CBSE Class 12 Mathematics Probability Worksheet Set 06

CONTENTS

• Definiation
• Terms
• Approaces Probability
• Types of experimental Probability

DEFINITIONS

Words 'chance' probably, or most probably etc. shows uncertainty in our statements.

The uncertainty of 'probably' etc. can be measured numerically by means of 'probability'.

TERMS

Definitions :

Trial and Event :

An experiment is called a trial if it results in anyone of the possible outcomes and all the possible outcomes are called events.

For Example

• Participation of player in the game to win a game, is a trial but winning or losing is an event.
• Tossing of a fair coin is a trial and turning up head or tail are events.
• Throwing of a dice is a trial and occurrence of number 1 or 2 or 3 or 4 or 5 or 6 are events.
• Drawing a card from a pack of playing cards is a trial and getting an ace or a queen is an event.

Favourable Events :

Those outcomes of a trial in which a given event may happen, are called favourable cases for that event.

For Example -

• If a coin is tossed then favourable cases of getting H is 1.
• If a dice is thrown then favourable case for getting 1 or 2 or 3 or 4 or 5 or 6, is 1.
• If two dice are thrown, then favourable cases of getting a sum of numbers as 9 are four i.e \( (4,5), (5,4), (3,6), (6,3) \).

Sample Space :

The set of all possible outcomes of a trial is called its sample space. It is generally denoted by S and each outcome of the trial is said to be a point of sample of S.

For example

• If a die is thrown once, then its sample space \( S = \{1, 2, 3, 4, 5, 6\} \)
• If two coins are tossed together then its sample space \( S = \{HT, TH, HH, TT\} \).

APPROACHES PROBABILITY

• (i) Experimental or Empirical or observed frequency approach.
• (ii) Classical approach
• (iii) Axiomatic approach

(i) Experimental Probability : Let there be \( n \) trials of an experiment and \( A \) be an event associated to it such that \( A \) happens in \( m \)-trials. Then the empirical probability of happening of event \( A \) is denoted by \( P(A) \) and is given by

\[ P(A) = \frac{m}{n} \]

i.e.,

\[ P(A) = \frac{\text{Number of trials in which the event happens}}{\text{Total number of trials}} \]

Clearly, \( 0 \le m \le n \). Therefore,

\[ 0 \le \frac{m}{n} \le 1 \implies 0 \le P(A) \le 1 \]

Thus, the probability of happening of an event always lies between 0 and 1. If \( P(A) = 1 \), then \( A \) is called a certain event and \( A \) is known as an impossible event, If \( P(A) = 0 \).

Further, if \( \bar{A} \) denotes negative of \( A \) i.e. event that \( A \) doesn’t happen, then for above cases \( m \), \( n \); we shall have

\[ P(\bar{A}) = \frac{n - m}{n} = 1 - \frac{m}{n} = 1 - P(A) \]

\[ \therefore P(A) + P(\bar{A}) = 1 \]

Playing Cards :

• (i) Total : 52 (26 red, 26 black)
• (ii) Four suits : Heart (♥) , Diamond (♦), Spade (♠), Club (♣) - 13 cards each
• (iii) Court Cards : 12 (4 Kings, 4 queens, 4 jacks)
• (iv) Honour Cards:16 (4 aces, 4 kings, 4 queens, 4 jacks)

TYPES OF EXPERIMENTAL PROBABILITY

1. Deterministic : Deterministic experiments are those experiments which when repeated under identical conditions produce the same result or outcome. For example, if we mark head (H) on both sides of a coin and it is tossed, then we always get the same outcome assuming that it does not stand vertically.

2. Random or probabilistic : If an experiment, when repeated under identical conditions, do not produce the same outcome every time but the outcome in a trial is one of the several possible outcomes, then it is known as a random or probabilistic experiment. For example, in the tossing of a coin one is not sure if a head (H) or tail (T) will be obtained, so it is a random experiment. Similarly, rolling an unbiased die is an example of a random experiment.

