These AP 10th Class Maths Chapter Wise Important Questions Chapter 13 Probability will help students prepare well for the exams.

## AP State Syllabus 10th Class Maths 13th Lesson Important Questions and Answers Probability

Question 1.

If P(E) = 3/4, what is the probability of “not E”?

Solution:

Probability P(E) = 3/4.

Prabability of “not E” P (\((\overline{\mathrm{E}})\)) = 1 – P(E)

= 1 – \(\frac{3}{4}=\frac{1}{4}\)

Question 2.

You are writing a test of 40 objective type questions. Each question carries 1 mark. What is the probability of marks you may get to be in multiple of 5?

Solution:

Total Number of questions = 40

Total Number of outcomes = 40

Number of Multiples of 5 upto 40 = 8

Favourable outcomes = 8

Probability for getting multiples of 5

Favourable outcomes for getting = \(\frac{\text { multiples of } 5}{\text { Total No. of possible outcomes }}\)

= \(\frac{8}{40}=\frac{1}{5}\)

Question 3.

A page is opened at random from a book containing 100 pages. Find the probability that the page number is a perfect square.

Solution:

Number of pages in given book = 100

The page numbers that will be perfectly a square number (If randomly that are selected) are 1,4,9, 16, 25, 36, 49, 64, 81 and 100.

∴ Number of favourable outcomes = 100

Number of all possible outcomes = its number of pages = 100

∴ Probability of getting a perfect number = \(\frac{10}{100}\) = 0.1

Question 4.

If P(E) = 0.546, what is the probability of ‘not E’ ?

Solution:

P(E) = 0.546

P(E) = 1 – P(E)

Probability of “not E” = 1 – 0.546

= 0.454

Question 5.

A box contains 3 blue and 4 red balls. What is the probability that the ball taken out randomly will be red ?

Solution:

Total number of balls in the box = 3 + 4 = 7

No. of favourable outcomes for picking a red ball = 4

∴ Probability of red ball

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

∴ P(E) = \(\frac{4}{7}\)

Question 6.

A three digit number is formed by the digits 2, 3 and 5 without repetition. What is the probability that the number is divisible by 5?

Solution:

Let ‘E’ be the event of choosing a three digit number divisible by 5.

All possible three digit numbers (without repetition) 235, 253, 325, 352, 523,532.

∴ n(S) = 6.

E = {235,325}

n(E) = 2

∴ P(E) = \(\frac{2}{6}=\frac{1}{3}\)

Question 7.

In a class – room, 32 students out of 60 can take tea. Find the probability of “The tea not taken”.

Solution:

Total Number of possible outcomes = 60

No. of students doesn’t take tea (No. of favourable outcomes) = 60 – 32

= 28

Probability of students not taken tea

= \(\frac{\text { No. of favourable outcomes for not taken tea }}{\text { Total No. of possible outcomes }}\)

= \(\frac{28}{60}=\frac{7}{15}\)

Question 8.

What are equally likely events ? Give one example.

Answer:

Equally likely events :

Two events are said to be equally likely events if the probability of occurrence of those events in that experiment is equal.

Question 9.

A bag contains 5 red and 8 white balls. If a ball is drawn at random from the bag, what is the probability that it will be

i) white ball

ii) not to be white ball

Solution:

Total number of balls present in bag = 5 (red) + 8 (white) = 13

Probability for taking out a white ball

P(E) = \(\frac{\text { No.of favourable out comes }}{\text { Total no.of out comes }}\) = \(\frac{8}{13}\)

Probability for not to be a white ball = p(\(\overline{\mathrm{E}}\))

We know P(E) + p(\(\overline{\mathrm{E}}\)) = 1

⇒ p(\(\overline{\mathrm{E}}\)) = 1 – p(E) = 1 – \(\frac{8}{13}=\frac{5}{13}\)

Question 10.

There are 5 cards in a box with numbers 1 to 5 written on them. If 2 cards are picked out from the box, write all the possible outcomes and find the probability of getting both even numbers.

Solution:

Total number of possible outcomes when 2 cards are picked out from the box = 10

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

∴ Number of favourable outcomes for getting both even numbers = 1, (2, 4)

∴ Probability of getting both even no.s

= \(\frac{\text { Number of favourable outcomes }}{\text { Total no. of possible outcomes }}\) = \(\frac{1}{10}\)

Question 11.

Solution:

A die is thrown once. Find the probability of getting

i) an even number

ii) an odd prime number.

