RS Aggarwal Class 10 Solutions Probability

The likeliness of an event occurring is termed Probability and it helps to improve upon the certainty of events occurring. It is a branch of mathematics which measures an event’s occurrence by quantifying it between two numbers 0 and 1. Here 1 indicates the absolute certainty of an event occurring and 0 tells us that an event is impossible.

Mathematically, probability can simply be expressed as: \(p(a)=\frac{p(a)}{[p(a)+p(b)]}\)

Some examples of where Probability has practical applications are:

  1. Throwing a Die
  2. Tossing a Coin
  3. Deck of Cards

Learn more about RS Aggarwal Class 10 Solutions Chapter 15 Probability below:

QUESTION 1: Fill in the blanks :

i) The probability an impossible event is _______.

ii) The probability of an sure event is________.

iii) For an event E, P(E) + P(not E)_________ .

iv) The probability of a possible but not a sure event lies between ______and ________.

v) The sum of the probability of all the outcomes of an experiment is________.

Solution:

i) The probability an impossible event is 0.

ii) The probability of a sure event is 1.

iii) For an event E, P(E) + P(not E) = 1.

iv) The probability of a possible but not a sure event lies between 0 and 1.

v) The sum of the probability of all the outcomes of an experiment is 1.

QUESTION 2: A coin is tossed once. What is the probability of getting a tail?

Solution:

When a coin is tossed outcomes are either head or tail

Let, the head is denoted by H

the tail is denoted by T

Sample space (S) = { H , T }

Hence, Probability of getting a tail = \(\frac{number\, of\, favorable \, outcomes}{total\, number\, of\, outcomes}\) = \(\frac{1}{2}\) = 0.5

QUESTION 3: 2 coins are tossed simultaneously. Find the probability of getting

i) exactly one head

ii) at most one head

iii) at least one head.

Solution:

When a coin is tossed outcomes are either head or tail

Let, head is denoted by H and tail is denoted by T

Sample space (S) = {HH , HT, TH, TT}

i) For exactly one head, favorable outcomes are {HT, TH }

Hence, Probability of getting exactly one head = \(\frac{number\, of\, favorable \, outcomes}{total\, number\, of\, outcomes}\) = \(\frac{2}{4}\) = 0.5

ii) For getting at most one head, favorable outcomes are {HT, TH, TT}

Hence, Probability of getting at most one head = \(\frac{number\, of\, favorable \, outcomes}{total\, number\, of\, outcomes}\) = \(\frac{3}{4}\) = 0.75

iii) For getting at least one head, favorable outcome are {HT, TH, HH}

Probability of getting at least one head = \(\frac{number\, of\, favorable \, outcomes}{total\, number\, of\, outcomes}\) = \(\frac{3}{4}\) = 0.75

QUESTION 4: A die is thrown once. Find the probability of getting

i) an even number

ii) a number less than 5

iii) a number greater than 2

iv) a number between 3 and 6

v) a number other than 3

vi) the number 5

Solution:

When a die is rolled possible outcomes are 1,2,3,4,5 and 6

Hence, sample space (S) = {1,2,3,4,5,6}

i) For an even number, favorable outcomes are {2,4,6}

Hence, Probability of getting even number = \(\frac{number\, of\, favorable \, outcomes}{total\, number\, of\, outcomes}\) = \(\frac{3}{6}\) = 0.5

ii ) For a number less than 5 , favorable outcomes are {1,2,3,4}

Hence, Probability of getting a number less than 5 = \(\frac{number\, of\, favorable \, outcomes}{total\, number\, of\, outcomes}\) = \(\frac{4}{6}\) = 0.667

iii) For a number greater than 2, favorable outcomes are { 3 , 4 , 5 , 6 }

Hence, Probability of getting a number greater than 2 = \(\frac{number\, of\, favorable \, outcomes}{total\, number\, of\, outcomes}\) = \(\frac{4}{6}\) = 0.667

iv) For a number between 3 and 6 , favorable outcomes are { 4 , 5 }

Hence, Probability of getting a number less than 5 = \(\frac{number\, of\, favorable \, outcomes}{total\, number\, of\, outcomes}\) = \(\frac{4}{6}\) = 0.667

v) For a number other than 3, favorable outcomes are {4 , 5}

Hence, Probability of getting a number other than 3 = \(\frac{number\, of\, favorable \, outcomes}{total\, number\, of\, outcomes}\) = \(\frac{4}{6}\) = 0.667

vi) The number 5

Hence, Probability of getting 5 = \(\frac{number\, of\, favorable \, outcomes}{total\, number\, of\, outcomes}\) = \(\frac{1}{6}\) = 0.167

QUESTION 5: A letter of the English alphabet is chosen at random. Determine the probability that the chosen letter is a consonant.

