Chapter 28.4 - More solved example on empirical probability

In the previous section we saw some problems on experimental probability. In this section we will see a few more solved examples.

Solved example 28.8
An insurance company selected 2000 drivers at random from a particular city to find a relationship between age and accidents. The data obtained is given in the table below:

Age of drivers (years)	Accidents in one year
	0	1	2	3	over 3
18-29	440	160	110	61	35
30-50	505	125	60	22	18
Above 50	360	45	35	15	9

Find the following:
(i) A driver is chosen at random from the city. What is the probability that, he is 18-29 years of age and has exactly 3 accidents in one year
(ii) A driver is chosen at random from the city. What is the probability that, he is 30-50 years of age and has one or more accidents in year 
(iii) A driver is chosen at random from the city. What is the probability that, he has no accidents in one year
Solution:
■ The given table was prepared by adopting the following procedure:
(i) Pick a driver at random. Find two points about him:
• His age
• No.of accidents in the past one year
(ii) Based on the two answers, make a tally mark in the appropriate place in the table.
Example:
• If the age of a driver is 32, the tally mark will fall somewhere in the row marked as '30-50'
    ♦ If he has made 2 accidents in the past year, the tally mark will fall in the column marked as '2'
When a tally mark is placed, one trial is complete. Note that 440+160+110+ . . . + 15+9 = 2000
■ If we calculate the probabilities based on the table, those probabilities will be applicable for all the drivers in that particular city.
1. Corresponding to 18-29 age and exact 3 accidents, we have 61  
• So probability = ^{Number of trials in which the event happened}⁄_{Total number of trials} = ⁶¹⁄₂₀₀₀ = 0.0305
2. Corresponding to 30-50 age, we have all values in that row. But all those values are not eligible. 
• The number of drivers in this age group, who made one or more accidents is 125+60+22+18 = 225
• So probability = ^{Number of trials in which the event happened}⁄_{Total number of trials} = ²²⁵⁄₂₀₀₀ = 0.1125
3. In this case, age is not considered. So the no. of trials which give the favourable outcome are 440+505+360 = 1305
• So probability = ^{Number of trials in which the event happened}⁄_{Total number of trials} = ¹³⁰⁵⁄₂₀₀₀ = 0.653

Solved example 28.9
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:

Bag	1	2	3	4	5
Number of seeds germinated	40	48	42	39	41

Find the following:
(i) Probability that more than 40 seeds in a bag will germinate  
(ii) Probability that 49 seeds in a bag will germinate
(iii) Probability that more than 35 seeds in a bag will germinate?
Solution:
■ There are 5 bags of seeds. From each bag, 50 seeds were selected. This selection of 50 seeds should be done carefully. Because those 50 seeds should give an overall representation of the bag from which they are selected. They should not be taken from the top of the bag. Neither should they be taken from the bottom or middle. The procedure to ensure 'randomness' should be strictly followed for each of the 5 bags
■ Further more, these 5 bags are selected at random from a stack of large number of bags. 'Randomness' should be ensured while picking these 5 bags also.   
■ When we calculate the probabilities, they will be applicable to a bag taken from the whole stack
■ If instead of taking 5 bags, 10 or 15 bags are tested (that is., increasing number of trials), the accuracy of the results will increase
1. No. of bags which give more than 40 seeds that would germinate = 3
• So probability = ^{Number of trials in which the event happened}⁄_{Total number of trials} = ³⁄₅ = 0.6
2. There are no bags which give 49 seeds that would germinate
• So probability = ^{Number of trials in which the event happened}⁄_{Total number of trials} = ⁰⁄₅ = 0
3. No. of bags which give more than 35 seeds that would germinate = 5
• So probability = ^{Number of trials in which the event happened}⁄_{Total number of trials} = ⁵⁄₅ = 1

Solved example 28.10
1500 families with 2 children were selected randomly, and the following data were recorded:

Number of girls in a family	2	1	0
Number of families	475	814	211

Compute the following:
(i) The probability that, a family chosen at random have two girls
(ii) The probability that, a family chosen at random have one girl
(iii) The probability that, a family chosen at random have no girl
(iv) Check whether the sum of all the above probabilities is 1
Solution:
■ 1500 families were selected at random from a region. There are thousands of more families in the region. When we calculate the probabilities based on the table, those probabilities, will be applicable to the whole region
1. 475 families have two girls
• So probability = ^{Number of trials in which the event happened}⁄_{Total number of trials} = ⁴⁷⁵⁄₁₅₀₀ = 0.3167
2. 814 families have one girl