Solution:

When a die is thrown once total number of possible outcomes = 6

1) For an even number, favourable outcomes = 3

Probability for an even number

= \(\frac{\text { No. of favourable outcomes }}{\text { Total No. of possible outcomes }}\)

2) For an odd prime number, favourable outcomes = 2

Probability for an odd prime number = \(\frac{2}{6}=\frac{1}{3}\)

Question 12.

A box contains 100 red cards, 200 yellow cards and 50 blue cards. If a card is drawn at random from the box, then find the probability that it will be

i) a blue card, ii) not a yellow card.

Solution:

Given,

Number of red cards =100

Number of yellow cards = 200

Number of blue cards = 50

Total number of cards in a box = 100 + 200 + 50 = 350

i) Let E_{1} = Event of selecting that a card drawn is blue.

Probability (E_{1})

= \(\frac{\text { Number of favourable outcomes }}{\text { Total Number of outcomes }}\)

= \(\frac{50}{350}=\frac{1}{7}\)

ii) Let E_{2} = Event of selecting that a card drawn is not a yellow card. Probability (E_{2})

= \(\frac{\text { Number of favourable outcomes }}{\text { Total Number of outcomes }}\)

= \(\frac{150}{350}=\frac{15}{35}=\frac{3}{7}\)

Question 13.

Prepare any two problems getting Probability using dice.

Solution:

1) Find the probability of getting an even number when the dice is thrown once?

2) What is the probability of getting an odd number when a dice is thrown once?

Question 14.

A bag contains 5 red, 5 green and 5 white balls of the same size. A ball is drawn at random from the bag. Is the probability of picking up a ball of any colour equally likely or not? Justify.

Solution:

Number of red balls = 5 = n(R)

No. of green balls = 5 = n(G)

No. of white balls = 5 = n(W)

Total balls = 15 = T(B)

Now probability of picking red balls

= P (R) = \(\frac{150}{350}=\frac{15}{35}=\frac{3}{7}\)

= \(\frac{5}{15}=\frac{1}{3}\)

Probability of picking green balls

= \(\frac{5}{15}=\frac{1}{3}\)

Probability of picking white balls

\(\frac{5}{15}=\frac{1}{3}\)

Question 15.

From a deck of 52 playing cards, King, Ace and 10 of Clubs were removed and remaining cards were well shuffled. If a card is drawn at random from the remaining, find the probability of

getting a card of

i) Club

ii) Ace

iii) Diamond king

iv) Club 5.

Solution:

Total number of possible outcomes

= 52 – 3 = 49

(i) Probability of getting a card of club Number of favourable outcomes

= \(\frac{\text { for getting a card of club }}{\text { Total No.of possible outcomes }}\)

= \(\frac{10}{49}\)

(ii) Probability of getting a card of ace = \(\frac{3}{49}\)

(iii) Probability of getting a card of diamond king = \(\frac{1}{49}\)

(iv) Probability of getting a card of club 5 = \(\frac{1}{49}\)

Question 16.

A bag contains 20 discs, which Are numbered from 1 to 20. If one disc is drawn at random from the bag, find the probability that it bears :

i) an even number,

ii) Prime number,

iii) Multiple of 5,

iv) Two digit odd number.

Solution:

Total number of possible outcomes = 20

i) For probability of the disc bears an even number

No. of favourable outcomes = 10 Probability

= \(=\frac{\text { No. of favourable outcomes }}{\text { Total No. of possible outcomes }}\) = \(\frac{10}{20}=\frac{1}{2}\)

ii) For probability of the disc bears a prime number

No.of favourable outcomes = 8

Probability = \(\frac{8}{20}=\frac{2}{5}\)

iii) For probability of the disc bears a multiple of 5

No. of favourable outcomes = 4

Probability = \(\frac{4}{20}=\frac{1}{5}\)

iv) For probability of the disc bears a two digit odd number

No. of favourable outcomes = 5

Probability = \(\frac{5}{20}=\frac{1}{4}\)

Question 17.

Two dice are thrown at the same time. What is the probability that the sum of two numbers appearing on the top of the dice is (a) 10, (b) less than or equal to 12, (c) a prime number, (d) multiple of ‘3’?