Solution:

Let E be the event of getting a consonant.

Out of 26 letters of English alphabets, there are 21 consonants.

Therefore,  P (getting a consonant) = P(E) = \(\frac{Number\, of\, outcomes\, favorable\, to\, E}{Number\, of\, all\, possible\, outcomes}\) = \(\frac{21}{26}\)

Thus, the probability of getting a consonant is \(\frac{21}{26}\).

QUESTION 6: A child has a die whose six faces shows the letter given below:

A B C A D A

The die is thrown once. What is the probability of getting?

i) A

ii) D

In a single throw of a die, the possible outcomes are:

A, B, C, A, D, A

Total number of possible outcomes = 6

(i) Let E1 be the event of getting A.

Number of favorable outcomes = 3

Therefore,  P(getting A) = P(E1) = \(\frac{3}{6}=\frac{1}{2}\)

(ii) Let E2 be the event of getting D.

Number of favorable outcomes = 1

Therefore,  P(getting B) = P(E2) = \(\frac{1}{6}\)

QUESTION 7: It is known that a box of 200 electric bulbs contains 16 defective bulbs .1 bulb taken out at random from the box. What is the probability that the bulb drawn is

i) defective

ii) nondefective

Solution:

Total number of possible outcomes = 200

(i) Number of detective bulbs = 16

Therefore, P (getting a defective bulb) = \(\frac{16}{200}=\frac{2}{25}\)

(ii) Number of non-defective bulbs = 200 – 16 = 184

Therefore, P (getting a non-defective bulb) = \(\frac{184}{200}=\frac{23}{25}\)

QUESTION 8: If the probability of winning a game is 0.7, what is the probability of losing it?

Solution:

For any event E, P (E) + P (not E) = 1

Let probability of winning a game = P (E) = 0.7

Therefore, P (winning a game) + P (losing a game) = 1

\(\Rightarrow\) P(losing a game) = 1 – 0.7 = 0.3

Thus, the probability of losing a game is 0.3.

QUESTION 9: There are 35 students In a class of whom 20 years boys and 15 are girls. From these students, one is chosen at random. What is the probability that the chosen student is a

i) boy

ii) girl

Solution:

Total number of students = 35

Number of boys = 20

Number of girls = 15

(i) Let E1 be the event that the chosen student is a boy.

Therefore,  P(choosing a boy) = P(E1) = \(\frac{Number\, of\, outcomes\, favorable\, to\, E_{1}}{Number\, of\, all\, possible\, outcomes}\) = \(\frac{20}{35}\) = \(\frac{4}{7}\)

Thus, the probability that the chosen student is a boy is \(\frac{20}{35}\).

(ii) Let E2 be the event that the chosen student is a girl.

Therefore,  P(choosing a girl) = P(E2) = \(\frac{Number\, of\, outcomes\, favorable\, to\, E_{2}}{Number\, of\, all\, possible\, outcomes}\) = \(\frac{15}{35}\) = \(\frac{3}{7}\)

Thus, the probability that the chosen student is a girl is \(\frac{3}{7}\).

QUESTION 10: In a lottery, there are 10 prizes and 25 blanks. What is the probability of getting a prize?

Solution:

Total number of lottery tickets = 10 + 25 = 35

Number of prizes = 10

Let E be the event of getting a prize.

Therefore,  P(getting a prize) = P(E) = \(\frac{Number\, of\, outcomes\, favorable\, to\, E}{Number\, of\, all\, possible\, outcomes}\) = \(\frac{10}{35}\) = \(\frac{2}{7}\)

Thus, the probability of getting a prize is \(\frac{2}{7}\).

QUESTION 11: 250 lottery tickets were sold and there are 5 prizes on these tickets. If Kunal has purchased 1 lottery ticket, what is the probability that he wins a prize?

