# Class 11 RD Sharma Solutions – Chapter 33 Probability – Exercise 33.3 | Set 1

• Last Updated : 08 May, 2021

### Question 1. Which of the following cannot be valid assignment of probability for elementary events of outcomes of sample space S={w1,  w2, w3, w4, w5, w6, w7} :

Elementary events are:

Solution:

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We know that  the value of probability of an event must be between 0 and 1 and that the sum of all probabilities of a event must be equal to one. Using this rationale,

1. This is valid since each probability of event w1 lies between 0 and 1 and the sum of all probabilities of event w1 is 1, or P{w1}= 1.

2. This is valid since each probability of event w2 lies between 0 and 1 and the sum of all probabilities of event w2  is 1, or P{w2}=2.

3. This is not valid as the sum of all probabilities is 2.8, which is more than one. P{wi} = 2.8

4. This is not valid since the probability of w7 is 15/14= 1.07 > 1.

Hence, the valid events are only 1 and 2.

### (i) a prime number

Solution:

Since a die is thrown, the number of outcomes in sample space must be 6. Thus, n{S}= 6.

Let E be the event of getting a prime number.

∴ E = {2,3,5} or n{E} = 3

We know, Probability of an event = Number of outcomes in that event / number of outcomes in sample space

∴ P{E} = n{E}/ n{S} = 3/6

∴ P{E}= 1/2

### (ii) 2 or 4

Solution:

Let N be the event of getting a 2 or 4.

Hence, N = (2,4} or n{N} = 2

Probability of event N or P{N} = n{N}/n{S}

or P{N} = 2/6

∴ P{N} = 1/3

### (iii) a multiple of 2 or 3

Solution:

Let R be the event of getting a multiple of 2 or 3 when a die is thrown.

∴ R = {2,3,4,6}

Thus, n{R} = 4

∴ P{R} = n{R}/ n{S} = 4/6

∴ P{R} = 2/3

### (i) 8 as the sum

Solution:

Since a pair of die have been thrown, the total number of outcomes in the sample space S become

6×6 = 62= 36.

Let R be the event of getting 8 as the sum when a pair of dice is thrown together.

∴ R = {(2,6), (3,5), (4,9), (5,3), (6,2)}

∴ n{R} = 5

Hence probability of R = n{R}/ n{S}

∴ P{R} = 5/36

### (ii) a doublet

Solution:

Let D be the event that a doublet appear on the dice when a pair of dice is thrown together.

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

Thus, n{D} = 6

Hence probability of D = n{D}/ n{S}

P{D} = 6/36 = 1/6

∴ P{D} = 1/6

### (iii) a doublet of prime numbers

Solution:

Let F be the event of getting a doublet of prime numbers when a pair of dice is thrown together.

∴ F = {(2,2), (3,3), (5,5)}

Thus, n{F} = 3

Hence probability of F = n{F}/ n{S}

P{F}= 3/36 = 1/12

∴ P{F} = 1/12

### (iv) a doublet of odd numbers

Solution:

Let Q be the event of getting a doublet of odd numbers when a pair of dice is thrown.

∴ Q = {(1,1), (3,3), (5,5)}

Thus, n{Q} = 3

Probability of Q = n{Q}/ n{S}

P{Q} = 3/36 = 1/12

∴ P{Q} = 1/12

### (v) a sum greater than 9

Solution:

Let K be the event of getting a sum greater than 9 when a pair of dice is thrown.

∴ K = {(4,6), (5,5), (5,6), (6,4), (6,5), (6,6)}

Thus, n{K} = 6

Probability of K = n{K}/n{S}

P{K}= 6/36 = 1/6

∴ P{K} = 1/6

### (vi) an even number on first

Solution:

Let W be the event of getting an even number on first dice when a pair of dice is thrown. If the first number has to be even, it means that any other number can appear on the second dice, be it even or odd. Using this rationale:

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

Thus, n{W} = 18

Probability of W = n{W}/ n{S}

= 18/36 = 1/2

∴ P{W} = 1/2

### (vii) an even number on one and a multiple of 3 on the other

Solution:

Let V be the event of that we get an even number on one and a multiple of 3 on the other die when a pair of dice is thrown simultaneously. In this case, it is not specified that the even number shall have to be on the face of first or the second dice.