EXAMPLES

Question. A coin is tossed 500 times with the following frequencies of two outcomes :
Head : 240 times, tail : 260 times
Find the probability of occurrence of each of these event.
Answer: It is given that the coin is tossed 500 times.
\( \therefore \) Total number of trials = 500
Let us denote the event of getting a head and of getting a tail by A and B respectively. Then,
Number of trials in which the event A happens = 240.
and, Number of trials in which the event B happens = 260.
\( \therefore P(A) = \frac{\text{Number of trials in which the event A happens}}{\text{Total number of trials}} = \frac{240}{500} = 0.48 \)
\( \therefore P(B) = \frac{\text{Number of trials in which the event B happens}}{\text{Total number of trials}} = \frac{260}{500} = 0.52 \)
Note : We note that \( P(A) + P(B) = 0.48 + 0.52 = 1 \). Therefore, A and B are the only two possible outcomes of trials.

Question. A die is thrown 1000 times with the following frequency for the outcomes 1, 2, 3, 4, 5 and 6 as given below :
Outcome : 1, 2, 3, 4, 5, 6
Frequency: 179, 150, 157, 149, 175, 190
Find the probability of happening of each outcome.
Answer: Let \( A_i \) denote the event of getting the outcome \( i \), where \( i = 1, 2, 3, 4, 5, 6 \). Then,
\( P(E_1) = \text{Probability of getting outcome 1} = \frac{\text{Frequency of 1}}{\text{Total number of times the die is thrown}} = \frac{179}{1000} = 0.179 \)
\( P(E_2) = \text{Probability of getting outcome 2} = \frac{\text{Frequency of 2}}{\text{Total number of times the die is thrown}} = \frac{150}{1000} = 0.15 \)
Similarly, we have,
\( P(E_3) = \frac{157}{1000} = 0.157 \),
\( P(E_4) = \frac{149}{1000} = 0.149 \),
\( P(E_5) = \frac{175}{1000} = 0.175 \),
and, \( P(E_6) = \frac{190}{1000} = 0.19 \)

The percentage of marks obtained by a student in the monthly unit tests are given below :

Question. Find the probability that the student gets a first class i.e. at least 60 % marks.
Answer: Total number of unit tests held = 5
Number of unit test in which the student gets a first class i.e. at least 60 % marks = 4.
\( \therefore \) Probability that the student gets a first class = \( \frac{4}{5} = 0.8 \)

Question. Find the probability that the student gets marks between 70 % and 80 %.
Answer: Total number of unit tests held = 5
Number of tests in which the student gets between 70 % and 80 % = 1.
\( \therefore \) Probability that a student gets marks between 70 % and 80 % = \( \frac{1}{5} = 0.2 \)

Question. Find the probability that the student gets a distinction i.e. 75 % or above.
Answer: Total number of unit tests held = 5
Number of tests in which the student gets distinction = 2.
\( \therefore \) Probability that a student gets distinction = \( \frac{2}{5} = 0.4 \)

Question. Find the probability that the student gets less than 65 % marks.
Answer: Total number of unit tests held = 5
Number of tests in which the student gets less than 65 % marks = 3
\( \dots \) Probability that a student gets less than 65 % marks = \( \frac{3}{5} = 0.6 \)

On one page of a telephone directory, there were 200 telephone numbers. The frequency distribution of their unit place digit (for example, in the number 25828573, the unit place digit is 3) is given in the table below :

Question. A number is chosen at random, find the probability that the digit at its unit's place is 6.
Answer: Total number of selected telephone numbers = 200
It is given that the digit 6 occurs 14 times at unit's place.
\( \therefore \) Probability that the digit at unit's place is 6 = \( \frac{14}{200} = 0.07 \)

Question. A number is chosen at random, find the probability that the digit at its unit's place is a non-zero multiple of 3.
Answer: Total number of selected telephone numbers = 200
A non-zero multiple of 3 means 3, 6 and 9.
Number of telephone number in which unit's digit is either 3 or 6 or 9 = 22 + 14 + 20 = 56.
\( \therefore \) Probability of getting a telephone number having a multiple of 3 at unit's place = \( \frac{56}{200} = 0.28 \)

Question. A number is chosen at random, find the probability that the digit at its unit's place is a non-zero even number.
Answer: Total number of selected telephone numbers = 200
Number of telephone number having an even number (2 or 4 or 6 or 8) at unit's place = 22 + 20 + 14 + 16 = 72
\( \therefore \) Probability of getting a telephone number having an even number at units place = \( \frac{72}{200} = 0.36 \)

Question. A number is chosen at random, find the probability that the digit at its unit's place is an odd number.
Answer: Total number of selected telephone numbers = 200
Number of telephone number having an odd digit (1 or 3 or 5 or 7 or 9) at units' place = 26 + 22 + 10 + 28 + 20 = 106
\( \therefore \) Probability of getting a telephone number having an odd numbr at unit's place = \( \frac{106}{200} = 0.53 \)

A tyre manufacturing company kept a record of the distance covered before a tyre to be replaced. Following table shows the resuts of 1000 cases.

