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(b) \(\frac{1}{462}\)

Let S be the sample space. 

Then, n(S) = Number of ways in which 6 boys and 6 girls can sit in a row = 12! 

Let E : Event of 6 girls and 6 boys sitting alternately. 

Then, the 6 girls and 6 boys can be arranged in alternate position in two ways. 

Ist way : We start with a boy. Then the arrangement is : B G    B G    B G    B G    B G    B G 

∴ Number of ways of arranging 6 boys in 6 places = 6! 

Number of ways of arranging 6 girls in 6 places = 6! 

∴ Number of ways of arranging 6 boys and 6 girls in alternate places = 6! × 6! 

Similarly, 

IInd way : Here we start with a girl. Then the arrangement is G B  G B    G B     G B     G B    G B 

∴ Number of ways of arranging 6 boys and 6 girls alternately this way 

= 6! × 6!

∴ n(E) = 6! × 6! + 6! × 6! 

= 2 × 6! × 6!

∴ n(E) = \(\frac{n(E)}{n(S)}\) = \(\frac{2\times6!\times6!}{12!}\)

= \(\frac{2\times6\times5\times4\times3\times2\times1}{12\times11\times10\times9\times8\times7}\) = \(\frac{1}{462}\).

(i) 6 boys and 6 girls sit in a row randomly. 

Total ways of their seating = 12! 

No. of ways in which all the 6 girls sit together = 6! x 7! (Considering all 6 girls as one person) 

∴ Probability of all girls sitting together = 6! x 7!/12! = 720/12 x 11 x 10 x 9 x 8 = 1/132 

(ii) Staring with boy, boys can sit in 6! Ways leaving one place between every two boys and two one a last. 

B _ B _ B _ B _ B _ B _

These left over places can be occupied by girls in 6! ways. 

∴ If we start, with boys. No. of ways of seating boys and girls alternately = 6! x 6! In the similar manner, if we start with girl, no. of ways of seating boys and girls alternately 

= 6! x 6! 

G _ G _ G _ G _ G _ G _ 

Thus total ways of alternate seating arrangements 

= 6! x 6! + 6! x 6! = 2 x 6! x 6! 

∴ Probability of making alternate seating arrangement for 6 boys and 6 girls 

= 2 x 6! x 6!/12! 

= 2 x 720/12 x 11 x 10 x 9 x 8 x 7 = 1/462 

Let p be the probability that B gets selected.

P (Exactly one of A, B is selected) = 0.6 (given)

P (A is selected, B is not selected; B is selected, A is not selected) = 0.6

P (A∩B′) + P (A′∩B) = 0.6

P (A) P (B′) + P (A′) P (B) = 0.6

(0.7) (1 – p) + (0.3) p = 0.6

p = 0.25

Thus the probability that B gets selected is 0.25.

Let E₁ be the event that A is selected in engineering college

and E₂ be the event that B is selected in engineering college

Then, we have

P(E₁) = 0.5, P(E₂) = 0.7 and P(E₁⋂ E₂) ≤ 0.3

Now, P(E₁) × P(E₂) ≤ 0.3

⇒ 0.5 × P(E₂) ≤ 0.3

⇒ P(E₂) ≤ 0.6

But P(E₂) = 0.7

Hence, the given statement is False.

Total no. of cards = 52

All jacks, queens & kings, aces of red colour are removed.

Total no. of possible outcomes = 52 – 2 – 2 – 2 – 2 = 44 {remaining cards}

(i) E ⟶ event of getting a black queen

No. of favourable outcomes = 2 {queen of spade & club}

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

P(E) = 2/44 = 1/22

(ii) E ⟶ event of getting a red card

No. of favourable outcomes = 26 – 8 = 18 {total red cards – jacks, queens, kings, aces of red colour}

P(E) = 18/44 = 9/22

(iii) E ⟶ event of getting a black jack

No. of favourable outcomes = 2 {jack of club & spade}

P(E) = 2/44 = 1/22

(iv) E ⟶ event of getting a picture card

No. of favourable outcomes = 6 {2 jacks, 2 kings & 2 queens of black colour}

P(E) = 6/44 = 3/22

We know that,

Total no. of cards = 52

All jacks, queens & kings, aces of red colour are removed.

