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i)  Total number of accidents = (440 + 160 + 110 + 61 + 35 + 505 + 125 + 60 + 22 + 18 + 360 + 45 + 35 + 15 + 9) = 2000 

Event: The driver being in the age group (18 – 29) years and having exactly 3 accidents = 61

P(E) = \(\frac {No.\,of\, favourables}{Total\, outcomes}\) = 61/2000

ii) Favourable outcomes = 125 + 60 + 22 + 18 = 225

Total outcomes = 2000

P(E) = 225/2000 = 9/80

iii) Favourable outcomes = 440 + 505 + 360 = 1305 

Total outcomes = 2000

P(E) = 1305/2000 = 261/400

Answer is (C) \(\frac { 4 }{ 7 }\)

Total number of marbles in the bag = 3 + 4 = 7

∴ The number of all possible outcomes = 7

The number of favourable outcomes of getting blue marbles = 4

∴ Required probability = \(\frac { 4 }{ 7 }\)

Three coins are tossed simultaneously the possible outcomes are.

HHH, HHT, HTH, THH, HTT, THT, TTH, TTT.

Total number of all possible outcomes = 8.

The favourable outcomes of getting 2 heads exactly 

= HHT, HTH, THH

∴ The number of favourable outcomes = 3

∴ The required probability = \(\frac { 3 }{ 8 }\).

Answer is (B) \(\frac { 6 }{ 23 }\)

The total number of students in the class = 23

The sum of the students of the houses A, B and C

= 4 + 8 + 5 = 17

The number of all possible outcomes = 23

The number of favourable outcomes = 17

The probability that the student choosen from the houses A, B and C = \(\frac { 17 }{ 23 }\)

The probability that the ‘student not’ from the houses if A, B and C.

= 1 – \(\frac { 17 }{ 23 }\) = \(\frac { 6 }{ 23 }\)

The sum of probabilities of all primary events of an experiment is 1.

Answer is (C) \(\frac { 1 }{ 4 }\)

When two coins are tossed simultanously the all possible outcomes are HH, HT, TH, TT

The number of all possible outcomes = 4

The number of favourable outcomes of getting two heads = 1

∴ The probability of getting two heads = \(\frac { 1 }{ 4 }\)

Number of balls in the bag = 3 red + 5 black.

∴ Total number of all possible outcomes = 8.

The number of favourable outcomes that the ball is not black = 3.

∴ The required probability = \(\frac { 3 }{ 8 }\)

Two balls are taken out at random which and of different colours. 

So, the sample set

S = {RB, RY, RW, BY, BW, YW}

Tossing two coins at the same time.

Total possible outcomes = HH, HT, TH, TT

∴ Total number of possible outcomes = 4

Favourable outcomes of getting maximum one tail are HT, TH and TT.

∴ Number of favourable outcomes = 3

∴ Required probability = \(\frac { 3 }{ 4 }\)

Total letters ;= 26 [A, B, C ….. Z]

Probability = \(\frac {No.\,of\, favourables}{Total\, No.\,of\,outcomes}\)

a) Vowels = a, e, i, o, u [5]

∴ (vowels) = 5/26

b) Letter after P = 10 

[Q, R, S, T, U, V, W, X, Y, Z]

Probability of a letter that comes after

’P’ = 10/26 = 5/13

c) A vowel or a consonant 

Vowel or consonants = 26 

[all letters from A to Z] 

P(vowel or consonant) = 26/26 =1

d) Not a vowel = 21 

[other than A, E, I, O, U]

∴ P(not a vowel) = 21/26

Answer is (A) \(\frac { 1 }{ 6 }\)

When a dice is thrown once then the number of all possible outcomes = 6

The odd number smaller than 3 = 1

The probability of getting an odd number smaller than 3 = \(\frac { 1 }{ 6 }\)

When a dice is thrown, the all possible outcomes = 6.

The number greater than 4, 5 and 6.

The number favourable outcomes = 2.

