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The given numbers are 00, 01, 02 . . . . 99. These are total 100 numbers, out of which the numbers, the product of whose digits is 18, are 29, 36, 63 and 92. 

∴ p = P (E) = 4/100 = 1/25 ⇒ q = 1 – p = 24/25 

From Binomial distribution 

P (E occurring at least 3 times) = P (E occurring 3 times) + P (E occurring 4 times) 

⁴C₃ p³q + ⁴C₄ p⁴ = 4 x (1/25)³ (24/25) + (1/25)⁴ = 97/(25)⁴ 

Given that A and B are independent events 

∴ P(A ∩B) = P (A). P(B) . . . . . . . . . . . . . . . . . . . . . . . . . (1) 

Also given that P (A ∩ B) = 1/6 . . . . . . . . . . . . . . . . . . . . . . . . . (2) 

And P(bar A ∩ B) = 1/3  ................(3)

Also P(bar A ∩ B) = 1 - P(A ∪ B)

⇒ P(A ∩ B) = 1 - P(A) - P(B)P(A ∩ B)

⇒ 1/3 = 1 - P(A) - P(B) + 1/6

⇒ P(A) + P(B) ........(4)

From (1) and (2) we get

Let P(A) = x and P(B) = y then eq's (4) (5) become

x + y = 5/6, xy = 1/6

⇒ x - y = (x + y)² - 4xy

= 25/36 - 4/6 = 1/6

∴ We get x = 1/2 and y = 1/3

Or x = 1/3 and y = 1/2

Thus  P(A) = 1/2 and P(B) = 1/3 or P(A) = 1/3 and P(B) = 1/2

Total number of all possible outcomes = 52

There are 26 red cards (including 2 queen) and apart from these, there are 2 more queens.

number of cards, each one of which is either a red card or a queen = 26 + 2 = 28

Let E be the event that the card drawn is neither a red card nor a queen.

Then. the number of favorable outcomes = (52 - 28) = 24

Therefore, P(getting neither a red card nor a queen) = P(E) = \(\frac{24}{52}\) =  \(\frac{6}{13}\)

Total number of cards in a well-shuffled pack = 52

Total number of queens and kings cards = 4 + 4 = 8

Therefore, there are 44 cards (52 – 8) that are neither king nor queen.

Total number of favorable outcomes = 44

Required probability = 44/52 = 11/13

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 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-

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 king, i.e.

P (Spade ‘or’ King) = 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

Total number cards in a deck = 52 

Number of Queens in a deck of cards = 4

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

∴ Probability of drawing a queen from the deck of cards P(queen)

= \(\frac{Number\,of\,queen\,in\,a\,deck\,of\,cards}{Total\,number\,of\,cards\,in\,a\,deck}\) = \(\frac{4}{52}=\frac{1}{13}\)

Total number of fishes = 13

P(probability that the fish are taken out is a male fish) = Male Fishes/Total Fishes = 5/13

There are 4 suits in a pack of 52 cards

13- Heart

13- Diamond

13- Spades

13- Clubs

Each suits has 1 ace , 1jack, 1king , 1queen

Face Cards are Jack, king ,queen. So there are 3 face cards.

Total number of face cards =12

26- red card , 26- black card

(i) Since There are 26 red Cards and a total of 2 kings are there.

So P(a king of red colour) = 2/26

(ii) Total number of face cards = 12

So P(a face card) = 12/52

(iii) Total number of red cards =26

Total number of red face cards = 6

P(a red face card) = 6/26

(iv) P(a jack of heart) = 1/13

(v) P(a spade) = 13/52

(vi) P(a queen of diamond) = 1/13

Total number of all possible outcomes = 52

(i) Total number of queen = 4

Therefore, P(getting a queen) = \(\frac{4}{52}\) = \(\frac{1}{13}\)

(ii) Number of diamond suits = 13

Therefore, P(getting a queen) = \(\frac{13}{52}\) = \(\frac{1}{4}\)

(iii) Total number of kings = 4

Let E be the event of getting a king or an ace card.

Then, the favorable outcomes = 4 + 4 = 8

Therefore, P(getting a queen) = P(E) = \(\frac{8}{52}\) = \(\frac{2}{13}\)

(iv) Number of red aces = 2

Therefore, P(getting a queen) = P(E) = \(\frac{2}{52}\) = \(\frac{1}{26}\)

Total number of cards = 52 

(i) P(One red king) = \(\frac{2}{52} = \frac{1}{26}\)

(ii) Number of a face card : 

4 king, 4 queen, 4 jack = Total 12 cards 

∴ P(1 face card) = \(\frac{12}{52} = \frac{3}{13}\)

(iii) One red colour face card = 6 

∴ P(1 red face card) = \(\frac{6}{52} = \frac{3}{13}\)

(iv) P(Heart Jack) = \(\frac{1}{52}\)

(v) P(1 spade) = \(\frac{13}{52} = \frac{1}{4}\)

(vi) P(Diamond card) = \(\frac{1}{52}\)

Number of good pens = 132 

Number of defective pens = 12 

Total number of pens = 132 + 12 = 144 

∴ Total number of outcomes in taking a pen at random = 144. 