• So probability = ^{Number of trials in which the event happened}⁄_{Total number of trials} = ⁸¹⁴⁄₁₅₀₀ = 0.54267
3. 211 families have no girl

• So probability = ^{Number of trials in which the event happened}⁄_{Total number of trials} = ²¹¹⁄₁₅₀₀ = 0.14067
4. Sum of the above probabilities = 0.3167 +0.54267 +0.14067 = 1.00004

Solved example 28.11
Three coins are tossed simultaneously 200 times with the following frequencies of different outcomes:

Outcome	3 heads	2 heads	1 head	No head
Frequency	23	72	77	28

If the three coins are simultaneously tossed again, compute the probability of 2 heads coming up
Solution:
No. of times 2 heads come up = 72
• So probability = ^{Number of trials in which the event happened}⁄_{Total number of trials} = ⁷²⁄₂₀₀ = ⁹⁄₂₅ = 0.36

Solved example 28.12
An organisation selected 2400 families at random and surveyed them to determine a relationship between income level and the number of vehicles in a family. The information obtained is given in a tabular form below:

Monthly income (Rs)	Vehicles per family
	0	1	2	Above 2
Less than 7000	10	160	25	0
7000 - 10000	0	305	27	2
10000 - 13000	1	535	29	1
13000 - 16000	2	469	59	25
16000 or more	1	579	82	88

Compute the following:
(i). A family chosen at random is earning 10000 - 13000 per month and owning exactly 2 vehicles
(ii). A family chosen at random is earning 16000 or more per month and owning exactly 1 vehicle
(iii). A family chosen at random is earning less than 7000 per month and does not own any vehicle
(iv). A family chosen at random is earning 13000 - 16000 per month and owning more than 2 vehicles
(v). A family chosen at random owns only 1 vehicle
Solution:
■ 2400 families were selected at random from a region. There are thousands of more families in the region. When we calculate the probabilities based on the table, those probabilities will be applicable to the whole region
1. Number of families in this income range owning exactly 2 vehicles = 29
• So probability = ^{Number of trials in which the event happened}⁄_{Total number of trials} = ²⁹⁄₂₄₀₀
2. Number of families in this income range owning exactly 1 vehicle = 579

• So probability = ^{Number of trials in which the event happened}⁄_{Total number of trials} = ⁵⁷⁹⁄₂₄₀₀
3. Number of families in this income range owning no vehicles = 10

• So probability = ^{Number of trials in which the event happened}⁄_{Total number of trials} = ¹⁰⁄₂₄₀₀
4. Number of families in this income range owning more than 2 vehicles = 25

• So probability = ^{Number of trials in which the event happened}⁄_{Total number of trials} = ²⁵⁄₂₄₀₀
5. In this case, income range is not required. Number of families who own exactly one vehicle is:
160 +305 +535 +469 + 579 = 2048

• So probability = ^{Number of trials in which the event happened}⁄_{Total number of trials} = ²⁰⁴⁸⁄₂₄₀₀

Solved example 28.13
To know the opinion of the students about the subject statistics, a survey of 200 students was conducted. The data obtained is tabulated below:

Opinion	Number of students
Like	135
Dislike	66

Compute the following:
(i) Probability that, a student chosen at random likes statistics
(ii) Probability that, a student chosen at random dislikes statistics
Solution:
1. Number of students who like statistics = 135
• So probability = ^{Number of trials in which the event happened}⁄_{Total number of trials} = ¹³⁵⁄₂₀₀ = ²⁷⁄₄₀
2. Number of students who dislike statistics = 66
• So probability = ^{Number of trials in which the event happened}⁄_{Total number of trials} = ⁶⁶⁄₂₀₀ = ³³⁄₁₀₀

Solved example 28.14
Eleven bags of wheat flour, each marked 5 kg, actually contained the following weights of flour:
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.
Solution:
No. of bags which contain more than 5 kg of flour = 7
• So probability = ^{Number of trials in which the event happened}⁄_{Total number of trials} = ⁷⁄₁₁

---

We have completed this discussion on experimental probability. Part III of this discussion can be seen in chapter 36.
In the next section, we will see Quadratic equations.