∴ The required probability = \(\frac { 2 }{ 6 }\) = \(\frac { 1 }{ 3 }\)

(i) The number of balls in the bag = 12

∴ The number of all possible outcomes = 12

The number of white balls = x

∴ The number of favourable outcomes = x

∴ Required probability = \(\frac { x }{ 12 }\)

(ii) After putting 6 more white balls in the bag, the total number of ball 

= 12 + 6 = 18

The number of white balls = (x + 6)

The new probability = \(\frac { x+6 }{ 18 }\)

According to the given problem,

\(\frac { x+6 }{ 18 }\) = \(\frac { x }{ 12 }\) × 2

⇒ \(\frac { x+6 }{ 18 }\) = \(\frac { x }{ 6 }\)

⇒ 6x + 36 = 18x

⇒ 18x – 6x = 36 

⇒ 12x = 6

⇒ x = \(\frac { 36 }{ 12 }\) = 3

Tossing the die all possible outcomes 

= 1, 2, 3, 4, 5, 6

∴ Total number of all possible outcomes =6

(i) The even digits are 2, 4, 6

∴ The number of favourable outcomes = 3

∴ Required probability = \(\frac { 3 }{ 6 }\) = \(\frac { 1 }{ 2 }\)

(ii) The digits more than 3 are 4, 5, 6

∴ The number of favourable outcomes = 3.

∴ Requred probability = \(\frac { 3 }{ 6 }\) = \(\frac { 1 }{ 2 }\)

(iii) The digit between 3 and 6 are 4 and 5.

∴ The number of favourable outcomes = 2

∴ Required probability = \(\frac { 2 }{ 6 }\) = \(\frac { 1 }{ 3 }\)

When a coin is tossed twice, the digits on the top faces may be

(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).

∴ Total number of all possible outcomes = 36

(i) The favourable outcomes of getting the sum 9 = (6, 3), (5, 4), (4, 5), (3, 6)

The number of favourable outcomes = 4

∴ Required probability = \(\frac { 4 }{ 36 }\) = \(\frac { 1 }{ 9 }\)

(ii) When the dice is thrown two times, then the favourable outcomes of getting a sum 13 = 0.

Hence the probability = 0.

The total number of marbles in the jar = 24

∴ Total number of all possible outcomes = 24.

Let the number of green marbles in jar be x.

∴ The number of favourable outcomes = x.

The probability that the marble drawn is green = \(\frac { x }{ 24 }\)

But we have the probability of drawing a green marble = \(\frac { 2 }{ 3 }\)

\(\frac { x }{ 24 }\) = \(\frac { 2 }{ 3 }\)

3x = 2 × 24

x = \(\frac { 2\times 24 }{ 3 }\)

x = 16.

Thus, the number of green marbles in the jar = 16

The number of blue marbles 

= 24 – 16 = 8

Hence, the number of blue marbles in the jar = 8

In the given game total number of figures = 8 (triangles) + 10 (squares) = 18

∴ Number of all possible outcomes = 18

(i) Let the event of occurring a triangle be (T).

∴ The number of favourable outcomes of T = 8.

⇒ P(T) = \(\frac { 8 }{ 18 }\) = \(\frac { 4 }{ 9 }\)

(ii) Let the event occurring a square be (S).

∴ The number of favourable outcomes of S = 10

⇒ P(S) = \(\frac { 10 }{ 18 }\) = \(\frac { 5 }{ 9 }\)

(iii) Let the event of occurring the blue squares be (B).

∴ There are 6 blue squares.

∴ The number of favourable outcomes of (B) = 6

⇒ P(A) = \(\frac { 6 }{ 18 }\) = \(\frac { 1 }{ 3 }\)

(iv) Let the event occurring red triangles be (P)

The number of red triangles = 8 – 3 = 5

∴ The number of favourable outcomes of (R) = 5

⇒ P(R) = \(\frac { 5 }{ 18 }\)

Answer is (A) \(\frac { 1 }{ 6 }\)

When two dice are thrown at the same time, then the total number of all possible outcomes 

= 6 × 6 = 36

The favourable outcomes of getting same digits on the top faces of two dice are (1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6).

∴ The number of favourable outcomes = 6.

The required probability = \(\frac { 6 }{ 36 }\) = \(\frac { 1 }{ 6 }\)

Total numbers of boys in a class = 15

Number of girls in a class = 5

Total number of students = 15 + 5 = 20

Probability of getting a copy of a boy is = Total number of boys/Total number of students

= 15/20

= 3/4

Probability of getting a copy of a girl is = Total number of girls/Total number of students

= 5/20

= 1/4

∴ The probability of getting a copy of a boy is more.