Total no. of possible outcomes = 52 – 2 – 2 – 2 – 2 = 44 (remaining cards)

(i) Let E = event of getting a black queen

No. of favourable outcomes = 2 (queen of spade & club)

Probability, P(E) = Number of favourable outcomes/ Total number of outcomes

P(E) = 1/22

(ii)  Let E = event of getting a red card

No. of favourable outcomes = 26 – 8

= 18 (total red cards jacks – queens, kings, aces of red colour)

Probability, P(E) = Number of favourable outcomes/ Total number of outcomes

P(E) = 18/44 = 9/22

(iii)  Let E = event of getting a black jack

No. of favourable outcomes = 2 (jack of club & spade)

Probability, P(E) = Number of favourable outcomes/ Total number of outcomes

P(E) = 2/44 = 1/22

(iv) Let E = event of getting a picture card

No. of favourable outcomes = 6 (2 jacks, 2 kings & 2 queens of black colour)

Probability, P(E) = Number of favourable outcomes/ Total number of outcomes

P(E) = 6/44 = 3/22

Let A denote the event that the candidate A is selected and B the event that B is selected. It is given that

P (A) = 0.5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (1) 

P (A ∩ B) ≤ 0.3 . . . . . . . . . . . . . . . . . . . . . . . . . . (2) 

Now, P (A) + P (B) - P (A ∩ B) = P (A ∪ B) ≤ 1 

Or 0.5 + P (B) - P (A ∩ B) ≤ 1 [Using (1)] 

Or P (B) ≤ 0.5 + P (A ∩ B) ≤ 0.5 + 0.3 [Using (2)] 

Or P (B) ≤ 0.8 ∴ P (B) cannot be 0.9

One vehicle is selected from 40 vehicles. 

Let event A: The selected vehicle is red. There are total of 18 red vehicles.

∴ P(A) =\(\frac {^{18}C_1}{^{40}C_1} = \frac {18} {40} = \frac {9}{20}\)

Let event B: The selected vehicle is a truck. There are total of 6 trucks. Since 2 trucks are red, they are common between A and B.

∴ P(A ∩ B) = \(\frac {^{2}C_1}{^{40}C_1} = \frac {2} {40} = \frac {1}{20}\)

∴ Probability that the selected vehicle is a truck under the condition that it is red = P(B/A)

= \(\frac {(PA∩B)}{P(A)}\)

= \(\frac {\frac{1}{20}}{\frac{9}{20}}\)

= 1/9.

Probability of an event cannot be negative. Here P(C) = – 0.06. 

∴ the above set of events are not possible. 

P(A) = 1/3, P(B) = 1/6, P(C) = 2/9, P(D) = 5/18

P(A) = 0.5, P(B) = 0.3 

Here A and B are mutually exclusive.

(i) P(A ∪ B) = P(A) + P(B) 

= 0.5 + 0.3 = 0.8 

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

= 0.5 + 0.3 – 0.8

P(A ∩ B) = 0 

P(A ∩ \(\bar{B}\)) = P(A) – P(A ∩ B) = 0.5 – 0 = 0.5 

(iii) P(\(\bar{A}\) ∩ B) = P(B) – P(A ∩ B) = 0.3 – 0 = 0.3

Total number of possible outcomes, n(S) = 52 – 2 – 2 – 2 – 2 = 44 

(i) Number of favorable outcomes, 

n(E) = 2

∴ P(E) = \(\frac{n(E)}{n(S)}\) = \(\frac{2}{44}\) = \(\frac{1}{22}\)

(ii) Number of favorable outcomes, n(E) = 26 – 8 = 18

∴ P(E) = \(\frac{n(E)}{n(S)}\)  = \(\frac{18}{44}\) = \(\frac{9}{22}\)

(iii) Number of favorable outcomes, 

n(E) = 2

∴ P(E) = \(\frac{n(E)}{n(S)}\)  = \(\frac{2}{44}\) = \(\frac{1}{22}\)

(iv) Number of favorable outcomes, 

n(E) = 6

∴ P(E) = \(\frac{n(E)}{n(S)}\)  = \(\frac{6}{44}\) = \(\frac{3}{22}\)

In a pack of 52 cards, there are 13 diamond cards. 

Let event A: The first card drawn is a diamond card.

∴ P(A) = \(\frac {^{13}C_1}{^{52}C_1} = \frac {13} {52} = \frac 1 4\)

(i) Let event B: The second card drawn is a diamond card. 

Since the first diamond card is kept aside, we now have 51 cards, out of which 12 are diamond cards. 

Probability that the second card is a diamond card under the condition that the first diamond card is kept aside in the pack 

=P(B/A) = \(\frac {^{12}C_1}{^{51}C_1} = \frac {4} {17}\)

∴ Required probability = P(A ∩ B)

= P(B/A) . P(A)

= 1/4 x 4/17

= 1/17

(ii) Let event B: The second card drawn is a diamond card. 