So long as an even number appears on any of the two dices, it is to be taken as an outcome of the given event. Same goes for multiple of 3 as well. Using this rationale:

∴ V = {(2,3), (2,6), (4,3), (4,6), (6,3), (6,6), (3,2), (3,4), (3,6), (6,2), (6,4)}

Thus, n{V} = 11

Probability of V = n{V}/ n{S}

∴ P{V} = 11/36

### (viii) neither 9 nor 11 as the sum of the numbers on the faces

Solution:

Let H be the event of getting neither 9 nor 11 as the sum of the numbers on the faces of two dice when they are thrown together.

Therefore, H’ shall be the event that either 9 or 11 appear as sum of the two numbers on the faces of the two dices.

H’ = {(3,6), (4,5), (5,4), (5,6), (6,3), (6,5)}

n{H’} = 6

P{H’} = n{H’}/ n{S} = 6/36 = 1/6

Thus, P{H} = 1−P{H’} = 1−1/6 = 5/6

### (ix) a sum less than 6

Solution:

Let E be the event of obtaining a sum less then 6 when a pair of die is thrown together.

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

∴ n{E} =10

Probability of E = P{E} = n{E}/ n{S} = 10/36 = 5/18

∴ P(E) = 5/18

### (x) a sum less than 7

Solution:

Let C be the event that a sum less than 7 appears on the faces of the two dice when thrown together.

∴ C = {(1,1), (1,2), (1,3), (1,4), (1,5), (2,1), (2,2),(2,3),(2,4),(2,5),(3,1),(3,2),(3,3).(4,1),(4,2),(5,1)}

Thus, n{C} = 15

Probability of C = P{C} = n{C}/ n{S}

= 15/36 = 5/12

∴ P(C) = 5/12

### (xi) a sum more than 7

Solution:

Let X be an event of getting a sum more than 7 from the numbers on the faces of the two dice when thrown together.

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

Thus n{X} = 15

Probability of X = P{X} = n{X}/ n{S} = 15/36 = 5/12

∴ P(X) = 5/12

### (xii) neither a doublet nor a total of 10

Solution:

Let A be the event of getting neither a doublet nor a total of 10 when a pair of dice is thrown simultaneously.

∴ A’ is the event of getting either a doublet or a total of 10 on the faces of the two dices.

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

Thus, n{A’} = 8

Probability of A’ = n{A’}/ n{S} = 8/36 = 2/9

Probability of A = P{A} = 1−P{A’}

= 1−2/9

∴ P(A’} = 7/9

### (xiii) odd number on the first and 6 on the second

Solution:

Let B be the event of getting an odd number on the first and 6 on the second face of the two dices.

∴ B = {(1,6), (3,6), (5,6)}

Thus n{B} = 3

Probability of B = P{B} = n{B}/ n{S} = 3/36 = 1/12

∴ P(B} = 1/12

### (xiv) a number greater than 4 on each side

Solution:

Let J be the event of getting a number greater than 4 on each side of dice when both dices are thrown simultaneously.

∴ J = {(5,5), (5,6), (6,5), (6,6)}

Thus n{J} = 4

Probability of J = P{J} = n{J}/ n{S} = 4/36 = 1/9

∴ P(J} = 1/9

### (xv) a total of 9 or 11

Solution:

Let I be the event of getting a total of 9 or 11 when two dice are thrown.

∴ I = {(3,6),(4,5),(5,4),(5,6),(6,3),(6,5)}

Thus n{I} = 6

Probability of I = P{I} = n{I}/ n{S} = 6/36 = 1/6

∴ P(I} = 1/6

### (xvi) a total greater than 8

Solution:

Let Z be the event of getting a total greater than 8.

∴ Z = {(3,6),(4,5),(4,6),(5,4),(5,5),(5,6),(6,3),(6,4),(6,5),(6,6)}

∴ n{Z} = 10

Probability of Z = n{Z}/ n{S} = 10/36 = 5/18

∴ P(Z} = 5/18

### Question 4. In a single throw of three dice, find the probability of getting a total of 17 or 18.

Solution:

Since three dice have been thrown simultaneously, the total number of outcomes in that sample space become 6×6×6= 63 = 216.

