Probability — Exercise 25(B)

A card is drawn from a well-shuffled pack of 52 cards. The probability of the card drawn to be king and jack is :

1. 1

2. 0

3. $\dfrac{1}{2}$

4. $\dfrac{3}{13}$

Answer

A card cannot be a king and a jack simultaneously.

P(drawing a king and jack) = $\dfrac{\text{No. of favourable outcomes}}{\text{Total no. of possible outcomes}} = \dfrac{0}{52}$ = 0.

Hence, Option 2 is the correct option.

The probability that a non-leap year has 53 Sundays is :

1. 1

2. $\dfrac{1}{7}$

3. $\dfrac{1}{365}$

4. $\dfrac{7}{365}$

Answer

There are 365 days in a non-leap year.

No. of weeks = $\dfrac{365}{7}$ = 52 weeks and 1 day.

Since, there is one sunday in a week, and one week has seven days.

So, in order to have 53 sundays the 1 day that remains should also be sunday.

P(that a non-leap year has 53 sundays)

= $\dfrac{\text{No. of favourable outcomes}}{\text{Total no. of possible outcomes}} = \dfrac{1}{7}$.

Hence, Option 2 is the correct option.

Out of 800 identical articles some are selected and remaining are rejected. If the probability of an article to be rejected is 0.425; the number of rejected articles is :

1. 340

2. 800

3. 430

4. 560

Answer

⇒ P(rejecting an item) = $\dfrac{\text{No. of favourable outcomes}}{\text{Total no. of possible outcomes}}$

⇒ 0.425 = $\dfrac{\text{No. of favourable outcomes}}{800}$

⇒ No. of favourable outcomes = 0.425 × 800 = 340.

Hence, Option 1 is the correct option.

A bag contains 24 balls of which 4 are red, 8 are white and remaining are blue. A ball is drawn at random, the probability of it to be blue is :

1. 0.5

2. 1

3. $\dfrac{1}{3}$

4. 2

Answer

No. of blue balls = 24 - 4 - 8 = 12.

P(drawing a blue ball)

= $\dfrac{\text{No. of favourable outcomes}}{\text{Total no. of possible outcomes}} = \dfrac{12}{24} = \dfrac{1}{2}$ = 0.5

Hence, Option 1 is the correct option.

An integer is chosen from integers between 0 and 51, the probability of it to be divisible by 6 is :

1. $\dfrac{3}{25}$

2. $\dfrac{4}{25}$

3. $\dfrac{1}{50}$

4. $\dfrac{1}{25}$

Answer

Integers between 0 and 51 are :

1, 2, 3, ......., 50.

No. of possible outcomes : 50

Integers divisible by 6 between 0 and 51 :

6, 12, 18, 24, 30, 36, 42, 48.

No. of favourable outcomes : 8

P(that a number is divisible by 6) : $\dfrac{\text{No. of favourable outcomes}}{\text{Total no. of possible outcomes}} = \dfrac{8}{50} = \dfrac{4}{25}$.

Hence, Option 2 is the correct option.

Nine cards (identical in all respects) are numbered 2 to 10. A card is selected from them at random. Find the probability that the card selected will be:

(i) an even number

(ii) a multiple of 3

(iii) an even number and a multiple of 3

(iv) an even number or a multiple of 3

Answer

We know that, there are totally 9 cards from which one card is drawn.

Total number of possible outcomes = 9

(i) From numbers 2 to 10, there are 5 even numbers i.e. 2, 4, 6, 8, 10

∴ Favorable number of outcomes = 5

P(selecting a card with an even number) = $\dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{5}{9}$.

Hence, the probability that card selected will be an even number = $\dfrac{5}{9}$.

(ii) From numbers 2 to 10, there are 3 numbers which are multiples of 3 i.e. 3, 6, 9.

∴ Favorable number of outcomes = 3

P(selecting a card with multiple of 3) = $\dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{3}{9} = \dfrac{1}{3}$.

Hence, the probability that card selected will be a multiple of 3 = $\dfrac{3}{9} = \dfrac{1}{3}$.

(iii) From numbers 2 to 10, there is one number which is an even number as well as multiple of 3 i.e. 6

∴ Number of favourable outcomes = 1

P(selecting a card with even number and multiple of 3) = $\dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{1}{9}$.

Hence, the probability of selecting a card with even number and multiple of 3 = $\dfrac{1}{9}$.