Since the first diamond card is replaced in the pack, we now again have 52 cards, out of which 13 are diamond cards. 

∴ Probability that the second card is a diamond card under the condition that the first diamond card is replaced in the pack =

P(B/A) = \(\frac {^{13}C_1}{^{52}C_1} = \frac {13} {52} = \frac 14\)

Required probability = P(A ∩ B) 

= P(B/A) . P(A)

= 1/4 x 1/4

= 1/16

Given A and B are mutually exclusive and P(A) = 0.5, P(B) = 0.3 

∴ P(A ∪ B) = P(A) + P(B) = 0.5 + 0.3 = 0.8 

So, P(A’ ∩ B’) = P{(A ∪ B)’} = 1 – P(A ∪ B) 

= 1 – 0.8 = 0.2

(i) Given P(A) = 0.36, P(A ∪ B) = 0.09, P(A ∩ B) = 0.25 

P(A ∪ B) = P(A) + P(B) – P(A ∩ B) 

(i.e.,) 0.90 = 0.36 + P(B) – 0.25 

0.90 = 0.11 + P(B) 

∴ P(B) = 0.90 – 0.11 = 0.79

(ii)  P(\(\bar{A}\) ∩ \(\bar{B}\)) = P{(A’ ∪ B)’} (Demorgan Law) 

P(A ∪ B)’ = 1 – P(A ∪ B) = 1 – 0.90 = 0.1

Given : P(A) = 0.60, P(A or B) = 0.85 and P(A and B) = 0.42 

To find : P(B) 

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

Substituting in the above formula we get,

0.85 = 0.60 + P(B) – 0.42 

0.85 = 0.18 + P(B) 

0.85 – 0.18 = P(B) 

0.67 = P(B) 

P(B) = 0.67

(c) 1/2

n(S) = 20 

A = {3, 6, 9, 12, 15, 18} ⇒ n(A) = 6

B = {4, 8, 12, 16, 20} ⇒ n(B) = 5 

A ∩ B = {12} ⇒ n(A ∩ B) = 1 

So P(A ∪ B) = P(A) + P(B) – P(A ∩ B)

= (6/20) + (5/20) - (1/20) = 10/20 = 1/2

n(S) = 52; P(J) = 4/52

(i) P(JJ) = (4/52) x (4/52) = (1/13) x (1/13) = 1/169

(ii) P(JJ) = (4/52) x (3/51) = (1/13) x (1/17) = 1/221

(i) P(A) = 0.5, P(B) = 0.6, P(A ∩ B) = 0.24 

P(A ∪ B) = P(A) + P(B) – P(A ∩ B) 

(i.e.,) P(A ∪ B) = 0.5 + 0.6 – 0.24 

= 1.1 – 0.24 = 0.86 

∴ P(A ∪ B) = 0.86

(ii) P(\(\bar{A}\) ∩ B) = P(B) – P(A ∩ B) 

= 0.6 – 0.24 = 0.36

(iii) P(A ∩ \(\bar{B}\)) = P(A) – P(A ∩ B) 

= 0.5 – 0.24 = 0.26

(iv) P(\(\bar{A}\) ∪ \(\bar{B}\)) = P {(A ∩ B)’} = 1 – P(A ∩ B) 

= 1 – 0.24 = 0.76

(v) P(\(\bar{A}\) ∩ \(\bar{B}\)) = P{A ∪ B)’} = 1 – P(A ∪ B) 

= 1 – 0.86 = 0.14.

Given A and B are independent. 

⇒ P(A ∪ B) = P(A).P(B) 

Here P(A ∪ B) = 0.6 and P(A) = 0.2 

To find P(B): 

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

(i.e.,) P(A ∪ B) = P(A) + P(B) – P(A) . P(B) 

(i.e.,) 0.6 = 0.2 + P(B) (1 – 0.2)

P(B) (0.8) = 0.4 

⇒ P(B) = 0.4/0.8 = 4/8 = 1/2 = 0.5.

Given P(A ∪ B) = 0.7, P(A ∩ B)= 0.2 and P(B) = 0.5 

To find P(A) 

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

(i.e.,) 0.7 = P(A) + 0.5 – 0.2 

⇒ 0.7 – 0.5 + 0.2 = P(A) 

(i.e.,) P(A) = 0.4 

Now P(A ∩ B) = 0.2 … (i) 

P(A) P(B) = 0.4 × 0.5 = 0.2 … (ii) 

(1) = (2) ⇒ P(A ∩ B) = P(A) P(B) 

⇒ A and B are independent.