Let E be the event of getting a total of 17 or 18. Thus E = {(6,6,5), (6,5,6), (5,6,6), (6,6,6)}

Hence, n{E} = 4

Probability of event E = P{E} = n{E}/ n{S} = 4/216 = 1/54

∴ P{E} = 1/54

### Question 5. Three coins are tossed together. Find the probability of getting:

Solution:

Since 2 coins have been tossed together, the total outcomes of the sample space are n{S} =

2×2×2 = 23 = 8.

Let A be the event of getting exactly 2 heads.

∴ A = {HHT, HTH, THH} or, n{A} = 3

Probability of event A = P{A} = n{A}/n{S} = 3/8

∴ P{A} = 3/8

### (ii) at least two heads

Solution:

Let B be the event of getting at least 2 heads when three coins are thrown together.

∴ B = {HHH, HHT, THH, HTH} or, n{B} = 4

Probability of event B = P{B} = n{B}/ n{S} = 4/8 = 1/2

∴ P{B} = 1/2

### (iii) at least one head and one tail

Solution:

Let C be the event of getting at least one head and one tail .

C= {(HHT, THT, HTT, TTH, HTH, THH} or, n{C}= 6

P{C}= n{C}/ N(S) = 6/8 = 3/4

∴ P{C}= 3/4

### Question 6. What is the probability that an ordinary year has 53 Sundays?

Solution:

We know that an ordinary year has 52 weeks and 1 day.

52 weeks in a year obviously implies 52 Sundays.

But in the question, we have to find the probability of having 53 Sundays which means that we have to fine the probability of the last 1 day of an ordinary year to be a Sunday.

Total number of days in a week = Number of outcomes of sample space = n(S) = 7

S= {MONDAY, TUESDAY, WEDNESDAY, THURSDAY, FRIDAY, SATURDAY, SUNDAY}

Hence, the probability of that one day being a Sunday = 1/7

Thus, the probability of having 53 Sundays in an ordinary year is 1/7.

### Question 7. What is the probability that a leap year has 53 Sundays and 53 Mondays?

Solution:

We know that in a leap year we have 52 weeks and 2 days (366 days)

The sample space for the last 2 days shall be

S= {(Monday, Tuesday), (Tuesday, Wednesday), (Wednesday, Thursday), (Thursday, Friday), (Friday, Saturday), (Saturday, Sunday), (Sunday, Monday)} or, n{S}=7

Let E denote the event of getting a Sunday and a Monday as the last 2 days in a leap year.

∴ E= {Sunday, Monday} or, n{E} = 1

Probability of E= P{E}= n{E}/ n{S} = 1/7

Thus, the probability that a leap year has 53 Sundays and 53 Mondays is 1/7.

### (a) all three balls are white

Solution:

Since there are 8+5= 13 balls in the bag and we are required to draw any three balls at random.

Total number of outcomes shall be= n{S}= 13C3 = 286.

Let A be the event that all the three balls drawn are white.

Hence the number of outcomes in event A= n{A} = 5C3= 10.

Probability of event A= n{A}/n{S} = 10/286= 5/143

Hence the probability of drawing all three white balls is 5/143.

### (b) all three balls are red

Solution:

Let R denote the event that all three balls drawn are red.

Since total number of red balls is 8, the number of ways of drawing 3 red balls out of 8 = 8C3 = 56 = n{R}.

We know, n{S}= 286

∴ Probability of R= n{R}/n{S}= 56/286 = 28/143

Hence, the probability of drawing all three red balls is 28/143.

### (c) one ball is red and two balls are white

Solution:

Let E be the event that one ball drawn is red and the other two balls are white.

Number of outcomes of E= 8C1 × 5C2 = 8 ×10 = 80.

∴ Probability of E= n{E}/n{S} = 80/286 = 40/143

Hence the probability of getting one red and other two balls as white is 40/143.

### Question 9. In a single throw of three dice, find the probability of getting same numbers on all the three dice.

Solution:

Since three dice have been rolled together, the total number of outcomes in the sample space= n{S}= 63= 6×6×6= 216.

Let E be the event of getting the same numbers on all the three dice.

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

Number of outcomes in E= n{E} = 6

Probability of E= n{E}/n{S}= 6/216 = 1/36

Hence the probability of getting the same numbers on all the three dice is 1/36.