(iv) From numbers 2 to 10, there are 7 numbers which are even numbers or a multiple of 3 i.e. 2, 3, 4, 6, 8, 9, 10

∴ Number of favourable outcomes = 7.

P(selecting a card with even number or multiple of 3) = $\dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{7}{9}$.

Hence, the probability of selecting a card with even number or multiple of 3 = $\dfrac{7}{9}$.

Hundred identical cards are numbered from 1 to 100. The cards are well shuffled and then a card is drawn. Find the probability that the number on the card drawn is:

(i) a multiple of 5

(ii) a multiple of 6

(iii) between 40 and 60

(iv) greater than 85

(v) less than 48

Answer

We know that, there are 100 cards from which one card is drawn.

Number of possible outcomes = 100

(i) From numbers 1 to 100, there are 20 numbers which are multiple of 5 i.e. {5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60, 65, 70, 75, 80, 85, 90, 95, 100}

∴ Number of favourable outcomes = 20.

P(selecting a card with a multiple of 5) = $\dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{20}{100} = \dfrac{1}{5}$.

Hence, the probability of selecting a card with a multiple of 5 = $\dfrac{1}{5}$.

(ii) From numbers 1 to 100, there are 16 numbers which are multiple of 6 i.e. {6, 12, 18, 24, 30, 36, 42, 48, 54, 60, 66, 72, 78, 84, 90, 96}

∴ Number of favourable outcomes = 16.

P(selecting a card with a multiple of 6) = $\dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{16}{100} = \dfrac{4}{25}$.

Hence, the probability of selecting a card with a multiple of 6 = $\dfrac{4}{25}$.

(iii) From numbers 1 to 100, there are 19 numbers which are between 40 and 60 i.e. {41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59}

∴ Number of favourable outcomes = 19.

P(selecting a card between 40 and 60) = $\dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{19}{100}$.

Hence, the probability of selecting a card between 40 and 60 = $\dfrac{19}{100}$.

(iv) From numbers 1 to 100, there are 15 numbers which are greater than 85 i.e. {86, 87, 88, 89, 90, 91, 92, 93, 94, 95, 96, 97, 98, 99, 100}

∴ Number of favourable outcomes = 15.

P(selecting a card with number greater than 85) = $\dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{15}{100} = \dfrac{3}{20}$.

Hence, the probability of selecting a card with number greater than 85 = $\dfrac{3}{20}$.

(v) From numbers 1 to 100, there are 47 numbers which are less than 48 i.e. {1, 2, ……….., 46, 47}

∴ Number of favourable outcomes = 47.

P(selecting a card with number less than 48) = $\dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{47}{100}$.

Hence, the probability of selecting a card with number less than 47 = $\dfrac{47}{100}$.

From 25 identical cards, numbered 1, 2, 3, 4, 5, ..... , 24, 25; one card is drawn at random. Find the probability that the number on the card drawn is a multiple of :

(i) 3

(ii) 5

(iii) 3 and 5

(iv) 3 or 5

Answer

We know that, there are 25 cards from which one card is drawn.

Number of possible outcomes = 25.

(i) From numbers 1 to 25, there are 8 numbers which are multiple of 3 i.e. {3, 6, 9, 12, 15, 18, 21, 24}

∴ Number of favourable outcomes = 8.

P(selecting a card with a multiple of 3) = $\dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{8}{25}$.

Hence, the probability of selecting a card with a multiple of 3 = $\dfrac{8}{25}$.

(ii) From numbers 1 to 25, there are 5 numbers which are multiple of 5 i.e. {5, 10, 15, 20, 25}

∴ Number of favourable outcomes = 5.

P(selecting a card with a multiple of 5) = $\dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{5}{25} = \dfrac{1}{5}$.

Hence, the probability of selecting a card with a multiple of 5 = $\dfrac{1}{5}$.

(iii) From numbers 1 to 25, there is only one number which is multiple of 3 and 5 i.e. {15}

∴ Number of favourable outcomes = 1.

P(selecting a card with a multiple of 3 and 5) = $\dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{1}{25}$.

Hence, the probability of selecting a card with a multiple of 3 and 5 = $\dfrac{1}{25}$.

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

∴ Number of favourable outcomes = 12.

P(selecting a card with a multiple of 3 or 5) = $\dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{12}{25}$.