Solution:

Total number of possible outcomes when rolling two dice at a time 6 × 6 = 36

Favourable outcomes of getting each sum is 10.

a) Sum be 10 = {(5, 5), (4, 6), (6, 4)}

No. of favourable outcomes = 3

∴ Required probability

= P(E) = \(\frac{3}{36}=\frac{1}{12}\)

b) The outcomes favourable to the event “Less than or equal to 12” be denotes by’F’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) (5,6)

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

No. of outcomes favourable to ‘F’ is n(F) = 36

∴ p(F) = \(\frac{\mathrm{n}(\mathrm{F})}{\mathrm{n}(\mathrm{S})}=\frac{36}{36}\) = 36 = 1

c) The outcomes favourable to the event “Sum of two numbers a prime number” be denoted by ‘G’ are (1, 1) (1, 2) (1,4) (1,6) (2, 1) (2, 3) (2, 5) (3, 2) (3, 4) (4, 1) (4, 3) (5, 2) (5, 6) (6, 1) (6, 5)

No. of outcomes favourable to ‘G’ is n(G) = 15

∴ p(G) = \(\frac{n(G)}{n(S)}=\frac{15}{36}=\frac{5}{12}\)

d) The outcomes favourable to the event “Sum of two numbers be multiple of 3” be denoted by ‘H’ are (1, 2) (1, 5) (2, 1) (2, 4) (3, 3) (3, 6) (4, 2) (4, 5) (5, 1) (5, 4) (6, 3) (6, 6)

No. of outcomes favourable to ‘H’ is n(H) =12

∴ p(H) = \(\frac{n(H)}{n(S)}=\frac{12}{36}=\frac{1}{3}\)

Question 18.

One card Is drawn from a well shuffled deck of 52 cards.

Find the probability of getting :

i) a King of red colour.

ii) the Jack of black.

iii) a black face card.

iv) the Queen of diamonds.

Solution:

Total no. of cards = 52

Total no. of possible outcomes = 52

A card is drawn randomly from a well shuffled deck of cards

i) No. of favourable outcomes forgetting a king of red colour = 2 Probability

= \(\frac{\text { No. of favourable outcomes }}{\text { Total No. of possible outcomes }}\)

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

ii) No. of favourable outcomes forgetting the jack of black = 2

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

iii) No. of favourable outcomes forgetting a black face card = 6

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

iv) No. of favourable outcomes forgetting the Queen of diamonds = 1

Probability = \(\frac{1}{52}\)

Question 19.

Two digit numbers are formed by the digits 0, 1, 2, 3, 4, where the digits are not repeated. Find the probability that

i) the number formed is greater than 42.

ii) the number formed is a multiple of 4.

Solution:

Two digit numbers formed by the digits 0, 1,2,3,4

where the digits are not repeated

(10, 12, 13, 14, 20, 21, 23, 24, 30, 31, 32, 34, 40, 41, 42, 43)

∴ Sample space = (10, 12, 13, 14, 20, 21,23,24,30,31,32,34,40,

41, 42, 43)

∴ n(S) = 16

i) Probability of getting the number formed is greater than 42

\(No. of possible outcomes $F$ $=$ Total outcomes\)

here no. of possible outcomes = 1 (that is 43 only)

Probability = (1)

\(\frac{1}{16}\) ……………….(1)

ii) In the sample space multiples of ‘4’ = 12, 20, 24, 32, 40

∴ No. of multiples of 4 = 5

Now probability for forming a multiple of ‘4’ = \(\frac{5}{16}\) …………….. (2)

Question 20.

From a pack of 52 playing cards; Jacks, Queens, Kings and Aces of red colour are removed. From the remaining, a card is drawn at random. Find the probability that the card drawn is

i) a black queen,

ii) a face card,

iii) a blackjack,

iv) a red card.

Solution:

Total number of cards = 52

Card removed = 2 + 24-2 + 2 = 8

∴ Remaining number of cards = 52 – 8 = 44

i) Number of black queens = 2

∴ Required probability = \(\frac{2}{44}=\frac{1}{22}\)

ii) Number of face cards left = 2 + 2 + 2 = 6

∴ Required probability = \(\frac{6}{44}=\frac{3}{22}\)

iii) Number of blackjacks = 2

∴ Required probability = \(\frac{2}{44}=\frac{1}{22}\)

iv) Number of red cards left = 26 – 8

= 18

∴ Required probability = \(\frac{18}{44}=\frac{9}{22}\)

Question 21.

There are 3 red and 4 white balls in a bag. If a ball is taken randomly then calculate the probability of it to be a) red ball b) white ball.