Correct option is: C) \(\frac {26}{31}\).

If 1st May is Monday then 8th, 15th, 22nd & 29th May are also Monday.

Total Mondays in month of May = 5.

Total number of days in month of May = 31.

\(\therefore\) Probability of choosing Monday in month of May whose 1st day is Monday = \(\frac 5{31}\) 

Probability of choosing non-Monday = \(1-\frac 5{31} = \frac {31-5}{31} = \frac {26}{31}\) 

Hence, required probability is \(\frac {26}{31}\).

Correct option is: C) \(\frac{26}{31}\)

Correct option is: D)\(\frac{3}{4}\)

When two coins tossed simultaneously then possible outcomes are (HH, HT, TH, TT)

\(\therefore\) Total possible outcomes = 4.

Favourable outcomes to event of getting at least one head are (HH, HT, TH)

\(\therefore\) Total fabourable outcomes = 3

\(\therefore\) Required probability = \(\frac 34\) 

Hence, probability of getting at least one head = \(\frac 34\).

Correct option is: D) \(\frac{3}{4}\)

Given: Two coins are tossed once.

We know, when two coins are tossed then the no. of possible outcomes are 2² = 4

So, the Sample spaces are {HH, HT, TT, TH}

∴ There are total 4 events associated with the given experiment.

E denotes the complement of an event E then \(\bar{E}\) denotes the affirmation aspect and the probability of any event`s occurring or non-occurring , together is always 1 

So P(E) + P(\(\bar{E}\) ) = 1

Total numbers of elementary events are 2+ 2 = 4 

Let E be the event of getting at least two heads 

The favorable outcomes are: HH, HT, TH 

Total numbers of favorable outcomes are 3 

P (ATLEAST one head) = P (E) = \(\frac{3}4\) 

Note: at least one head means the outcome with one head or more will be considered favorable and the outcome with no head at all will be ruled out.

The number of sample points in the random experiment of tossing five balanced coins is n = 2⁵ = 32.

Total numbers of elementary events are 52 

Let E be the event of getting an ace 

The numbers of favorable events are: 4 

P (an ace) = P (E) = \(\frac{4}{52}\) = \(\frac{1}{13}\)

A = Event that an item is defective

∴ P(A) = 2 % = \(\frac{2}{100 }\)= 0.02

A’ = Event that an item is non-defective

∴ P(A’) = 1 – P(A) = 1 – 0.02 = 0.98

From 8 workers 3 workers are excellent in efficiency. So the remaining 5 workers are moderate in efficiency. Total number of primary outcomes of selecting 2 workers at random from 8 workers is

n = ⁸C₂ =\( \frac{8×7}{2×1}\) = 28

(1) A = Event that the selected both the workers have excellent efficiency

∴ Favourable outcomes for the event A is m = ³C₂ × ⁵C₂ = 3 × 1 = 3.

Hence, P(A) = \(\frac{m}{n} = \frac{3}{28}\)

(2) B = Event that the selected both the workers have moderate efficiency

∴ Favourable outcomes for the event B is m = ⁵C₂ × ³C₀ = 10 × 1 = 10.

Hence, P(B) =\( \frac{m}{n} = \frac{10}{28} = \frac{5}{14}\)

(3) C = Event that in the selected two workers one worker is excellent and ope worker is moderate in efficiency.

∴ Favourable outcomes for the event C is

m = ³C₁ × ⁵C₁ = 3 × 5 = 15.

Hence. P(C) = \(\frac{m}{n }= \frac{15}{28}\)

3 bulbs are defective in a box of 10 bulbs.

∴ 7 bulbs are non-defective.

Total number of primary outcomes of selecting 2 bulbs randomly from 10 bulbs is,

n = ¹⁰C₂ = \(\frac{10×9}{2×1}\) = 45A = Event that the room will be lighted after 

starting the electric supply.

There are two options of occurring event A:

If 2 bulbs are non-defective and are fixed in the first bulb holder.

OR

If in 2 bulbs one bulb is non-defective and one bulb Is defective and are fixed in the second bulb holder.

∴ Favourable outcomes for event A is,

m = ⁷C₂ + ⁷C₁ × ³C₁

= \(\frac{7×C}{2×1} \)+ ( 7 × 3)

= 21 + 21 = 42

∴ p(A) = \(\frac{m}{n} = \frac{42}{45} = \frac{14}{15}\)

Total number of primary outcomes of drawing two cards from a well shuffled pack of 52 cards is

n = ⁵²C₂ = \(\frac{52×51}{2×1} =1326\)

(1) A = Event that both the cards drawn are of different colours, i.e., one card is of black colour and one is of red colour.