Hence, the probability of selecting a card with a multiple of 3 or 5 = $\dfrac{12}{25}$.

A dice is thrown once. Find the probability of getting a number :

(i) less than 3

(ii) greater than or equal to 4

(iii) less than 8

(iv) greater than 6

Answer

We know that,

In throwing a dice, total possible outcomes = {1, 2, 3, 4, 5, 6}

No. of favourable outcomes = 6.

(i) On a dice, numbers less than 3 are = {1, 2}

∴ No. of favourable outcomes = 2.

P(getting a number less than 3) = $\dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{2}{6} = \dfrac{1}{3}$.

Hence, the probability of getting a number less than 3 = $\dfrac{1}{3}$.

(ii) On a dice, numbers greater than or equal to 4 = {4, 5, 6}

∴ No. of favourable outcomes = 3.

P(getting a number greater than or equal to 4) = $\dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{3}{6} = \dfrac{1}{2}$.

Hence, the probability of getting a number greater than or equal to 4 is $\dfrac{1}{2}$.

(iii) On a dice, numbers less than 8 = {1, 2, 3, 4, 5, 6}

∴ No. of favourable outcomes = 6.

P(getting a number less than 8) = $\dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{6}{6} = 1$.

Hence, the probability of getting a number less than 8 = 1.

(iv) On a dice, numbers greater than 6 = 0

∴ No. of favourable outcomes = 0.

P(getting a number greater than 6) = $\dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{0}{6}$ = 0.

Hence, the probability of getting a number greater than 6 = 0.

A book contains 85 pages. A page is chosen at random. What is the probability that the sum of the digits on the page is 8?

Answer

There are 85 pages.

∴ No. of possible outcomes = 85.

Page numbers whose sum of digits equal to 8 are {08, 17, 26, 35, 44, 53, 62, 71, 80}.

∴ No. of favourable outcomes = 9

P(getting a page with sum of digits equal to 8) = $\dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{9}{85}$.

Hence, the probability of getting a page with sum of digits equal to 8 is $\dfrac{9}{85}$.

A pair of dice is thrown. Find the probability of getting a sum of 10 or more, if 5 appears on the first dice.

Answer

When two dice are thrown simultaneously;

Total number of possible outcomes = 6 × 6 = 36.

For obtaining a sum of 10 or more, if 5 appears on the first dice, the favourable outcomes are :

(5, 5) and (5, 6).

∴ No. of favourable outcomes = 2.

P(getting a sum of 10 or more) = $\dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{2}{36} = \dfrac{1}{18}$.

Hence, the probability of getting a sum of 10 or more, if 5 appears on the first dice = $\dfrac{1}{18}$.

If two coins are tossed once, what is the probability of getting :

(i) 2 heads ?

(ii) at least one head ?

(iii) both heads or both tails ?

Answer

When two coins are tossed,

Possible outcomes = {HH, HT, TH, TT}.

No. of possible outcomes = 4.

(i) Favourable outcomes for getting 2 heads = {HH}.

No. of favourable outcomes = 1.

P(getting 2 heads) = $\dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{1}{4}$.

Hence, the probability of getting 2 heads = $\dfrac{1}{4}$.

(ii) Favourable outcomes for getting at least one head = {HT, TH, HH}.

No. of favourable outcomes = 3.

P(getting at least one head) = $\dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{3}{4}$.

Hence, the probability of getting at least one head = $\dfrac{3}{4}$.

(iii) Favourable outcomes for getting both heads or both tails = {HH, TT}.

No. of favourable outcomes = 2.

P(getting both heads or both tails) = $\dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{2}{4} = \dfrac{1}{2}$.

Hence, the probability of getting both heads or both tails = $\dfrac{1}{2}$.

Two dice are rolled together. Find the probability of getting :

(i) a total of at least 10.

(ii) a multiple of 2 on one die and an odd number on the other die.

Answer

When two dice are thrown simultaneously;

Total number of possible outcomes = 6 × 6 = 36.

(i) For obtaining a sum of at least 10, the favourable outcomes are :

(4, 6), (5, 5), (5, 6), (6, 4), (6, 5) and (6, 6).

∴ No. of favourable outcomes = 6.

P(getting a total of at least 10) = $\dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{6}{36} = \dfrac{1}{6}$.

Hence, the probability of getting a total of at least 10 = $\dfrac{1}{6}$.