No. of favourable outcomes in taking a good pen = 132. 

∴ Probability of taking a good pen

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

= \(\frac{132}{144}=\frac{11}{12}\)

Number of male fish = 5 

Number of female fish = 8 

Total number of fish = 5 m + 8 f = 13 fishes. 

∴ Number of total outcomes in taking a fish at random from the aquarium =13. 

Number of male fish = 5 

∴ Number of outcomes favourable to male fish = 5. 

∴ The probability of taking a male fish =

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

= 5/18

= 0.38

Total number of cards = 52. 

∴ Number of all possible outcomes in drawing a card at random = 52. 

i) Number of outcomes favourable to the king of red colour = 2(♥ K, ♦ K)

∴ Probability of getting the king of red colour

P(E) = \(\frac{No.\,of\,favourable\,outcomes}{Total\,no.\,of\,outcomes}\)

\(=\frac{2}{52}=\frac{1}{26}\)

ii) Number of face cards in a deck of cards = 4 × 3 = 12 (K, Q, J) 

Number of outcomes favourable to select a face card = 12. 

∴ Probability of getting a face card

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

\(=\frac{12}{52}=\frac{3}{13}\)

iii) Number of red face cards = 2 × 3 = 6. 

∴ Number of outcomes favourable to select a red face card = 6. 

∴ Probability of getting a red face

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

\(=\frac{6}{52}=\frac{3}{26}\)

iv) Number of outcomes favourable to the jack of hearts = 1. 

∴ Probability of getting jack of hearts

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

= \(\frac{1}{52}\)

v) Number of spade cards = 13 

∴ Number of outcomes favourable to ‘a spade card’ = 13. 

∴ Probability of drawing a spade

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

\(=\frac{13}{52}=\frac{1}{4}\)

vi) Number of outcomes favourable to the queen of diamonds = 1. 

∴ Probability of drawing the queen of diamonds

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

= \(\frac{1}{52}\)

Let E and F denote respectively the events that first and second ball drawn are black. We have to find P(E ∩ F) or P (EF).

Now P(E) = P (black ball in first draw) = 10/15

Also given that the first ball drawn is black, i.e., event E has occurred, now there are 9 black balls and five white balls left in the urn. Therefore, the probability that the second ball drawn is black, given that the ball in the first draw is black, is nothing but the conditional probability of F given that E has occurred.

i.e. P(F|E)  = 9/14

By multiplication rule of probability, we have

P (E ∩ F) = P(E) P(F|E) = 10/15 x 9/14 = 3/7

Total number of balls = 18, number of red balls = 8 and number of black balls = 10

∴ Probability of drawing a red ball = 8/18

Similarly, probability of drawing a black ball = 10/18

(i) Probability of getting both red balls = P (both balls are red) = P (a red ball is drawn at first draw and again a red ball at second draw) = 8/18 x 8/18 = 16/81

(ii) P (probability of getting first ball is black and second is red) = 10/18 x 8/18 = 20/81

(iii) Probability of getting one black and other red ball = P(first ball is black and second is red) + P (first ball is red and second is black) 

= 10/18 x 8/18 + 8/18 x 10/18 

= 20/81 + 20/81 = 40/81

Let H : Set of students reading Hindi newspaper and E : set of students reading English newspaper.

Let n(S) = 100 Then, n(H) = 60 

n(E) = 40 and n(H ∩ E) = 20

∴ P(H) = 60/100 = 3/5, P(E) = 40/100 = 2/5 and P(H ∩ E) = 20/100 = 1/5

(i) Required probability = P (student reads neither Hindi nor English newspaper)

=  P(H' ∩ E') = P(H ∪ E)' = 1 - P(H ∪ E)

= 1 - [P(H) + P(E) - P(H E)] = 1 - [3/5 + 2/5 - 1/5] = 1 - 4/5 = 1/5

(ii) Required probability = P(a randomly chosen student reads English newspaper, if he/she reads

Hindi newspaper) = P(E/H) = (P(E ∩ H))/P(H) = (1/5)/(3/5) = 1/3

(iii) Required probability = P (student reads Hindi newspaper when it is given that reads English newspaper)

= P(H/E) = (P(H ∩ E))/P(E) = (1/5)/(2/5) = 1/2

(a) P (both all red) \(=\frac{8}{18} \times \frac{8}{18}=\frac{16}{81}\)

(b) P (one of them is black and other red) = P(First ball black, second red) or P (First red, second black) 

\(=\frac{10}{18} \times \frac{8}{18}+\frac{8}{18} \times \frac{10}{18}=\frac{20}{81}+\frac{20}{81}=\frac{40}{81}\)

Probability of solving the problem by A, P(A) =  1/2

Probability of solving the problem by B, P(B) = 1/3

Probability of not solving the problem by A = P(A’) = 1 – P(A) = 1 - 1/2 = 1/2

and probability of not solving the problem by B = P(B’) = 1 – P(B) = 1 - 1/3 = 2/3