Question. If you buy a tyre of this company, what is the probability that it will need to be replaced before it has covered 400 km ?
Answer: Total number of trials = 1000
\( \therefore \) Probability that a tyre will need to be replaced before it has covered 400 km = \( \frac{210}{1000} = 0.21 \)

Question. If you buy a tyre of this company, what is the probability that it will last more that 900 km ?
Answer: Total number of trials = 1000
The number of tyres that last more than 900 km = 385 + 80 = 465
\( \therefore \) Probability that a tyre will last more than 900 km = \( \frac{465}{1000} = 0.465 \)

Question. If you buy a tyre of this company, what is the probability that it will need to be replaced after it has covered somewhere between 400 km and 1400 km ?
Answer: Total number of trials = 1000
The number of tyres which require replacement after covering distance between 400 km and 1400 km = 325 + 385 = 710.
\( \dots \) Probability that a tyre require replacement 400 km and 1400 km = \( \frac{710}{1000} = 0.71 \)

Question. If you buy a tyre of this company, what is the probability that it will not need to be replaced at all ?
Answer: Total number of trials = 1000
The number of tyres that do not need to be replaced at all = 0
\( \therefore \) Probability that a tyre does not need be replaced = \( \frac{0}{1000} = 0 \)

Question. If you buy a tyre of this company, what is the probability that it will need to be replaced ?
Answer: Total number of trials = 1000
Since all the tyres we have considered to be replaced, so Probability that a tyre needs to be replaced = \( \frac{1000}{1000} = 1 \)

Fifty seeds were selected at random from each of 5 bags of seeds, and were kept under standardised conditions favourable to germination. After 20 days the number of seeds which had germinated in each collection were counted and recorded as follows :

Question. What is the probability germinated of more than 40 seeds is a bag ?
Answer: Total number of bags = 5
Number of bags in which more than 40 seeds germinated out of 50 seeds = 3.
\( \therefore \) Probability of germinated of more than 40 seeds in a bag = \( \frac{3}{5} \).

Question. What is the probability germinated of 49 seeds in a bag ?
Answer: Total number of bags = 5
Number of bags in which 49 seeds germinated = 0.
\( \therefore \) Probability of germination of 49 seeds = \( \frac{0}{5} = 0 \)

Question. What is the probability germinated of more than 35 seeds in a bag ?
Answer: Total number of bags = 5
Number of bags in which more than 35 seeds germinated = 5.
\( \therefore \) Probability of germination of more than 35 seeds = \( \frac{5}{5} = 1 \).

Question. What is the probability germinated of at least 40 seeds in a bag ?
Answer: Total number of bags = 5
Number of bags in which at least 40 seeds germinated = 4
\( \therefore \) Probability of germination of at least 40 seeds = \( \frac{4}{5} \)

Question. What is the probability germinated of at most 40 seed in a bag ?
Answer: Total number of bags = 5
Number of bags in which at most 40 seeds germinated = 2.
\( \therefore \) Probability of germination of at most 40 seeds = \( \frac{2}{5} \)

The distance (in km) of 40 female engineers from their residence to their place of work were found as follows -

Question. Find the probability that an engineer lives : less than 7 km from her place of work ?
Answer: Total number of female engineers = 40
Number of female engineers living at a distance less than 7 km from their place of work = 10.
\( \therefore \) Probability that a female engineer lives at a distance less than 7 km from her place of work = \( \frac{10}{40} = \frac{1}{4} = 0.25 \)

Question. Find the probability that an engineer lives : at least 7 km from her place of work ?
Answer: Total number of female engineers = 40
Number of female engineers living at least 7 km away from their place of work = 30
\( \dots \) Probability that a female engineer lives at least 7 km away from her place of work = \( \frac{30}{40} = 0.75 \)

Question. Find the probability that an engineer lives : within \( \frac{1}{2} \) km from her place of work ?
Answer: Total number of female engineers = 40
Since there is no engineer living at a distance less than \( \frac{1}{2} \) km from her place of work.
\( \therefore \) Probability that an engineer within \( \frac{1}{2} \) km from her place of work = \( \frac{0}{40} = 0 \).