(ii) For obtaining a multiple of 2 on one die and an odd number on the other die, the favourable outcomes are :

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

∴ No. of favourable outcomes = 18.

P(getting a multiple of 2 on one die and an odd number on the other die) = $\dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{18}{36} = \dfrac{1}{2}$.

Hence, the probability of getting a multiple of 2 on one die and an odd number on the other die = $\dfrac{1}{2}$.

A card is drawn from a well-shuffled pack of 52 cards. Find the probability that the card drawn is :

(i) a spade.

(ii) a red card.

(iii) a face card.

(iv) 5 of heart or diamond.

(v) Jack or queen.

(vi) ace and king.

(vii) a red and a king.

(viii) a red or a king.

Answer

There are 52 cards in a deck which are divided into 4 suits of 13 cards each.

∴ No. of possible outcomes = 52.

(i) There are 13 spades in a deck of playing cards.

∴ No. of favourable outcomes = 13.

P(drawing a spade) = $\dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{13}{52} = \dfrac{1}{4}$.

Hence, the probability of drawing a spade = $\dfrac{1}{4}$.

(ii) There are 26 red cards (13 hearts and 13 diamonds).

∴ No. of favourable outcomes = 26.

P(drawing a red card) = $\dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{26}{52} = \dfrac{1}{2}$.

Hence, the probability of drawing a red card = $\dfrac{1}{2}$.

(iii) There are 12 face cards (4 kings, 4 queens, 4 jacks) in a deck.

∴ No. of favourable outcomes = 12.

P(drawing a face card) = $\dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{12}{52} = \dfrac{3}{13}$.

Hence, the probability of drawing a face card = $\dfrac{3}{13}$.

(iv) There are 2 cards one of each heart and diamond with no. 5.

∴ No. of favourable outcomes = 2.

P(drawing a 5 of heart or diamond) = $\dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{2}{52} = \dfrac{1}{26}$.

Hence, the probability of drawing a 5 of heart or diamond = $\dfrac{1}{26}$.

(v) There are 4 jacks and 4 queens in a deck.

∴ No. of favourable outcomes = 8.

P(drawing a jack or queen) = $\dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{8}{52} = \dfrac{2}{13}$.

Hence, the probability of drawing a jack or queen = $\dfrac{2}{13}$.

(vi) An ace and a king cannot be in a single card.

∴ No. of favourable outcomes = 0.

P(drawing an ace and a king) = $\dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}}$ = 0.

Hence, the probability of drawing an ace and a king = 0.

(vii) The king of diamond and heart is red in colour.

∴ No. of favourable outcomes = 2.

P(drawing a red and a king) = $\dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{2}{52} = \dfrac{1}{26}$.

Hence, the probability of drawing a red and a king = $\dfrac{1}{26}$.

(viii) There are 26 red (13 hearts + 13 diamonds) and 1 king of each (club and spade).

∴ No. of favourable outcomes = 26 + 1 + 1 = 28.

P(drawing a red or a king) = $\dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{28}{52} = \dfrac{7}{13}$.

Hence, the probability of drawing a red or a king = $\dfrac{7}{13}$.

A bag contains 16 coloured balls. Six are green, 7 are red and 3 are white. A ball is chosen, without looking into the bag. Find the probability that the ball chosen is :

(i) red

(ii) not red

(iii) white

(iv) not white

(v) green or red

(vi) white or green

(vii) green or red or white

Answer

There are 16 coloured balls.

∴ No. of possible outcomes = 16.

(i) There are 7 red balls.

∴ No. of favourable outcomes = 7

P(drawing a red ball) = $\dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{7}{16}$.

Hence, the probability of drawing a red ball = $\dfrac{7}{16}$.

(ii) There are 9 non red balls (6 green + 3 white).

∴ No. of favourable outcomes = 9

P(not drawing a red ball) = $\dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{9}{16}$.

Hence, the probability of not drawing a red ball = $\dfrac{9}{16}$.

(iii) There are 3 white balls.

∴ No. of favourable outcomes = 3

P(drawing a white ball) = $\dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{3}{16}$.

Hence, the probability of drawing a white ball = $\dfrac{3}{16}$.

(iv) There are 13 non white balls (6 green + 7 red).

∴ No. of favourable outcomes = 13

P(not drawing a white ball) = $\dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{13}{16}$.

Hence, the probability of not drawing a white ball = $\dfrac{13}{16}$.

