5}{25}+\frac{6}{25}+\frac{7}{25} \)= 1

Thus, conditions of probability distribution of discrete random variable are satisfied. So the given distribution is a probability distribution of a discrete random variable X.

(i) Here, let E be the event ‘getting a number greater than 4’. The number of possible outcomes is six : 1, 2, 3, 4, 5 and 6, and the outcomes favourable to E are 5 and 6. Therefore, the number of outcomes favourable to E is 2. So,

P(E) = P(number greater than 4) =2/6 =1/3

(ii) Let F be the event ‘getting a number less than or equal to 4’.

Number of possible outcomes = 6

Outcomes favourable to the event F are 1, 2, 3, 4.

So, the number of outcomes favourable to F is 4.

Therefore, P(F) =4/6 =2/3

Are the events E and F in the example above elementary events? No, they are not because the event E has 2 outcomes and the event F has 4 outcomes.

Correct option is: D) 0.41

P (Krishma wins) + P(Reshma wins) = 1

\(\Rightarrow\) 0.59 + P(Reshma wins) =1 (Given)

\(\Rightarrow\) P(Reshma wins) = 1- 0.59 = 041

Hence, the probability of Reshma winning the chess game is 0.41

Correct option is: D) 0.41

Well-shuffling ensures equally likely outcomes.

(i) There are 4 aces in a deck. Let E be the event ‘the card is an ace’. 

The number of outcomes favourable to E = 4

The number of possible outcomes = 52 

Therefore, P(E) = 4/52 = 1/13

(ii) Let F be the event ‘card drawn is not an ace’.

The number of outcomes favourable to the event F = 52 – 4 = 48

The number of possible outcomes = 52

Therefore,

P(E)=48/52=12/13

Let S and R denote the events that Sangeeta wins the match and Reshma

wins the match, respectively.

The probability of Sangeeta’s winning = P(S) = 0.62 (given)

The probability of Reshma’s winning = P(R) = 1 – P(S)

[As the events R and S are complementary]

= 1 – 0.62 = 0.38

There are 30 cards from which one card is drawn .

Total number of elementary events = n(S) = 30

(i) From numbers 1 to 30, there are 10 numbers which are multiple of 4 or 6 i.e. {4, 6, 8, 12, 16, 18, 20, 24, 28, 30}

Favorable number of events = n(E) = 10

Probability of selecting a card with a multiple of 4 or 6 = n(E)/n(S) = 10/30 = 1/3

(ii) From numbers 1 to 30, there are 2 numbers which are multiple of 3 and 5 i.e. {15, 30}

Favorable number of events = n(E) = 2

Probability of selecting a card with a multiple of 3 and 5 = n(E)/n(S) = 2/30 = 1/15

(iii) From numbers 1 to 30, there are 14 numbers which are multiple of 3 or 5 i.e. {3, 5, 6, 9, 10, 12, 15, 18, 20, 21, 24, 25, 27, 30}

Favorable number of events = n(E) = 14

Probability of selecting a card with a multiple of 3 or 5 = n(E)/n(S) = 14/30 = 7/15

We know that

Number of tests which he gets more than 60% is 2

The total number of tests is 6

So we get

Required probability = Number of tests which he gets more than 60%/ total number of tests

By substituting the values

Required probability = 2/6

On further calculation

Required probability = 1/3

Let,

(i) E be the event of getting more than 70% marks

The number of times E happens = 3

P(A) = \(\frac{3}{5}=0.6\)

(ii) F be the event of getting less than 70% marks

The number of times F happens = 2

 P(B) = \(\frac{2}{5}=0.4\)

(iii) G be the event of getting a distinction

The number of times G happens = 1

P(C) = \(\frac{1}{5}=0.2\)

Total number of unit tests taken = 5 

We know, Probability of an event = (Number of Favorable outcomes) / (Total Numbers of outcomes)

(i) Number of times student got more than 70% = 3 

Probability (Getting more than 70%) 

= \(\frac{3}{5}\) 

= 0.6 

(ii) Number of times student got less than 70% = 2 

Probability (Getting less than 70%) 

= \(\frac{2}{5}\)

= 0.4 

(iii) Number of times student got a distinction = 1 

[Marks more than 75%] 

Probability (Getting a distinction) 

= \(\frac{1}{5}\) 

= 0.2

Answer: (a) 4.04(0.8)⁸

Solution:

Given that the probability that a customer wants a replacement of a guaranteed product after 9 months is 0.2 .

Therefore, p = 0.2 and q = 1 − p = 1 − 0.2 = 0.8 .

Let X denote the number of customers who wants replacement of their guaranteed product in out of those 10 customers.

Since, each customer have independent opinion, therefore, the trials are Bernoulli trails.

Clearly, X has a binomial distribution with n = 10 and p = 0.2 .

Now, the probability that out of 10 customers at most 2 customers will desire for replacement after 9 months is P(X ≤ 2) = P(0) + P(1) + P(2)

\(^{10}C_0p^0q^{10}\) + \(^{10}C_1pq^{9}\) + \(^{10}C_2p^2q^{8}\)

= \((0.8)^{10}+10(0.2)(0.8)^9+45(0.2)^2(0.8)^8\)

= \((0.8)^8(0.64+1.6+1.8)=4.04(0.8)^8\)

given: bag which contains 4 red, 5 black and 7 white balls

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

two balls are drawn at random, therefore 

total possible outcomes are ¹⁶C₂ 

therefore n(S)=120 

(i) let E be the event of getting both white balls 

E= {(W) (W)} 

n(E)= 7C₂ = 21

P(E) = \(\frac{n(E)}{n(S)}\)

P(E) = \(\frac{21}{120}\) = \(\frac{7}{40}\) 

(ii) let E be the event of getting one black and one red ball 

E= {(B) (R)} 

n(E)= ⁵C₁ ⁴C₁ = 20

P(E) = \(\frac{n(E)}{n(S)}\)

P(E) = \(\frac{20}{120}\) = \(\frac{1}{6}\) 

(iii) let E be the event of getting both balls of same colour 

E= {(B) (B)} or {(W) (W)} or {(R) (R)} 

n(E)= ⁷C₂ + ⁵C₂ + ⁴C₂ = 37

P(E) = \(\frac{n(E)}{n(S)}\)

P(E) = \(\frac{37}{120}\) 

Given: If Coin is tossed twice times. 

To Find: Write the sample space associated to this experiment. 

Explanation: Here, two coins are tossed, that means two probability will occur at same time 

So, The sample space will be 

S={HT, TH, HH, TT} 

Hence, Sample space is {HT, HH, TT, TH}

Given: A coin is tossed once. 

To Find: Write its sample space? 

Explanation: Here, the coin is tossed only once, 

Then, there are two probability either Head(H) or Tail(T) 

So, Sample will be 

S = {H, T} 

Where, H denotes Head and T denotes Tail 

Hence, The sample is {H, T}