**ML Aggarwal ICSE Solutions Class 10 Math 22th Chapter Probability**

ML Aggarwal ICSE Solutions Class 10 Math chapter 22 probability**Chapter 22 – ****Probability**

**(1) A box contains 600 Screws, one – tenth are rusted. One screw is taken out at random from this box. Find the Probability that it is a good screw.**

**Solution:**

Given that,

A box contains 600 screws, one – tenth are rusted. And one screw is taken out at random from this box.

Then, total no. of screws = 600

Rusted screws = 1/10 of 600 = 60

Thus, good screws remained = 600 – 60 = 540

Thus, Probability to get a good screw is found to be

P (E) = no. of good screws/ total screws = 540/600.

P (E) = 9/10 is the required Probability.

**(2) In a lottery, there are 5 prized tickets and 995 blank tickets. A person buys a lottery ticket. Find the Probability of his wining a pried.**

**Solution:**

Given that,

In a lottery, there are 5 prized tickets and 995 blank tickets.

A person buys a lottery ticket.

Then, no. of prized tickets = 5

Number of blank tickets = 995

Then, total no. of tickets available = 5 + 995 = 1000

Thus, Probability of Prized ticket is found to be

P (E) = no. of prized ticket/ total tickets = 5/1000 = 1/200

P (E) = 1/200 is the required Probability.

**(3) 12 defective pens are accidently mixed with 132 good ones. It is not possible to just look at a pen and tell whether or not it is defective. One pen is taken out at random from this lot. Determine the Probability that the pen taken out is a good one.**

**Solution:**

Given that,

12 defective pens are accidentally mixed with 132 good ones.

It is not possible to just look at a pen not it is defective.

And one pen is taken out at random from this lot.

Here, no. of defective pens = 12

No. of good pens = 132

Then, total no. of pens = 12 + 132 = 144

Thus, Probability to get a good pen is found to be

P (E) = no. of good pens/ total no. of pens = 132/144 = 11/12

P (E) = 11/12 is the required Probability.

**(4) Two players, Sonia and Somali play a tennis match. It is known that the Probability of Sonia winning the match is 0.69. What is the Probability of Somali winning?**

**Solution:**

Given that,

Two players, Sonia and Sonali play a tennis match. It is given that, the Probability of Sonia winning is found to be 0.69.

Let us consider, the Probability of Sonia’s winning the game is P (E).

and P (E) is the Probability of Sonia’s loosing the game.

Thus

P (E) + P(Ē) = 1

0.69 + P(Ē) = 1

P(Ē) = 1 – 0.69 = 0.31

P(Ē) = 0.31

Thus, the Probability of winning the game for Sonali is found to be 0.31.

**(5) A bag contains 3 red balls and 5 black balls. A ball is drawn at random from the bag. What is the Probability that the ball drawn is**

**(i) Red?**

**(ii) Not red?**

**Solution:**

Given that,

A bag contains 3 red balls and 5 black balls. And a ball is drawn at random from the bag.

No. of red balls = 3

No. of black balls = 5

Total no. of balls = 3 + 5 = 8

Let us consider, the Probability of finding the red ball is found to be P (E)

Then P(Ē) is the probability of finding the ball which is not red

Thus P(E) + P(Ē) = 1

(i) P(E) = no of red balls / total no. of balls = 3/8

The probability to get red ball is found to be 3/8

(ii) Now, P (Ē) = 1 – P(E)

= 1 – 3/8

= 8-3/8

P (Ē) = 5/8 is the probability to get the ball which is not red

**(6) A letter is chosen from the word ‘TRIANGLE’. What is the Probability that it is a Vowel?**

**Solution:**

Given that,

A letter is chosen from the word ‘TRIANGLE’. In given word there are three vowels i.e. I, A, E.

Total no. of letters present in the given word = 8

No. of vowels present in the given word = 3

Then, Probability of Choosing a letter should be a vowel is found to be,

P (E) = no. of vowels/total no. of letters = 3/8

P (E) = 3/8 is the required Probability.

**(7) A letter of English alphabet is chosen at random. Determine the Probability that the letter is a consonant.**

**Solution:**

Given that,

A letter of English alphabet is chosen at random.

Total no. of letters of English alphabet = 26

No. of Vowels = 5

Then, no. of consonant = 26 – 5 = 21

Thus, Probability to get letter which is consonant is given by

P (E) = no. of consonant/total no. of letters = 21/26

P (E) = 21/26

**(9) A bag contains 6 red balls, 8 white balls, 5 green balls and 3 black balls. One ball is drawn at random from the bag. Find the Probability that the ball is**

**(i) White**

**(ii) red or black**

**(ii) not green**

**(iv) neither white nor black.**

**Solution:**

Given that,

A bag contains 6 red balls, 8 white balls, 5 green balls and 3 black balls.