(v) Since, there are only 3 coloured (red, green and white) balls.

We can say that,

P(drawing a green or red ball) = P(not drawing a white ball) = $\dfrac{13}{16}$.

Hence, the probability of drawing a green or red ball = $\dfrac{13}{16}$.

(vi) Since, there are only 3 coloured (red, green and white) balls.

We can say that,

P(drawing a white or green ball) = P(not drawing a red ball).

From part (ii),

P(not drawing a red ball) = $\dfrac{9}{16}$.

∴ P(drawing a white or green ball) = $\dfrac{9}{16}$.

Hence, the probability of drawing a white or green ball = $\dfrac{9}{16}$.

(vii) There are 6 green, 7 red and 3 white balls.

∴ No. of favourable outcomes = 16.

P(drawing a green or red or white ball) = $\dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{16}{16}$ = 1.

Hence, the probability of drawing a green or red or white ball = 1.

A ball is drawn at random from a box containing 12 white, 16 red and 20 green balls. Determine the probability that the ball drawn is :

(i) white

(ii) red

(iii) not green

(iv) red or white

Answer

Box contains 12 white, 16 red and 20 green balls.

∴ No. of possible outcomes = 48.

(i) There are 12 white balls.

∴ No. of favourable outcomes = 12

P(drawing a white ball) = $\dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{12}{48} = \dfrac{1}{4}$.

Hence, the probability of drawing a white ball = $\dfrac{1}{4}$.

(ii) There are 16 red balls.

∴ No. of favourable outcomes = 16

P(drawing a red ball) = $\dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{16}{48} = \dfrac{1}{3}$.

Hence, the probability of drawing a red ball = $\dfrac{1}{3}$.

(iii) There are 28 non green balls (12 white + 16 red).

∴ No. of favourable outcomes = 28

P(not drawing a green ball) = $\dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{28}{48} = \dfrac{7}{12}$.

Hence, the probability of not drawing a green ball = $\dfrac{7}{12}$.

(iv) Since, there are only 3 different colour balls.

We can say that,

P(drawing a red or white ball) = P(not drawing a green ball) = $\dfrac{7}{12}$.

Hence, the probability of drawing a red or white ball = $\dfrac{7}{12}$.

A card is drawn from a pack of 52 cards. Find the probability that the card drawn is :

(i) a red card

(ii) a black card

(iii) a spade

(iv) an ace

(v) a black ace

(vi) ace of diamonds

(vii) not a club

(viii) a queen or a jack.

Answer

In a deck of 52 cards, there are 4 suits of 13 cards each.

No. of possible outcomes = 52.

(i) There are 26 red cards (13 hearts + 13 diamonds).

∴ No. of favourable outcomes = 26.

P(drawing a red card) = $\dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{26}{52} = \dfrac{1}{2}$.

Hence, probability of drawing a red card = $\dfrac{1}{2}$.

(ii) There are 26 black cards (13 clubs + 13 spades).

∴ No. of favourable outcomes = 26.

P(drawing a black card) = $\dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{26}{52} = \dfrac{1}{2}$.

Hence, probability of drawing a black card = $\dfrac{1}{2}$.

(iii) There are 13 spades in a deck of playing cards.

∴ No. of favourable outcomes = 13.

P(drawing a spade) = $\dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{13}{52} = \dfrac{1}{4}$.

Hence, probability of drawing a spade = $\dfrac{1}{4}$.

(iv) There are 4 ace in a deck of playing cards.

∴ No. of favourable outcomes = 4.

P(drawing an ace) = $\dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{4}{52} = \dfrac{1}{13}$.

Hence, probability of drawing an ace = $\dfrac{1}{13}$.

(v) There are 2 black ace (1 of club + 1 of spade).

∴ No. of favourable outcomes = 2.

P(drawing a black ace) = $\dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{2}{52} = \dfrac{1}{26}$.

Hence, probability of drawing a black ace = $\dfrac{1}{26}$.

(vi) There is 1 ace of diamond.

∴ No. of favourable outcomes = 1.

P(drawing an ace of diamond) = $\dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{1}{52}$.

Hence, probability of drawing an ace of diamond = $\dfrac{1}{52}$.

(vii) There are 13 clubs so, there are 39 (52 - 13) non-club cards.

∴ No. of favourable outcomes = 39.

P(drawing a non-club card) = $\dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{39}{52} = \dfrac{3}{4}$.