Solution:

Total number of tickets = 250

Kunal wins a prize if he gets a ticket that assures a prize.

Number of tickets on which prizes are assured = 5

Therefore, P(Kunal wins a prize) = \(\frac{5}{250}\) = \(\frac{1}{50}\)

QUESTION 12: 17 cards numbered 1, 2, 3, 4, 5, . . . . . . . . ,17 are put in a box and mixed thoroughly. A card is drawn at random from the box. Find the probability that the card is drawn bears

i) an odd number

ii) a number divisible by 5.

Solution:

Total number of cards = 17

(i) Let E1 be the event of choosing an odd number.

These numbers are 1, 3, 5, 7, 9, 11, 13, 15 and 17.

Therefore,  P(getting an odd number) = P(E1) = \(\frac{Number\, of\, outcomes\, favorable\, to\, E_{1}}{Number\, of\, all\, possible\, outcomes}\) = \(\frac{9}{17}\)

Thus, the probability that the card drawn bears an odd number is \(\frac{9}{17}\).

(ii) Let E2 be the event of choosing a number divisible by 5.

These numbers are 5, 10 and 15.

Therefore,  P(getting a number divisible by 5) = P(E2) = \(\frac{Number\, of\, outcomes\, favorable\, to\, E_{2}}{Number\, of\, all\, possible\, outcomes}\) = \(\frac{3}{17}\)

Thus, the probability that the card drawn bears a number divisible by 5 is \(\frac{3}{17}\).

QUESTION 13: A game of chance consists of spinning an arrow, which comes to rest pointing at one of the numbers 1, 2, 3, 4, 5, 6, 7, 8 and these are equally likely outcomes. Find the probability that the arrow will point at any factor of 8.

Solution:

Number of all possible outcomes = 8

Let E be the event of getting any factor of 8.

These numbers are 1, 2, 4 and 8.

Therefore, P(arrow will point at any factor of 8) = P(E) = \(\frac{Number\, of\, outcomes\, favorable\, to\, E}{Number\, of\, all\, possible\, outcomes}\) = \(\frac{4}{8}\) = \(\frac{1}{2}\)

Thus, the probability that the arrow will point at any factor of 8 is \(\frac{1}{2}\).

QUESTION 14: In a family of three children, find the probability of having at least one boy.

Solution:

All possible outcomes are BBB, BBG, BGB, GBB, BGG, GBG, GGB and GGG.

Number of all possible outcomes = 8

Let E be the event of having at least one boy.

Then the outcomes are BBB, BBG, BGB, GBB, BGG, GBG and GGB.

Number of possible outcomes = 7

Therefore, P(Having at least one boy) = P(E) = \(\frac{Number\, of\, outcomes\, favorable\, to\, E}{Number\, of\, all\, possible\, outcomes}\) = \(\frac{7}{8}\)

Thus, the probability of having at least one boy is \(\frac{7}{8}\).

QUESTION 15: A bag contains 4 white balls, 5 red balls, 2 black balls and 4 green balls. A ball is drawn at random from the bag. Find the probability that it is

i) Black

ii) Not green

iii) Red or white

iv) Neither red nor green

Solution:

Total number of balls = 15

(i) Number of black balls = 2

Therefore, P(getting a black ball) = \(\frac{Number\, of\, favorable\, outcomes }{Number\, of\, all\, possible\, outcomes}\) = \(\frac{2}{15}\)

Thus, the probability of getting a black ball is \(\frac{2}{15}\).

(ii) Number of balls which are not green = 4 + 5 + 2 = 11

Therefore, P(getting a ball which is not green) = \(\frac{Number\, of\, favorable\, outcomes }{Number\, of\, all\, possible\, outcomes}\) = \(\frac{11}{15}\)

Thus, the probability of getting a ball which is not green is \(\frac{11}{15}\).

(iii) Number of balls which are either red or white = 4 + 5 = 9

Therefore,  P(getting a ball which is red or white) = \(\frac{Number\, of\, favorable\, outcomes }{Number\, of\, all\, possible\, outcomes}\) = \(\frac{9}{15}\) = \(\frac{3}{5}\)

Thus, the probability of getting a ball which is red or white is \(\frac{3}{5}\).