And one ball is drawn at random from the bag.

No. of red balls = 6

No. of white balls = 8

No. of green balls = 5

No. of black balls = 3

Total no. of balls in a bag = 22

(i) The Probability to get the ball which is white is given by

P (E) = no. of white balls/ total no. of balls = 8/22 = 4/11

P (E) = 4/11

(ii) The Probability to get the ball which is red or black is given by

P (E) = no. of red balls + no of black balls/ total no of balls

P (E) = 6+3/22 = 9/22

P (E) = 9/22

(iii) The Probability to get the ball which is not green (means it may be red or white or black) is given by

P (E) = no. of red balls + no of black balls + no of black balls / total no. of balls

P (E) = (6 + 8 + 3)/22 = 17/22

P(E) = 17/22

(iv)The Probability to get the ball which is neither white nor black is given by

P (E) = no of red balls + no. of green balls / total no. of balls

P (E) = 1/2

**(11) A die is thrown once. What is the Probability that the**

**(i) Number is even**

**(ii) Number is greater than 2**

**Solution:**

Given that,

A die is thrown once then sample space is given by S = { 1, 2, 3, 4, 5, 6} = h (s) = 6

(i) The numbers which are even E = {2, 4, 6} = h (E) = 3

Then, Probability to get the no. which is even is given by

P (E) = n (E) / n (s) = 3/6 = 1/2

(ii) The no. which are greater than 2

E = {3, 4, 5, 6} and n (E) = 4

Then, Probability to get the no. which is greater than 2 by

P (E) = n (E) / n (s) = 4/6 = 2/3

P (E) = 2/3

** **

** (13) A die has 6 faces marked by the given numbers as Shown below:**

1 |
2 |
3 |
-1 |
-2 |
-3 |

** **

**The die is thrown once. What is the Probability of getting**

**(i) A Positive integer**

**(ii) An integer greater than -3**

**(iii) The Smallest** **integer**

**Solution:**

Given that, A die has 6 faces marked by the given number as

1 | 2 | 3 | -1 | -2 | -3 |

Total no. of outcomes n(s) = 6

(i) Positive metiers (E) = {1, 2 ,3} = n (E) = 3

Thus, the Probability of getting a positive integer is found to be

P (E) = n (E) / n (s) = 3/6 = 1/2

P (E) = 1/2

(ii) an integer greater than – 3 = E = {-2, -1, 3, 2, 1} and n (E) = 5

Thus, the Probability of getting an integer greater than (-3) is found to be

P (E) = n (E) / n (s) = 5/6

P (E) = 5/6

(iii) The smallest integer = E = {-3} and (E) = 1

Thus, the Probability of getting the smallest integer is found to be

P (E) = n (E) / n (s) = 1/6

**(15) Find the Probability that the month of January, may, have 5 Mondays is **

**(a) a leap year**

**(b) a non leap year**

**Solution:**

We Know that,

In the month of January, there are 31 days in an ordinary Year.

In an ordinary year 365 days but in a leap year 366 days are there.

(i) In the month of January of an ordinary year there are 31 days i.e. 4 weeks and 3 days.

Then, P (E) = 3/7

(ii) In the month of January of a leap year, there are 31 days i.e. 4 weeks and 3 days.

P (E) = 3/7

**(16) Find the Probability that the month of February, May have 5 Wednesdays in**

**(i) a leap year**

**(ii) a non leap year**

**Solution:**

In the month of February, there are 29 days in a leap year while 28 days in a non leap year.

(i) In a leap year, there are 4 complete weeks and 1 day.

Then Probability of Wednesday = P (E) = 1/7

(ii) In a non leap year, there are 4 complete weeks and days in the month of February.

Then Probability of Wednesday = P (E) = 0/7 = 0

**(17) Sixteen Cards are labelled as a, b, c, m, n, o, p. They are put in a box and shuffled. A boy is asked to draw a card from the box. What is the Probability that the **

**(i) a Vowel**

**(ii) a consonant**

**(iii) none of the letters of the word median**

**Solution:**

Given that,

Sixteen cards are labelled as a, b, c, ….. m, n, o, p. They are put in a box and shuffled.

And a boy is asked to draw a card from the box.