Question. Find the probability that an engineer lives : at most 15 km from her place of work ?
Answer: Total number of female engineers = 40
Number of engineers living at a distance of 15km or less away from her place of work = 30.
\( \therefore \) Probability that an engineer lives at most 15 km away from her place of work = \( \frac{30}{40} = 0.75 \)

An insurance company selected 2000 drivers at random in a particular city to find a relationship between age and accidents. The data obtained are given in the following table:

Question. Find the probability of a driver chosen at random form the life city: being 18-29 years of age and having exactly 3 accidents in one year.
Answer: Total number of drivers = 2000
The number of drivers who are 18-29 years old and have exactly 3 accidents in one year is 61.
\( \therefore \) Probability of a driver being 18-29 years of age and has exactly 3 accidents = \( \frac{61}{2000} = 0.0305 \)

Question. Find the probability of a driver chosen at random form the life city: being 30-50 years of age and having one or more accidents in a year.
Answer: Total number of drivers = 2000
The number of drivers 30-50 years of age and having one or more accidents in one year = 125 + 60 + 22 + 18 = 225.
\( \therefore \) Probability of a driver being 30-50 years of age and having one or more accidents = \( \frac{225}{2000} = 0.1125 \)

Question. Find the probability of a driver chosen at random form the life city: having no accidents in one year.
Answer: Total number of drivers = 2000
The number of drivers having no accidents in one year = 440 + 505 + 360 = 1305
\( \therefore \) probability of a driver having no accident in one year = \( \frac{1305}{2000} = 0.653 \)

Question. Find the probability that a number selected at random from the numbers 1 to 25 is not a prime number when each of the gievn number is equally likely to be selected.
Answer: Here \( S = \{1, 2, 3, 4, \dots, 25\} \)
Let \( E \) = event of getting a prime number = \( \{2, 3, 5, 7, 11, 13, 17, 19, 23\} \).
Then, \( n(E) = 9 \).
\( \therefore P(E) = \frac{n(E)}{n(S)} = \frac{9}{25} \).
Required probability = \( 1 - P(E) = 1 - \frac{9}{25} = \frac{16}{25} \).

Question. Eleven bags of wheat flour, each marked 5 kg. actually contained the following weights of flour (in kg.) :
4.97 5.05 5.08 5.03 5.00 5.06 5.08 4.98 5.04 5.07 5.00
Find the probability that any of these bags chosen at random contains more than 5 kg of flour.
Answer: Total number of bags = 11
Number of bags containing more than 5 kg of flour = 7
Therefore, probability of bags containing more then 5 kg of flour = \( \frac{\text{Number of bags containing more than 5kg flour}}{\text{Total number of bags}} = \frac{7}{11} \)

The record of a weather station shows that out of the past 250 consecutive days, its weather forecasts were correct 175 times.

Question. What is the probability that on a given day it was correct ?
Answer: The total number of days for which the record is available = 250
P(correct forecast) = \( \frac{\text{Number of days when the forecast was correct}}{\text{Total number of days for which the record is available}} = \frac{175}{250} = 0.7 \)

Question. What is the probability that it was not correct on a given day ?
Answer: The total number of days for which the record is available = 250
The number of days when the forecast was not correct = 250 – 175 = 75.
P(not correct forecast) = \( \frac{75}{250} = 0.3 \)

Question. If the probability of winning a game is 0.3, what is the probability of lossing it ?
Answer: Probability of winning a game = 0.3.
Probability of losing it = q (say).
\( \implies 0.3 + q = 1 \implies q = 1 – 0.3 \implies q = 0.7 \)

Two coins are tossed simultaneously.

Question. Find the probability of getting two heads.
Answer: Let H denotes head and T denotes tail. On tossing two coins simultaneously, all the possible outcomes are \( \{HH, HT, TH, TT\} \).
The probability of getting two heads = P(HH) = \( \frac{\text{Event of occurrence of two heads}}{\text{Total number of possible outcomes}} = \frac{1}{4} \)

Question. Find the probability of getting at least one head.
Answer: Let H denotes head and T denotes tail. On tossing two coins simultaneously, all the possible outcomes are \( \{HH, HT, TH, TT\} \).
The probability of getting at least one head = P(HT or TH or HH) = \( \frac{\text{Event of occurrence of at least one head}}{\text{Total number of possible outcomes}} = \frac{3}{4} \)