(iv) Number of balls which are neither red nor green = 4 + 2 = 6

Therefore,  P(getting a ball which is neither red nor green) = \(\frac{Number\, of\, favorable\, outcomes }{Number\, of\, all\, possible\, outcomes}\) = \(\frac{6}{15}\) = \(\frac{2}{5}\)

Thus, the probability of getting a ball which is neither red nor green is \(\frac{2}{5}\).

QUESTION 16: A card is drawn at random from a well-shuffled pack of 52 cards. Find the probability of getting

i) a red king

ii) a queen or a jack.

Solution:

Total number of cards = 52

(i) Number of red kings = 2

Therefore,  P(getting a red king) = \(\frac{Number\, of\, favorable\, outcomes }{Number\, of\, all\, possible\, outcomes}\) = \(\frac{2}{52}\) = \(\frac{1}{26}\)

Thus, the probability of getting a red king is \(\frac{1}{26}\).

(ii) Number of queens or jacks = 4 + 4 = 8

Therefore,  P(getting a queen or a jack) = \(\frac{Number\, of\, favorable\, outcomes }{Number\, of\, all\, possible\, outcomes}\) = \(\frac{8}{52}\) = \(\frac{2}{13}\)

Thus, the probability of getting a queen or a jack is \(\frac{2}{13}\).

QUESTION 17: A card is drawn at random from a well-shuffled pack of 52 cards. Find the probability that the drawn card neither a king nor a queen.

Solution:

Favorable outcomes = 52 – 4kings – 4queens = 44

Total outcomes = 52

P(E) = \(\frac{Number\, of\, outcomes\, favorable\, to\, E}{Number\, of\, all\, possible\, outcomes}\) = \(\frac{44}{52}\) = \(\frac{11}{13}\)

Thus, the probability that the drawn card is neither a king nor a queen is \(\frac{11}{13}\).

QUESTION 18: A card is drawn at random from a well-shuffled pack of 52 cards. Find the probability of getting

(i) a red face card

(ii) a black king.

Solution:

(i) Favorable outcomes = 2red kings + 2red queens + 2red jack = 6

Total outcomes = 52

P(E) = \(\frac{Number\, of\, outcomes\, favorable\, to\, E}{Number\, of\, all\, possible\, outcomes}\) = \(\frac{6}{52}\) = \(\frac{3}{26}\)

Thus, the probability of getting a red face card is \(\frac{3}{26}\).

(ii) Favorable outcomes = 2 (because there are 4 kings, 2 black and 2 red)

Total outcomes = 52

P(E) = \(\frac{Number\, of\, outcomes\, favorable\, to\, E}{Number\, of\, all\, possible\, outcomes}\) = \(\frac{2}{52}\) = \(\frac{1}{26}\)

Thus, the probability of getting a black king is \(\frac{1}{26}\).

QUESTION 19: Two different dice tossed together. Find the probability that

i) The number on each die is even,

ii) The sum of the numbers appearing on the two dice is 5.

Solution:

When two different dice are thrown, then the total number of outcomes = 36.

(i) Let E1 be the event of getting an even number on each die.

These numbers are (2, 2), (2, 4), (2, 6), (4, 2), (4, 4), (4, 6), (6, 2), (6, 4), (6, 6).

Number of favorable outcomes = 9

Therefore,  P(getting an even number on both dice) = P(E1) = \(\frac{Number\, of\, outcomes\, favorable\, to\, E_{1}}{Number\, of\, all\, possible\, outcomes}\) = \(\frac{9}{36}\) = \(\frac{1}{4}\)

Thus, the probability of getting an even number on both dice is \(\frac{1}{4}\).

(ii) Let E2 be the event of getting the sum of the numbers appearing on the two dice is 5.

These numbers are (1, 4), (2, 3), (3, 2) and (4, 1)

Number of favorable outcomes = 4

Therefore, P(getting the sum of the numbers appearing on the both dice is 5) = P(E2) = \(\frac{Number\, of\, outcomes\, favorable\, to\, E_{2}}{Number\, of\, all\, possible\, outcomes}\) = \(\frac{4}{36}\) = \(\frac{1}{9}\)

Thus, the probability of getting the sum of the numbers appearing on the two dice is 5 is \(\frac{1}{9}\).


Practise This Question

If 35 is the upper limit of the class-interval of class-size 10, then the lower limit ofthe class- interval is :