Here, Sample space (S) = {a, b, c, d, e, f, g, h, i, j, k, l, m, n, o, p}

n (s) = 16

(i) no. of vowels E = { a, e, i, o} = n (E) = 4

Then, Probability that the card drawn is vowel is given by

P (E) = n (E) / n (s) = 4/16 = 1/4

(ii) No. of consonants E = { b, c, d, f, g, h, j, k, l, m, n, p}

n (E) = 12

Then, Probability that the card drawn is consonant is given by

P (E) = n (E) / n (s) = 12/16 = 3/4

(iii) None of the letters of the} E = {b, c, f, g, h, j, k, l, o, p} word median

n (E) = 10

Then, Probability that the card drawn is none of the letters of the word median

P (E) = n (E) / n (s) = 10/16 = 5/8

**(18) An integer is chosen between 0 and 100. What is the Probability that it is**

**(i) divisible by 7**

**(ii) not divisible by 7**

**Solution:**

Given that,

An integer is chosen between 0 and 100.

Total no. of integers between of 100 = 99

n (s) = 99

(i) The no. which are divisible by 7:

E = {7, 14, 21, 28, 35, 42, 56, 63, 70, 77, 24, 91, 98}

N (E) = 14

Then, Probability to get a no. divisible by 7 is given by

P (E) = n (E) / n (s) = 14/99

(ii) No. which are not divisible by 7

N (E) = 99 – 14 = 85

Then, Probability of getting the no. which is not divisible by 7 is found to be

P (E) = n (E) / n (s) = 85/99

**(20) There are 25 discs numbered 1 to 25. They are Put in a closed box and shaken thoroughly. A disc is drawn at random from the box.**

**Find the Probability that the no. on the disc is**

**(i) an odd no.**

**(ii) divisible by 2 and 3 both**

**(iii) a number less than 16**

**Solution:**

Given that,

There are 25 discs numbered 1 to 25.

They are put in a closed box and shaken thoroughly. And a disc is drawn at random from the box.

Here, Sample space S = {1, 2, 3, ….. 25} and n (s) = 25

(i) odd no. bet 1 to 25: E = {1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25}

n (E) = 13

Then, Probability to get no. on the disc on odd no. is found to be

P (E) = n (E) / n (s) = 13/25

(ii) No. which are divisible by 2 and 3 both:

E = {6, 12, 18, 24} and n (E) = 4

Then, the Probability to get no. on the disc which is divisible by 2 and 3 is found to be

P (E) = n (E) / n (s) = 4/25

(iii) No. less than 16 = E = {1, 2, 3, …..15} and n (E) = 15

Then, Probability to get a no. on the disc than 16 is found to be

P (E) = n (E) / n (s) = 15/25 = 3/5

** (22) Cards bearing numbers 2, 4, 6, 8, 10, 12, 14, 16, 18 and 20 are kept in a bag. A card is drawn at random from the bag. Find the Probability of getting a card which is **

**(i) a Prime number**

**(ii) a number divisible by 4**

**(iii) a no. that is a multiple of 6**

**(iv) an odd no.**

**Solution:**

Given that,

Cards bearing numbers 2, 4, 6, 8, 10, 12, 14, 16, 18 and 20 are kept in a bag.

A card is drawn at random from the bag.

Here, Sample Space S = {2, 4, 6, 8, 10, 12, 14, 16, 18, 20}

n (S) = 10

(i) Prime no E = {2} = n (E) = 1

Then, Probability of getting a card which is a prime number is found to be

P (E) = n (E) / n (s) = 1/10

P (E) = 1/10

(ii) No. divisible by 4: E = {4, 8, 12, 16, 20} and n (E) = 5

Then, Probability of getting a card which is a no. divisible by 4 is found to be

P (E) = n (E) / n (s) = 5/10 = 1/2

(iii) No. which are divisible by 6 or multiplies of 6:

E = {6, 12, 18} and n (E) = 3

Then, Probability of getting the no. which is multiple of 6 is found to be

P (E) = n (E) / n (s) = 5/10 = 3/10

(iv) odd no. E = {.} and n (E) = 0

Then, Probability of getting the no. which is odd no. is found to be

P (E) = n (E) / n (s) = 0/10

** (23) Cards marked with numbers 13, 14, 15, ….60 are placed in a box and mixed thoroughly. One card is drawn at random from the box. Find the Probability that the number on the card drawn is **

**(i) divisible by 5**

**(ii) a perfect square no.**

**Solution:**

Given that,

Cards marked with numbers 13, 14, 15,….., 60 are placed in a box and mixed thoroughly.