Solution:

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

∴ Total outcomes = 7

a) Number of favourable outcomes for picking a red ball = 3

∴ Probability of red ball = \(\frac{\text { No. of favourable outcomes }}{\text { No. of total outcomes }}\) = \(\frac{3}{7}\)

outcomes 7

P(E) = \(\frac{3}{7}\)

Now number of favourable outcomes for picking a white ball = 4

∴ Probability for picking a white ball = \(\frac{4}{7}\)

Question 22.

Find the probability of existing 53 Sundays in a common year.

Solution:

In a common year, no. of days = 365

So number of weeks = 365 = (52) 7 + (1)

That means there are 52 weeks and 1 more day in a year.

So this 1 day may be any day of the week,

i.e., from Sunday to Saturday.

So number of possible outcomes = 7

Number of favourable outcomes for Sunday = 1

Hence the probability for 53 Sundays = \(\frac{1}{7}\)

Question 23.

In a skinner numbered from 1-20, find the probability of getting the following

(i) prime number

(ii) composite number

(iii) multiple of three.

Solution:

i) Numbers in the skinner is 1 to 20.

So the number of all possible outcomes = 20

List of prime numbers in the above skinner = 2, 3, 5, 7, 11, 13, 17, 19

So, number of favourable outcomes for a prime number = 8

∴ Probability of a prime number = \(\frac{8}{20}=\frac{2}{5}\)

ii) List of composite numbers in above skinner = 4,6,8,9,10,12,14,15,16,18,20

∴ Number of favourable outcomes for a composite number = 11

∴ Probability of getting composite number = \(\frac{11}{20}\)

iii) List of multiples of 3 in above skinner = 3, 6, 9, 12, 15 and 18

∴ Number of favourable outcomes for a multiple of 3 = 6

∴ Probability of getting a multiple of 3 = \(\frac{6}{20}=\frac{3}{10}\)

Question 24.

When a six face die is rolled find the probability of getting the following

i) getting less than five

ii) getting more than five.

Solution:

Faces on the dice = 6

∴ number of total possible outcomes = 6

i) List of possible out comes less than five = 1, 2, 3, 4

So number of favourable outcomes for less than five = 4

∴ Probability of getting a number less than five = \(\frac{4}{6}=\frac{2}{3}\)

ii) List of possible outcomes for getting more than five = (6)

∴ Number of possible outcomes for getting more than five = 1

∴ Probability of getting more than five = \(\frac{1}{6}\)

Question 25.

When two dice are rolled.

Fill the probability of the sum on their faces in the table given below.

Draw the bar graph for the above.

Solution:

Now on a graph paper

Consider “Sum on faces” on x-axis and respective probabilities on y-axis to draw bar graphs from the above table.

Question 26.

In a 50 marks examination, there is 80% possibility to pass in that exam. So find the probability for pass in exam.

Solution:

Possibility for pass in Exam = 80% = \(\frac{80}{100}\)

that means in out of 100 chances (total possible out comes = 100) 80 chances are favourable to pass (number of favourable out comes = 80)

∴ Probability for pass = \(=\frac{\text { number of favourable outcomes }}{\text { number of total possible outcomes }}\)

= \(\frac{80}{100}=\frac{4}{5}\)

then its probability = \(\frac{4}{5}\)

Question 27.

In a class 32 students out of 60 take tea. So find the probability of choosing randomly at student who doesn’t take tea.

Solution:

Total number of students = 60 = Number of possible outcomes.

Number of students take tea = 32

So number of students who don’t take tea = 60 – 32 = 28

Number of favourable outcomes for choosing a boy who doesn’t take tea = 28

∴ Its probability = \(\frac{28}{60}=\frac{7}{15}\)

Question 28.

In the following table, temperature and rainfall conditions are stated. Then find the possibility of rainfall and no rainfall.

Solution:

Position – I:

i) Rainfall possibility is high when temperature is more than 35°C.

Position – II :

Rainfall possibility is low when temperature is below 31°C.

Question 9.

Find the probability of winning of each team from the given table.

Solution:

i) Probability of winning to India when bats first = \(\frac{20}{30}\left(\frac{\text { Won }}{\text { Played }}\right)=\frac{2}{3}\) = 0.66

ii) Probability of Pakistan ~ 28 = \(\frac{12}{28}=\frac{3}{7}\) = 0.43

iii) Probability of Australia = \(\frac{22}{44}=\frac{1}{2}\) = 0.50

iv) Probability of Sri Lanka = \(\frac{20}{50}=\frac{2}{5}\) = 0.40

So highest probability of winning is with INDIA i.e., = \(\frac{140}{210}=\frac{2}{3}\) = 0.66