### Question 10. Two unbiased dice are thrown together. Find the probability that the total of numbers on the dice is greater than 10.

Solution:

Since two dice have been rolled together, the total number of outcomes in the sample space= n{S}= 62= 6×6= 36.

Let E be the event of getting a total of numbers on the dice greater than 10.

∴E= {(5,6), (6,5), (6,6)}

Number of outcomes in E= n{E}= 3

Since n{S}= 52, Probability of E= n{E}/n{S}= 3/36 = 1/12

Hence the probability of getting a total of numbers on the dice greater than 10  is 1/12.

### (i) a black king

Solution:

Since  a card is drawn from a pack of 52 cards, so number of elementary events in the sample space = n {S} = 52C1 = 52.

Let E be the event of drawing a black king. Since there are two black kings, one of spade and other of club,

∴n {E} = 2C1 = 2

Since n{S}= 52, Probability of event E = n{E}/n{S} = 2/52 = 1/26.

Hence the probability of drawing a black king is 1/26.

### (ii) either a black card or a king

Solution:

Let A be the event of drawing a black card or a king. We know that there are 26 black cards and 4 kings out of which 2 kings are black.

Hence, total number of outcomes of event A= n{E} = 26C1 + 4C12C1= 28

In this case, 2 needs to be subtracted from total because there are two black kings which are already counted in black cards and to avoid double counting.

Since n{S}= 52, Probability of A= n{A}/n{S} = 28/52 = 7/13.

Hence, the probability of getting either a black card or a king is 7/13.

### (iii) black and a king

Solution:

Let W be the event of drawing a black card and a king. We know there are two black kings, one of spade and other of club.

Hence number of outcomes of W= n{W} = 2C1 = 2.

Since n{S}=52, Probability of W= n{W}/n{S} = 2/52 = 1/26.

Hence the probability of drawing a black card and a king is 1/26.

### (iv) a jack, queen, or a king

Solution:

Let B be the event of drawing a jack, queen or king. We know there are 4 kings, 4 queens and 4 jacks in a deck of cards.

Hence total number of outcomes of event B = n{B} = 4C1 + 4C1 + 4C1 = 12.

Since n{S}= 52, Probability of B = P{B} = n{B}/n{S} = 12/52 = 3/13

Hence the probability of drawing a jack, queen or king is 3/13.

### (v) neither a heart nor a king

Solution:

Let L be the event of drawing neither a heart nor a king and let L′ as the event that either a heart or king appears.

Since there are 13 hearts and 4 kings total number of outcomes = n (L′) = 6C1 + 4C1 – 1=16 ,deducting the king of hearts.

Probability of L’ = P{L′} = n{L’}/n{S} = 16/52 = 4/13

So, P{L} = 1 – P{L’} = 1 – 4/13 = 9/13

Hence the probability of drawing  neither a heart nor a king is 9/13.

### (vi) spade or an ace

Solution:

Let D be the event of drawing a spade or king. We know there are 13 spades and 4 kings, but 1 king is already included in the 4 kings.

Number of outcomes in D = n{D} = 13C1 + 4C1 – 1=16

Since N{S} = 52, Probability of event D = P{D} = n{D}/n{S} = 16/52 = 4/13

Hence the probability of drawing either a spade or an ace is 4/13.

### (vii) neither an ace nor a king

Solution:

Let K be the event of drawing neither an ace nor a king and K′ as the event that either an ace or king appears.

Number of outcomes in K’ = n{K’} = 4C1 + 4C1 = 8

Since n{S}= 52, Probability of K’ = P{K’} = n{K’}/n{S} = 8/52= 2/13

So, P{K} = 1 – P{K’} = 1 – 2/13= 11/13

Hence the probability of drawing neither an ace nor a king is 11/13.

### (viii) a diamond card

Solution:

Let X be the event of drawing a diamond card. We know there are 13 diamond cards.

Hence number of outcomes of event X = n{X} = 13C1 = 13.

Since n{S} = 52, Probability of event X = P{X} = n{X}/n{S} = 13/52 = 1/4

Hence the probability of drawing a diamond card is 1/4.