(i) P (the problem is solved) = 1 – P(none of them solve the problem) 

=  1 (A' ∩ B') = 1 - P(A')P(B')

(since A and B are independent A’ and B’ are independent)

= 1 - (1/2 x 2/3) = 1 - 1/3 = 2/3

(ii) P (exactly one of them solve the problem) = P(A) P(B’) + P(A’) P(B)

= 1/2 x 2/3 + 1/2 x 1/3 = 1/3 + 1/6 = (2 + 1)/6 = 3/6 = 1/2

Given: P(A/B) = 0.4 

 \(\frac{P(A∩ B)}{P(A)} = 0.4\\=\frac{P(A∩ B)}{0.8}=0.4\\=P(A∩ B)=0.32\\=\\P(\frac{A}{B}) = \frac{P(A∩ B)}{P(B)}\\=\frac{0.32}{0.5}\\=0.64\)

n(S) = 6³ = 216

E = {( 1, 1, 4), (1, 2, 4), (1, 3, 4)……….(1, 6, 4),

(2, 1, 4), (2, 2, 4), (2, 3, 4)……..(2, 6, 4),

(3, 1, 4), (3, 2, 4), (3, 3, 4)……..(3, 6, 4),

(4, 1, 4), (4, 2, 4), (4, 3, 4)…….(4, 6, 4),

(5, 1, 4), (5, 2, 4), (5, 3, 4)……..(5, 6, 4),

(6, 1, 4), (6, 2, 4), (6, 3, 4)……..(6, 6, 4)}

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

⇒ E ∩ F = {(6, 5, 4)}

P(F) = \(\frac{6}{216}\) and P(E ∩ F) = \(\frac{1}{216}\)

Then, P(E/F) = \(\frac{P(E ∩ F)}{P(F)} = \frac{\frac{1}{216}}{{\frac{6}{216}}} = \frac{1}{6}\)

TRUE

As A and B are independent

⇒ P(A ∩ B) = P(A)P(B)

As, P(A’ ∩ B’) = P(A ∪ B)’ {using De morgan’s law}

As, P(A ∪ B)’ = 1 – P(A ∪ B)

As we know P(A ∪ B) = P(A) + P(B) – P(A ∩ B)

⇒ P(A ∪ B)’ = 1 –[ P(A) + P(B) – P(A ∩ B)]

⇒ P(A’ ∩ B’) = 1 - P(A) - P(B) + P(A)P(B) {as A & B are independent}

= [1 – P(A) ]– P(B) (1-P(A)]

⇒ P(A’ ∩ B’) = (1 – P(A))(1 – P(B))

= P(A’)P(B’)

hence proved

TRUE

If A and B are independent events. It implies-

P(A ∩ B) = P(A)P(B)

P(A’ ∩ B) = P(A’)P(B)

And P(A ∩ B’) = P(A)P(B’)

P(exactly one of A, B occurs) = P(A’ ∩ B) + P(A ∩ B’)

⇒ P(exactly one of A, B occurs) = P(A’)P(B)+ P(A)P(B’)

∴ Statement is true.

It is given that out of 30 bulbs, 6 are defective. 

Number of non-defective bulbs = 30 – 6 = 24 

4 bulbs are drawn from the lot with replacement. 

Let p = P(obtaining a defective bulb when a bulb is drawn) = 6/30 = 1/5

and q = P(obtaining a non-defective bulb when a bulb is drawn) = 24/30 = 4/5

Using Binomial distribution, we have

P(X = 0) = P (no defective bulb in the sample) = ⁴C₀p⁰q⁴ = (4/5)⁴ = 256/625

P(X = 1) = P (one defective bulb in the sample) = ⁴C₁p¹q³ = 4(1/5)¹(4/5)³ = 256/625

P(X = 2) = P (two defective and two non-defective bulbs are drawn) = ⁴C₂p²q² = 6(1/5)²(4/5)² = 96/625

P(X = 3) = P (three defective and one non-defective bulbs are drawn) = ⁴C₃p³q¹ = 4(1/5)³(4/5) = 16/625

 P(X = 4) = P (four defective bulbs are drawn) = ⁴C₄p⁴q⁰ = (1/5)⁴ = 1/625

Therefore, the required probability distribution is as follows.

p = 1/10

As n = 5 {representing no. of trials}

p = probability of success

As it is a binomial distribution.

∴ probability of failure = q = 1 – p

Given,

P(X = 2) = 9.P(X = 3)

The binomial distribution formula is:

P(x) = _nC_x P^x (1 – P)^{n – x}

Where:

x = total number of “successes.”

P = probability of success on an individual trial

n = number of trials

Using binomial distribution,

⇒ ⁵C₂p²q^5-2 = 9⁵C₃p³q^5-3

⇒ 10p²q³ = 9×10p³q²

⇒ 10q = 90p {As , p≠0 and q ≠ 0}

⇒ q = 9p

⇒ 1-p = 9p ⇒ 10p = 1

∴ p = 1/10