One card is drawn at random from the box.

Number of card S = {13, 14, 15,…., 60} = n (s) = 48

(i) Cards on which no, which is divisible by S

E = {15, 20, 25, 30, 35, 40, 45, 50, 55, 60} and n (E) = 10}

Thus, Probability that the number on the card drawn is divisible by S is found to be

P (E) = n (E) / n (s) = 10/48 = 5/24

(ii) Perfect Square no. E = {16, 25, 36, 49} and n (E) = 4

Thus, Probability that the no. on the card drawn is a perfect square no. is found to be

P (E) = n (E) / n (s) = 4/48 = 2/24 = 1/12

** (24) Tickets numbered 3, 5, 7, 9, ….., 29 are placed in a box and mixed thoroughly. One ticket is drawn at random from the box. Find the Probability that the no. on the ticket is **

**(a) Prime no.**

**(b) a no less than 16**

**(c) a no divisible by 3**

**Solution:**

Given that,

Tickets numbered 3, 5, 7,9,….., 29 are placed in a box and mixed thoroughly.

One ticket is drawn at random from the box.

No. on tickets S = {3, 5, 7, 9,…., 29} and n (s) = 14

S = {3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29}

(i) Prime no. E = {3, 5, 7, 11, 13, 17, 19, 23, 29}

n (E) = 9

Thus, Probability that the no. on the ticket is a prime no. is found to be

P (E) = n (E) / n (s) = 9/14

P (E) = 9/14

(ii) Numbers less than 16 E = {3, 5, 7, 9, 11, 13, 15} & n (E) = 7

Thus, Probability that the no. on the ticket is a no. less than 16 is found to be

P (E) = n (E) / n (s) = 7/14 = 1/2

(iii) Numbers divisible by 3: E = {3, 9, 15, 21, 23} and n (E) = 5

Thus, Probability that the no. on the ticket is divisible by 3 is found to be

P (E) = n (E) / n (s) = 5/14

P (E) = 5/14

**(26) A bag contains is balls of which some are white and others are red. If the Probability of drawing a red ball is twice that of a white ball, find the number of white balls in the bag.**

**Solution:**

Given that,

A bag contains is balls of which some are white and other are red.

And Probability of drawing a red ball is twice that of a white ball.

Let us consider, the no. of white balls in a bag be ‘x’.

Then, no. of red balls in a bag = (15 – x)

From given condition,

2 x ( 15 – x/ 15) = x/15

2 ( 15 – x ) = x

30 – 2x = x

3x = 30

X = 10

Thus,

{The no. of white balls = 10

The no. of red balls = 15 – 10 = 5}

**(27) A bag contains 6 red balls and some blue. If the Probability of drawing a blue ball is twice that of a red ball, find the no. of balls in the bag.**

**Solution:**

Given that,

A bag contains 6 red balls & some blue balls. And the Probability of drawing a blue ball is twice that of a red ball.

Let us Consider, blue balls in no. = x

Red balls in no. = 6

Total no. of balls = x + 6

P (E_{b}) = 2 P (E_{r})

x/ (x + 6) = 2 [6/x + 6]

x = 12 are the blue balls in no.

Then, total no. of balls in the bag = x + 6 = 12 + 6 = 18

**(28) A card is drawn from a well – shuffled pack of 52 cards.**

**Find the Probability of getting**

**(i) 2 of Spades**

**(ii) a jack**

**(iii) a king of red colour**

**(iv) a card of diamond**

**(v) a king or a queen**

**(vi) a non – face card**

**Solution:**

Given that,

A card is drawn from a well – shuffled pack of 52 cards.

Number of possible out wines = 52

(i) The Probability of getting 2 of spades is given by.

P (E) = no. of favourable outcomes/ total outcomes = 1/52

P (E) = 1/52

(ii) There are total 4 jack Present in a card suit.

Then, Probability of getting a jack is found to be

P (E) = no. of favourable outcomes/ total outcomes = 4/52 = 1/13

P (E) = 1/13

(iii) There are 2 king of red colour in cards.

Then, Probability of getting a king of red colour is found to be

P (E) = no. of favourable outcomes/ total outcomes = 2/52 = 1/26

P (E) = 1/26

(iv) There are total 13 cards of diamond.