Hence, probability of not drawing a club = $\dfrac{3}{4}$.

(viii) There are 4 queens and 4 jacks.

∴ No. of favourable outcomes = 8.

P(drawing a queen or a jack) = $\dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{8}{52} = \dfrac{2}{13}$.

Hence, probability of drawing a queen or a jack = $\dfrac{2}{13}$.

Thirty identical cards are marked with numbers 1 to 30. If one card is drawn at random, find the probability that it is :

(i) a multiple of 4 or 6.

(ii) a multiple of 3 and 5.

(iii) a multiple of 3 or 5.

Answer

Since, there are 30 identical cards.

No. of possible outcomes = 30.

(i) Cards numbered 4, 6, 8, 12, 16, 18, 20, 24, 28, 30 are multiple of either 4 or 6.

∴ No. of favourable outcomes = 10.

P(getting a card which is a multiple of 4 or 6) = $\dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{10}{30} = \dfrac{1}{3}$.

Hence, the probability of drawing a card which is a multiple of 4 or 6 = $\dfrac{1}{3}$.

(ii) Cards numbered 15, 30 are multiple of 3 and 5.

∴ No. of favourable outcomes = 2.

P(getting a card which is a multiple of 3 and 5) = $\dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{2}{30} = \dfrac{1}{15}$.

Hence, the probability of drawing a card which is a multiple of 3 and 5 = $\dfrac{1}{15}$.

(iii) Cards numbered 3, 5, 6, 9, 10, 12, 15, 18, 20, 21, 24, 25, 27, 30 are multiple of either 3 or 5.

∴ No. of favourable outcomes = 14.

P(getting a card which is a multiple of 3 or 5) = $\dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{14}{30} = \dfrac{7}{15}$.

Hence, the probability of drawing a card which is a multiple of 3 or 5 = $\dfrac{7}{15}$.

In a single throw of two dice, find the probability of :

(i) a doublet

(ii) a number less than 3 on each dice

(iii) an odd number as a sum

(iv) a total of atmost 10

(v) an odd number on one dice and a number less than or equal to 4 on other dice.

Answer

When two dice are rolled simultaneously;

Total number of possible outcomes = 6 × 6 = 36.

(i) For obtaining a doublet, the favourable outcomes are : (1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6).

∴ Number of favourable outcomes = 6

P(getting a doublet) = $\dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{6}{36} = \dfrac{1}{6}$.

Hence, the probability of getting a doublet = $\dfrac{1}{6}$.

(ii) For obtaining a number less than 3 on each dice, the favourable outcomes are : (1, 1), (1, 2), (2, 1), (2, 2).

∴ Number of favourable outcomes = 4

P(getting a number less than 3 on each dice) = $\dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{4}{36} = \dfrac{1}{9}$.

Hence, the probability of getting a number less than 3 on each dice = $\dfrac{1}{9}$.

(iii) For obtaining an odd number as a sum, the favourable outcomes are : (1, 2), (1, 4), (1, 6), (2, 1), (2, 3), (2, 5), (3, 2), (3, 4), (3, 6), (4, 1), (4, 3), (4, 5), (5, 2), (5, 4), (5, 6), (6, 1), (6, 3), (6, 5).

∴ Number of favourable outcomes = 18

P(getting an odd number as a sum) = $\dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{18}{36} = \dfrac{1}{2}$.

Hence, the probability of getting an odd number as the sum = $\dfrac{1}{2}$.

(iv) For obtaining a total of atmost 10, the favourable outcomes are : (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), (6, 1), (6, 2), (6, 3), (6, 4).

∴ Number of favourable outcomes = 33

P(getting a total of atmost 10) = $\dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{33}{36} = \dfrac{11}{12}$.

Hence, the probability of getting a total of atmost 10 = $\dfrac{11}{12}$.

(v) For obtaining an odd number on one dice and a number less than or equal to 4 on other dice, the favourable outcomes are : (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (2, 1), (2, 3), (2, 5), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (4, 1), (4, 3), (4, 5), (5, 1), (5, 2), (5, 3), (5, 4).

∴ Number of favourable outcomes = 20

P(getting an odd number on one dice and a number less than or equal to 4 on other dice) = $\dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{20}{36} = \dfrac{5}{9}$.

Hence, the probability of getting an odd number and a number less than or equal to 4 on other dice = $\dfrac{5}{9}$.