### (ix) not a diamond card

Solution:

Let X be the event of drawing not a diamond card and X′ as the event that diamond card appears. We know there are 13 diamond cards.

Hence number of outcomes of event X = n{X} = 13C1 = 13.

Since n{S} = 52, Probability of event X = P{X} = n{X}/n{S} = 13/52 = 1/4

So, P{X’} = 1 – P{X’} = 1 – 1/4 = 3/4

Hence the probability of not drawing a diamond card is 3/4.

### (x) a black card

Solution:

Let E be the event of drawing a black card. We know there are 26 black cards (spades and clubs).

Hence total number of outcomes of event E = n{E} = 26C1 = 26

Since n{S} = 52, Probability of event E = P{E} = n{E}/n{S} = 26/52= 1/2

Hence the probability of drawing a black card is 1/2.

### (xi) not an ace

Solution:

Let E be the event of drawing not an ace and  E′ as the event that ace card appears. We know that there are 4 ace cards.

Number of outcomes in E’ = n{E’} = 4C1 = 4.

Since n{S} = 52, Probability of event E’ = P{E’} = n{E’}/n{S} = 4/52 = 1/13

So, P (E) = 1 – P (E′) = 1 – 1/13 =12/13

Hence the probability of not drawing an ace is 12/13.

### (xii) not a black card

Solution:

Let E be the event of not drawing a black card. We know there are 26 cards other than black cards (red cards of hearts and diamonds)

Hence number of outcomes of drawing a red card = n{E} = 26C1 = 26 .

Since n{S} = 52, Probability of event E = P{E} = n{E}/n{S} = 26/52 = 1/2.

Hence the probability of not drawing a black card is 1/2.

### Question 12. In shuffling a pack of 52 playing cards, four are accidentally dropped. Find the chance that the missing cards should be one from each suit.

Solution:

A pack of 52 cards from which 4 are dropped. We now have to find the probability that the missing cards should be one from each suit.

We know that, from well shuffled pack of cards, 4 cards missed out total possible outcomes = n{S} = 52C4 = 270725

Let E be the event that four missing cards are from each suite

Hence total number of outcomes of event E = n{E} = 13C1 × 13C1 × 13C1 × 13C1 = 134

Probability of event E =P{E} = n{E}/n{S} = 134/270725 = 2197/20825

Hence the probability that the missing cards should be one from each suit is 2197/20825.

### Question 13.  From a deck of 52 cards, four cards are drawn simultaneously. Find the chance that they will be the four honors of the same suit.

Solution:

We have to find the probability that all the face cards  drawn from a pack of 52 cards are of same suits.

Total possible outcomes in sample space =n{S} = 52C4

Let E be the event that all the cards drawn are face cards of same suit.

Hence number of outcomes of event E = n (E)= 4 × 4C4 = 4 × 1 = 4.

Probability of event E =P{E} = n{E}/n{S}= 4/270725

Hence the probability that the face cards  drawn from a pack of 52 cards are of same suits is 4/270725.

### Question 14. Tickets numbered from 1 to 20 are mixed up together and then a ticket is drawn at random. What is the probability that the ticket has a number which is a multiple of 3 or 7?

Solution:

We have to find the probability that the ticket has a number which is a multiple of 3 or 7.

Total possible outcomes in sample space =n{S} = 20C1 = 20.

Let E be the event of getting ticket which has number that is multiple of 3 or 7.

Hence outcomes of event E are {3,6,9,12,15,18,7,14}. Number of outcomes of event E are n{E} = 8.

Probability of event E =P{E} = n{E}/n{S}= 8/20 = 4/5

Therefore, the probability that the ticket has a number which is a multiple of 3 or 7 is 4/5.

### Question 15. A bag contains 6 red, 4 white, and 8 blue balls. If three balls are drawn at random, find the probability that one is red, one is white and one is blue.

Solution:

We have to find the probability that the one is red, one is white and one is blue.

Since three balls are drawn so, total number of outcomes for drawing 3 balls is n{S} = 18C3 = 816

Let E be the event that one red, one white and one blue ball is drawn.

n{E} = 6C1 × 4C1 × 8C1 = 192

Probability of event E =P{E} = n{E}/n{S} = 192 / 816 = 4/17

Therefore, the probability that one is red, one is white and one is blue is 4/17.

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