Total number of possible outcomes = Total number of alphabets = 26

We know,

Probability of occurrence of an event = (Total number of favorable outcomes) / (Total number of outcomes)

(i) a vowel

Favorable outcomes are a, e, i, o, u

Total number of favorable outcomes = 5

Therefore, the probability that the letter is chosen is a vowel = 5/26

(ii) a consonant

Total number of consonant = 26 – 5 = 21

Therefore, the probability that the letter is chosen is a consonant = 21/26

All possible outcomes of throwing the dice = 1, 2, 3, 4, 5, 6.

∴ The number of all possible outcomes = 6.

The favourable outcomes of getting a number grater than 3 = 4, 5, 6.

∴ Number of favourable outcomes = 3.

∴ Required probability = \(\frac { 3 }{ 6 }\) = \(\frac { 1 }{ 2 }\)

A deck of 52 playing cards contains 13 diamond and 3 other kings.

∴ Total number of diamonds cards and 3 kings 

= 13 + 3 = 16.

The probability that the drawing card is a diamond or a king 

= \(\frac { 16 }{ 52 }\) = \(\frac { 4 }{ 13 }\)

Total number of notes in the boxes A and B 

= 25 + 50 = 75.

∴ Total number of all possible outcomes = 75

The number of notes on which Rs 1 is marked 

= 19 + 45 = 64

Number of notes on which Rs 1 is not marked 

= 75 – 64 = 11

Let the event that the note contains some thing else but Rs 1 be (A).

∴ The number of favourable outcomes of event (A)

= 11.

⇒ P(A) = \(\frac { 11 }{ 75 }\)

Hence P(A) = \(\frac { 11 }{ 75 }\)

Two dices contain the numbers 0, 1, 1, 1, 6, 6 on the faces each.

The all possible outcomes will be:

(0, 0), (0, 1), (0, 1), (0, 1), (0, 6), (0, 6)

(1, 0), (1, 1), (1, 1), (1, 1), (1, 6), (1, 6)

(1, 0), (1, 1), (1, 1), (1, 1), (1, 6), (1, 6)

(1,0), (1, 1), (1, 1), (1, 1), (1, 6), (1, 6)

(6, 0), (6, 1), (6, 1), (6, 1), (6, 6), (6, 6)

(6, 0), (6, 1), (6, 1), (6, 1), (6, 6), (6, 6)

(i) The possible numbers which have different outcomes are (0, 0), (0, 1), (0, 6), (1, 0), (1, 1), (1, 6), (6, 6)

Hence, the possible outcomes are : (0, 1, 2, 6, 7, 12)

(ii) Total number of all possible outcomes = 36

The numbers that have a sum 7 are

(1,6), (1,6), (1,6), (1,6), (1,6), (1,6)

(6, 1), (6, 1), (6, 1), (6, 1), (6, 1), (6, 1)

The number of favourable outcomes = 12

∴ The probability of getting a sum of 7 = \(\frac { 12 }{ 36 }\) = \(\frac { 1 }{ 3 }\)

Total numbers of consonants = 21

Number of white vowels = 5

Total number of alphabets = 21 + 5 = 26

Probability of getting a consonant is = Total number of consonants/Total number of alphabets

= 21/26

Probability of getting a vowel is = Total number of vowels/Total number of alphabets

= 5/26

∴ The probability of getting a consonant is more.

Answer is (C) \(\frac { 1 }{ 9 }\)

The numbers 6 to 50 are printed on the cards in the bag.

∴ The number of all possible outcomes = 45

The perfect square numbers printed on the cards are 9, 16, 25, 36, 49

The number of favourable outcomes = 5

The required probability = \(\frac { 5 }{ 45 }\) = \(\frac { 1 }{ 9 }\)

The deck of 52 playing cards has 13 cards of spades.

But it is not an ace, then number of spades = 13 – 1 = 12

∴ Required probability = \(\frac { 12 }{ 52 }\) = \(\frac { 3 }{ 13 }\)

The number of playing cards in a deck = 52.

(i) The number of diamonds cards in the deck = 13.

∴ The probability that the card drawn is a diamonds 

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

(ii) The number of non-diamonds cards 

= 52 – 13 = 39

∴ The probability that the card drawn is not diamonds 

= \(\frac { 39 }{ 52 }\) = \(\frac { 3 }{ 4 }\)

(iii) The number of black cards in the deck = 13 spades + 13 clubs = 26.