In a pack of 52 cards, 26 cards are black and 26 cards are red.

∴ Favourable outcomes for the event A is

m = ²⁶C₁ × ²⁶C₁ = 26 × 26 = 676

Hence, P(A) = \(\frac{m}{n} = \frac{676}{1326} = \frac{26}{51}\)

(2) B = Event that both the cards are face cards. In a pack of 52 cards, 12 cards are face cards.

∴ Favourable outcomes for the event B is

m = ¹²C₂ = \(\frac{12×11}{2×1} = 66.\)

Hence, p(B) = \(\frac{m}{n} = \frac{66}{1326} = \frac{11}{221}\)

(3) C = Event that one of the two cards is a king. In a pack of 52 cards, 4 cards are of king and other cards are 48.

∴ Favourable outcomes for the event C is

m = ⁴C₁ × ⁴⁸C₁ = 4 × 48 = 192

Hence, P(C) =\(\frac{ m}{n} = \frac{192}{1326} = \frac{32}{221}\)

(i) W.k.t P(0) + P(1) + P(2) + P(3) = 1

⇒ k + k/2 + k/4 + k/8 = 1

(8k + 4k + 2k + k)/8 = 1

15k = 8

Hence, k = 8/15

(ii) P(X ≤ 2) = P(X = 0) + P(X = 1) + P(X = 2)

= k + k/2 + k/4 = 7k/4 = 7/4 x 8/15 = 14/15

And P(X > 2) = P(X = 3) = k/8 = 1/8 x 8/15 = 1/15

(iii) P(X ≤ 2) + P(X ≥ 2) = 14/15 + 1/15 = (14 + 1)/15 = 15/15 = 1

Given: Total number of watches = 100 and number of defective watches = 10

So, the probability of selecting a detective watch = 10/100 = 1/10

Now,

n = 8, p = 1/10 and q = 1 – 1/10 = 9/10, r ≥ 1

P(X ≥ 1) = 1 – P(x = 0) = 1 – ⁸C₀ (1/10)⁰(9/10)^{8 – 0} = 1 – (9/10)⁸

Therefore, the required probability is 1 – (9/10)⁸.

Let X represent the number of spade cards among the five cards drawn. Since, the drawing card is with replacement, the trials are Bernoulli trials. 

In a well-shuffled deck of 52 cards, there are 13 spade cards.

p = P (success) = P (a spade card is drawn) = 13/52 = 1/4

and q = 1 - p = 1 - 1/4 = 3/4

X has a binomial distribution with n = 5, p = 1/4 and q = 3/4

Therefore, by Binomial distribution 

P(X = r) = ⁿC_r p^rq^{n – r}, where r = 0, 1, 2,...,n

P(X = r) = ⁵C_r(1/4)^r(3/4)^{5 - r}

(i) P (all the five cards are spades) = P( X = 5) = ⁵C₅P⁵q⁰ = P⁵ = (1/4)⁵ = 1/1024

(ii) P (only three cards are spades) = P( X = 3) = ⁵C₃P³q² = 10(1/4)³(3/4)² = 90/1024 = 45/512

(iii) P (none is a spade) = P (X = 0) = ⁵C₀P⁰q⁵ = (3/4)⁵ = 243/1024

The repeated tosses of a die are Bernoulli trials. 

Let X denote the number of successes of getting odd numbers in an experiment of 6 trials. 

p = P (success) = P (getting an odd number in a single throw of a die)

∴ p= 3/6 = 1/2 and q = P (failure) = 1 - p = 1 - 1/2 = 1/2

Therefore, by Binomial distribution

P(x = r) = ⁿC_rp^n-rq^r, where r = 0, 1, 2,...., n

P(X = r) = ⁶C_r(1/2)^6-r(1/2)^r = ⁶C_r(1/2)⁶ 

(i) P(5 successes)= ⁶C₅(1/2)⁶ = 6 x 1/2⁶ = 6/64 = 3/32

(ii) P (at least 5 successes) = P (5 successes) + P (6 successes) = ⁶C₅(1/2)⁶ + ⁶C₆(1/2)⁶ = 6/64 + 1/64 = 7/64

(iii) P (at most 5 successes) = 1 – P (6 successes) = 1 - ⁶C₆(1/2)⁶ = 1 - 1/64 = 63/64