Question. Find the probability of getting no head.
Answer: Let H denotes head and T denotes tail. On tossing two coins simultaneously, all the possible outcomes are \( \{HH, HT, TH, TT\} \).
The probability of getting no head = P(TT) = \( \frac{\text{Event of occurrence of no head}}{\text{Total number of possible outcomes}} = \frac{1}{4} \)

Question. On tossing three coins at a time, find -
(i) All possible outcomes.
(ii) events of occurrence of 3 heads, 2 heads, 1 head and 0 head.
(iii) probability of getting 3 heads, 2 heads, 1 head and no head.
Answer: Let H denotes head and T denotes tail. On tossing three coins at a time,
(i) All possible outcomes = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}. These are the 8 possible outcomes.
(ii) An event of occurrence of 3 heads = {HHH} = 1
An event of occurrence of 2 heads = {HHT, HTH, THH} = 3
An event of occurrence of 1 head = {HTT, THT, TTH} = 3

SOLVED EXAMPLES

Question. A bag contains 12 balls out of which \( x \) are white,
(i) If one ball is drawn at random, what is the probability that it will be a white ball ?
(ii) If 6 more white balls are put in the bag, the probability of drawing a white ball will double than that in (i). Find \( x \).
Answer: Random drawing of balls ensures equally likely outcomes
Total number of balls = 12
\( \therefore \) Total number of possible outcomes = 12
Number of white balls = \( x \)
(i) Out of total 12 outcomes, favourable outcomes = \( x \)
\( \text{P(White ball)} = \frac{\text{Number of favourable outcomes}}{\text{Total number of possible outcomes}} = \frac{x}{12} \)
(ii) If 6 more white balls are put in the bag, then
Total number of white balls = \( x + 6 \)
Total number of balls in the bag = \( 12 + 6 = 18 \)
\( \text{P(White ball)} = \frac{\text{Number of favourable outcomes}}{\text{Total number of possible outcomes}} = \frac{x + 6}{18} \)
According to the question,
Probability of drawing white ball in second case = \( 2 \times \) probability of drawing white ball in first case
\( \Rightarrow \frac{x + 6}{18} = 2 \left(\frac{x}{12}\right) \)
\( \Rightarrow \frac{x + 6}{18} = \frac{x}{6} \)
\( \Rightarrow 6x + 36 = 18x \)
\( \Rightarrow 12x = 36 \)
\( \Rightarrow x = 3 \)
Hence, number of white balls = 3

Question. What is the probability that a leap year, selected at random will contain 53 Sundays ?
Answer: Number of days in a leap year = 366 days
Now, 366 days = 52 weeks and 2 days
The remaining two days can be
(i) Sunday and Monday
(ii) Monday and Tuesday
(iii) Tuesday and Wednesday
(iv) Wednesday and Thursday
(v) Thursday and Friday
(vi) Friday and Saturday
(vii) Saturday and Sunday
For the leap year to contain 53 Sundays, last two days are either Sunday and Monday or Saturday and Sunday.
\( \therefore \) Number of such favourable outcomes = 2
Total number of possible outcomes = 7
\( \therefore \text{P(a leap year contains 53 sundays)} = \frac{2}{7} \)

Question. Three unbiased coins are tossed together. Find the probability of getting :
(i) All heads,
(ii) Two heads
(iii) One head
(iv) At least two heads.
Answer: Elementary events associated to random experiment of tossing three coins are
HHH, HHT, HTH, THH, HTT, THT, TTH, TTT
\( \therefore \) Total number of elementary events = 8.
(i) The event "Getting all heads" is said to occur, if the elementary event HHH occurs i.e. HHH is an outcome. Therefore,
\( \dots \) Favourable number of elementary events = 1
Hence, required probability = \( \frac{1}{8} \)
(ii) The event "Getting two heads" will occur, if one of the elementary events HHT, THH, HTH occurs.
\( \therefore \) Favourable number of elementary events = 3
Hence, required probability = \( \frac{3}{8} \)
(iii) The events of getting one head, when three coins are tossed together, occurs if one of the elementary events HTT, THT, TTH happens.
\( \dots \) Favourable number of elementary events = 3
Hence, required probability = \( \frac{3}{8} \)
(iv) If any of the elementary events HHH, HHT, HTH and THH is an outcome, then we say that the event "Getting at least two heads" occurs.
\( \therefore \) Favourable number of elementary events = 4
Hence, required probability = \( \frac{4}{8} = \frac{1}{2} \)