∴ The probability that the drawn cards is black 

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

(iv) The probability of getting not black card 

= 1 – \(\frac { 1 }{ 2 }\) = \(\frac { 1 }{ 2 }\)

The numbers printed on the cards are 3, 5, 7, 9, 35, 37.

∴ Total number of all possible outcomes = 18

The prime numbers marked on the cards are 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37.

The number of favourable outcomes = 11

∴ The required probability = \(\frac { 11 }{ 18 }\)

The deck contains 52 playing cards and a card can be drawn by 52 different ways.

∴ Total number of all possible outcomes = 52.

(i) Let the event of getting red king be R.

∴ The number of favourable outcomes of R = 2

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

(ii) Let the event of getting a queen or jack be A.

∴ The number of favourable outcomes of A = 8.

∴ P(A) = \(\frac { 8 }{ 52 }\) = \(\frac { 2 }{ 13 }\)

Hence (i) P(R) = \(\frac { 1 }{ 26 }\) and (ii) P(A) = \(\frac { 2 }{ 13 }\).

A deck of playing cards has 52 cards.

Shuffling the deck properly and one card is drawn, then the total possible outcomes = 52.

(i) There are four jack in the deck of which the jack of hearts arid diamonds are red.

∴ The number of favourable outcomes of drawing a red jack = 2

∴ Probability that the card drawn is a red jack = \(\frac { 2 }{ 52 }\)

= \(\frac { 1 }{ 26 }\)

(ii) The deck contains 13 cards of hearts and 13 cards of diamonds. 

So the number of favourable outcomes that the card is red = 26.

∴ The probability that the card drawn is red 

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

(iii) There is only one hearts ace in the deck.

So, the number of favourable outcomes = 1

∴ The probability that the drawn card is a hearts ace = \(\frac { 1 }{ 52 }\)

(iv) There is only one diamonds queen in the deck,

So, the number of favourable outcomes of getting a diamond queen = 1

∴ The probability that the card drawn is a diamonds queen = \(\frac { 1 }{ 52 }\)

(v) There are 13 spades cards in the deck 

so, the favourable outcomes = 13

∴ The probability that the card drawn is a spades

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

The number of playing cards in deck = 52.

∴ Number of all possible outcomes = 52.

The number of aces in deck = 4.

∴ The number of favourable outcomes = 4

∴ The required probability = \(\frac { 4 }{ 52 }\) = \(\frac { 1 }{ 13 }\)

The probability that the card drawn is not an ace 

= 1 – \(\frac { 1 }{ 13 }\) = \(\frac { 12 }{ 13 }\)

Answer is (D) 51

The number of playing cards in a deck = 52.

One card is drawn from this well shuffled deck.

The probability that the card drawn is an ace of diamond is E.

∴ The number of favourable outcomes of E = 52 – 1 

= 51

Total number of all favorable cases is n(S) = 52

Let A be the event that first card drawn is a king. There are four kings in the pack. Hence, the probability of the first card is a king is

P(A) = \(\cfrac{4}{52}\)

Let B be the event that second card is an ace without replacement. Then there are 4 aces in the pack as the cards are not replaced. Therefore, the probability of the second card is an ace is

P(B|A) = \(\cfrac{4}{51}\)

Then the probability of getting first is a king and the second is an ace without replacement is

= P(A)P(B|A)

⇒ \(\cfrac{4}{52}\times\cfrac{4}{51}\) (as there are 4 kings out of 52 cards in first draw, and 4 aces out of 51 cards in the second draw as the cards are not replaced)

⇒ \(\cfrac{4}{663}\)

The probability that first is a king and the second is an ace without replacement is \(\cfrac{4}{663}\) 

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

As each horse has equal chance of winning the race, Number of ways in which one of the five horses wins the race = ⁵C₁ 

∴ n(S) =⁵C₁  = \(\frac{|\underline5}{|\underline4|\underline1}\) 5

To find the chance that Mr A selects the winning horses, it is essential that one of the two horses selected by him wins the race. 

E : Mr A selecting the winning horse. 

⇒ n(E) = ²C₁ = 2 

∴ Required probability = \(\frac{n(E)}{n(S)}\) = \(\frac{2}{5}\).