Then, Probability of getting a card of diamond is found to be

P (E) = 13/52= 1/4

(v) Number of Kings or queen = 4 + 4 = 8

Then, Probability of getting a king or queen is found to be

P (E) = 8/52 = 4/26 = 2/13

P (E) = 2/13

(vi) There are non – face cards = 52 – (3 × 4) = 52 – 12 = 40

Then, Probability of getting a non – face card is found to be

P (E) = 40/50 = 20/26 = 10/13

P (E) = 10/13

**(30) All the three face cards of spades are removed from a well – shuffled pack of 52 cards. A card is then drawn at random from the remaining pack. Find the Probability of getting**

**(i) a black face card**

**(ii) a queen**

**(ii) a black card**

**(iv) a heart**

**Solution:**

Given that,

All the three face cards of spades are removed from a well – shuffled pack of 52 cards.

And a card is drawn at random from the remaining pack.

No. of remaining cards = 52 – 3 = 49 = n (s) = 49

(i) Total black face cards = 6 – 3

Then, Probability of getting a black face card is found to be

P (E) = n (E) / n (s) = 3/49

P (E) = 3/49

(ii) Number of queens = 4 – 1 = 3

Then, Probability of getting a queen is found to be

P (E) = n (E) / n (s) = 3/49

(iii) Number of black cards = 26 – 3 = 23

Then, Probability of getting a black card is found to be

P (E) = n (E) / n (s) = 23/49

P (E) = 13/49

(iv) Number of heart = 13

Then, Probability of getting a heart card is found to be

P (E) = n (E) / n (s) = 13/49

P (E) = 13/49

**(31) From a pack of 52 cards, black jack, a red queen and two black kings fell down. A card was then drawn from the remaining pack at random. Find the Probability that the card drawn is**

**(i) a black card**

**(ii) a king**

**(iii) a red queen**

**Solution:**

Given that,

From a pack of 52 cards, a black jack, a red queen and two black kings fell down.

And a card is drawn from the remaining pack at random.

The no. of remaining cards = 52 – (1 + 1 + 2) = 48

n (s) = 48

(i) Probability of getting a black card is found to be

P (E) = n (E) / n (s) = 23/48

P (E) = 23/48

(ii) No. of king = 4 – 2 = 2

Then, Probability of getting a king is found to be

P (E) = n (E) / n (s) = 2/48 = 1/24

P (E) = 1/24

(iii) No. of red queens = 2 – 1 = 1

Then, Probability of getting a red queen is found to be

P (E) = n (E) / n (s) = 1/48

P (E) = 1/48

**(32) Two coins are tossed once. Find the Probability of getting**

**(i) 2 heads**

**(ii) at least one tail**

**Solution:**

Given that,

Two coins are tossed.

Sample Space S = {HH, HT, TT, TH} and n (s) = 4

(i) Probability of getting at least 2 heads is found to be

P (E) = n (E) / n (s) = 1/4

P (E) = 1/4

(ii) At least one tail E = {HT, TT, TH} and n (E) = 3

Then, Probability of getting at least one tail is found to be

P (E) = n (E) / n (s) = 3/4

P (E) = 3/4

**(34) Two different dice are thrown simultaneously.**

**Probability of getting**

**(i) a number greater than 3 on each dice**

**(ii) an odd number on both dice**

**Solution:**

Given that,

Two different dice are thrown simultaneously.

Sample Space S = { (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)}

n (s) = 36

(i) Numbers greater than 3 on each dice

E = {(4,4), (4,5), (4,6), (5,4), (5,5), (5,6), (6,4), (6,5), (6,6)}

n = (E) = 9

Then, Probability of getting a number greater than 3 on each dice is found to be

P (E) = n (E) / n (s) = 9/36 = 1/4

P (E) = 1/4

(ii) odd no. on both dice: E = {(1,1), (1,3), (1,5), (3,1), (3,3), (3,5), (5,1), (5,3), (5,5)}

n (E) = 9

Then, Probability of getting a odd no. on both dice is found to be

P (E) = n (E) / n (s) = 9/36 = 1/4

**(35) Two different dice are thrown at the same time find the Probability of getting**

**(i) a doublet**

**(ii) a sum of 8**

**Solution:**

Given that,

Two different dice are thrown at the same time.

n (s) = 36

(i) a doublet = E = {(1,1), (2,2), (3,3), (4,4), (5,5), (6,6)}

and n (E) = 6

Then, Probability of getting a doublet is found to be

P (E) = n (E) / n (s) = 6/36 = 1/6

(ii) Sum of 8: E = {(2,6), (3,5), (4,4), (5,3), (6,2)}

n (E) = s

Then, Probability of getting a sum of 8 is found to be

P (E) = n (E) / n (s) = 5/36

P (E) = 5/36