Question. A piggy bank contains hundred 50 p coins, fifty Re 1 coins, twenty Rs 2 coins and ten Rs 5 coins. If it is equally likely that one of the coins will fall out when the bank is turned upside down, what is the probability that the coin
(i) will be a 50 p coin ?
(ii) will not be a Rs 5 coin ?
Answer: Number of 50 p coins = 100
Number of Re 1 coins = 50
Number of Rs 2 coins = 20
Number of Rs 5 coins = 10
(i) The number of favourable outcomes of 50 p coin to fall = 100
Total number of coins = \( 100 + 50 + 20 + 10 = 180 \)
Total number of possible outcomes = 180
\( \text{P} = \frac{\text{Number of favourable outcomes}}{\text{Total number of possible outcomes}} \)
\( \text{P(50 p)} = \frac{100}{180} = \frac{5}{9} \)
(ii) Number of favourable outcomes of Rs 5 coin to not fall = \( 180 - 10 = 170 \)
\( \text{P} = \frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}} \)
\( \text{P(not Rs 5)} = \frac{170}{180} = \frac{17}{18} \)

Question. A box contains 20 balls bearing numbers, 1, 2, 3, 4, ... 20. A ball is drawn at random from the box. What is the probability that the number on the balls is
(i) An odd number
(ii) Divisible by 2 or 3
(iii) Prime number
(iv) Not divisible by 10
Answer: Total number of possible outcomes = 20
\( \text{Probability} = \frac{\text{Number of favourable outcomes}}{\text{Total number of possible outcomes}} \)
(i) Number of odds out of first 20 numbers = 10
Favourable outcomes of odd = 10
\( \text{P(odds)} = \frac{\text{Favourable outcomes of odd}}{\text{Total number of possible outcomes}} = \frac{10}{20} = \frac{1}{2} \)
(ii) The numbers divisible by 2 or 3 are 2, 3, 4, 6, 8, 9, 10, 12, 14, 15, 16, 18, 20.
Favourable outcomes of numbers divisible by 2 or 3 = 13
\( \text{P(numbers divisible by 2 or 3)} = \frac{\text{Favourable outcomes of divisible by 2 or 3}}{\text{Total number of possible outcomes}} = \frac{13}{20} \)
(iii) Prime numbers out of first 20 numbers are 2, 3, 5, 7, 11, 13, 17, 19
Favourable outcomes of primes = 8
\( \text{P(primes)} = \frac{\text{Favourable outcomes of primes}}{\text{Total number of possible outcomes}} = \frac{8}{20} = \frac{2}{5} \)
(iv) Numbers not divisible by 10 are 1, 2, .. 9, 11, ... 19
Favourable outcomes of not divisible by 10 = 18
\( \text{P(not divisible by 10)} = \frac{\text{Favourable outcomes of not divisible by 10}}{\text{Total number of possible outcomes}} = \frac{18}{20} = \frac{9}{10} \)

IMPORTANT POINTS TO BE REMEMBERED

• In the experimental approach to probability, we find the probability of the occurrence of an event by actually performing the experiment a number of times and adequate recording of the happening of event.
• In the theoretical approach to probability, we try to predict what will happen without actually performing the experiment.
• An outcome of a random experiment is called an elementary event.
• An event associated to random experiment is a compound event if it is obtained by combining two or more elementary events associated to the random experiment.

EXERCISE # 1

Question. 1000 families with 2 children were selected randomly, and the following data were recorded :

If a family is chosen at random, find the probability that it has (i) No boy (ii) one boy (iii) 2 boys (iv) at least one boy (v) at most one boy.
Answer: (i) 0.14, (ii) 0.56, (iii) 0.3, (iv) 0.86, (v) 0.7

Question. The percentage of marks obtained by a student in the monthly unit tests are given below :

Find the probability that the student gets :
(i) a first class i.e. at least 60% marks
(ii) marks between 70% and 80%
(iii) a distinction i.e. 75% or above
(iv) less than 65% marks.
Answer: (i) 0.8, (ii) 0.2, (iii) 0.4, (iv) 0.4

Question. Three coins are tossed simultaneously 100 times with the following frequencies of different outcomes :

If the three coins are simultaneously tossed again, compute the probability of -
(i) 2 heads coming up
(ii) 3 heads coming up
(iii) at least one head coming up
(iv) getting more heads than tails
(v) getting more tails than heads.
Answer: (i) 0.36, (ii) 0.12, (iii) 0.86, (iv) 0.48, (v) 0.52

Question. 1500 families with 2 children were selected randomly and the following data were recorded :

If a family is chosen at random, compute the probability that it has -
(i) No girl
(ii) 1 girl
(iii) 2 girls
(iv) at most one girl
(v) more girls than boys.
Answer: (i) 0.1406, (ii) 0.5426, (iii) 0.3166, (iv) 0.6833, (v) 0.3166

Question. It is known that a box of 600 electric bulbs contains 12 defective bulbs. One bulb is taken out at random from this box. What is the probability that it is non-defective bulb ?
Answer: \( \frac{49}{50} \)

Question. A number is chosen at random among the first 100 natural numbers. Find the probability that the number chosen being a multiple of 5.
Answer: \( \frac{1}{5} \)

Question. In a cricket match, a batsman hits a boundary 6 times out of 30 balls he plays. Find the probability that on a ball played : [NCERT]
(i) he hits boundary
(ii) he does not hit a boundary
Answer: (i) 0.2, (ii) 0.8

Question. 17 cards numbered 1, 2, 3, ...., 16, 17 are put in a box and mixed thoroughly. One person draws a card from the box. Find the probability that the number on the card is -
(i) odd
(ii) a prime
(iii) divisible by 3
(iv) not divisible by 3 and 2 both
Answer: (i) \( \frac{9}{17} \), (ii) \( \frac{7}{17} \), (iii) \( \frac{5}{17} \), (iv) \( \frac{15}{17} \)

A company selected 2400 families at random and surveyed them to determine a relationship between income level and the number of vehicles in home. The information gathered is listed in the table below : [NCERT]

If a family is chosen, find the probability that the family is :

Question. earning Rs 10000-13000 per month and owning exactly 2 vehicles.
Answer: \( \frac{29}{2400} \)

Question. earning Rs 16000 or more per month and owning exactly 1 vehicle.
Answer: \( \frac{29}{120} \)

Question. earning less than Rs 7000 per month and does not own any vehicle.
Answer: \( \frac{1}{240} \)

Question. earning Rs 13000-16000 per month and owning more than 2 vehicles.
Answer: \( \frac{1}{96} \)

Question. owning not more than 1 vehicle.
Answer: \( \frac{1031}{1200} \)

Question. owning at least one vehicle.
Answer: \( \frac{589}{600} \)

The blood groups of 30 students of class IX are recorded as follows :
A, B, O, O, AB, O, A, O, B, A, O, B, A, O, O
A, AB, O, A, A, O, O, AB, B, A, O, B, A, B, O
A student is selected at random from the class for blood donation. Find the probability that the blood group of the student chosen is 

Question. A
Answer: 0.3

Question. B
Answer: 0.2

Question. AB
Answer: 0.1

Question. O
Answer: 0.4

Over the past 200 working days, the number of defective parts produced by a machine is given in the following table :

Determine the probability that tomorrow's output will have :

Question. no defective part
Answer: \( \frac{1}{4} \)

Question. atleast one defective part
Answer: \( \frac{3}{4} \)

Question. not more than 5 defective parts
Answer: 0.73

Question. more than 13 defective parts.
Answer: 0

Three coins are tossed simultaneously 100 times with the following frequencies of different outcomes :

Find the probability of getting :

Question. No head
Answer: \( \frac{1}{5} \)

Question. Two heads
Answer: \( \frac{33}{100} \)

A bag contains 6 black, 7 red and 2 white balls. A ball is drawn from the bag at random. Find the probability that the ball drawn is :

Question. red
Answer: \( \frac{7}{15} \)

Question. black or white
Answer: \( \frac{8}{15} \)

Question. no black
Answer: \( \frac{3}{5} \)

A die is thrown 400 times, the frequency of the outcomes of the events 1, 2, 3, 4, 5 and 6 are noted in the table given below :

Find the probability of occurrence of :

Question. an odd number
Answer: \( \frac{13}{25} \)

Question. a prime number
Answer: \( \frac{193}{400} \)

Question. A coin is tossed 1000 times, if the probability of getting a tail is \( \frac{3}{8} \), how many times head is obtained ?
Answer: 625