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chapter 28.3 - Solved examples on Experimental probability

In the previous section we saw the results when the number of times 'two coins are tossed simultaneously' is increased. In this section we will see how those findings can be used to solve practical problems. Before that, we will just recall what the following terms mean:

1. Trial 2. Outcome 3. Favourable outcome 4. Event

1. Trial:

• In experiment I, each toss of the coin is a trial

• In experiment II, each roll of the die is a trial

• In experiment III, each 'simultaneous toss of two coins' is a trial      

2. Outcome:

• In experiment I, consider any one trial. It has two possible outcomes:

H or T

• In experiment II, consider any one trial. It has six possible outcomes:

1, 2, 3, 4, 5 or 6

• In experiment III, consider any one trial. It has three possible outcomes:

TT, HT or HH

3. Favourable outcome:

■ In experiment I, consider any one trial. It has two possible outcomes:

H or T

• Let the outcome in that trial be H. 

    ♦ If we already wanted 'H' before the beginning of the trial, then it is a favourable outcome

    ♦ If what we wanted was 'T' before the beginning of the trial, then it is not a favourable outcome

• Let the outcome in that trial be T. 

    ♦ If we already wanted 'T' before the beginning of the trial, then it is a favourable outcome

    ♦ If what we wanted was 'H' before the beginning of the trial, then it is not a favourable outcome

■ In experiment II, consider any one trial. It has six possible outcomes:

1, 2, 3, 4, 5 or 6

• Let the outcome in that trial be 5. 

    ♦ If we already wanted '5' before the beginning of the trial, then it is a favourable outcome

   ♦ If what we wanted was a number other than '5' before the beginning of the trial, then it is not a favourable outcome

• Let the outcome in that trial be 4. 

    ♦ If we already wanted 'an even number' before the beginning of the trial, then it is a favourable outcome. (In that case, the out comes 2 and 6 will also be favourable)

   ♦ If what we wanted was an odd number, before the beginning of the trial, then it is not a favourable outcome

■ In experiment III, consider any one trial.It has three possible outcomes:

TT, HT or HH

• Let the outcome in that trial be HT. 

    ♦ If we already wanted 'HT' before the beginning of the trial, then it is a favourable outcome

   ♦ If what we wanted was TT or HH before the beginning of the trial, then it is not a favourable outcome

4. Event:

■ In experiment I, consider any one trial. 

• At the end of that trial, we will get an outcome. This outcome may or may not be favourable

• If what we get is a favourable outcome, we say: An event has occurred

Example:

Suppose we decided that we want H. This decision was made before carrying out the trial.

At the end of the trial, the outcome is H

Then an event has occurred

■ In experiment II, consider any one trial.

Suppose we decided that we want 2. This decision was made before carrying out the trial. 

At the end of the trial, the outcome is 2

Then an event has occurred

If at the end of the trial the outcome is 1, 3, 4, 5 or 6, we don't have an event

Another example:

Suppose we decided that we want an even number. This decision was made before carrying out the trial. 

At the end of the trial, the outcome is 6

Then an event has occurred

If at the end of the trial the outcome is  2, 4 or 6, we have an event

If at the end of the trial the outcome is 1, 3 or 5, we don't have an event

■ In experiment III, consider any one trial.

Suppose we decided that we want HT. This decision was made before carrying out the trial. 

At the end of the trial, the outcome is HT

Then an event has occurred

If at the end of the trial the outcome is HH or TT, we don't have an event

---

• We know that Probability = ^{Number of favourable outcomes}⁄_{Total number of all possible outcomes}. 

We have seen the details in an earlier chapter 1.5.

■ In the present chapter we did some experiments

• In each experiment, we did trials

    ♦ At the end of each trial, we got an outcome

    ♦ This outcome may be favourable or unfavourable

• If it is favourable, we have an event

---

■ So in this chapter we will be dealing with 'Experimental probability'.

• 'Experimental probability' is also called 'Empirical probability'

• Empirical probability of an event E to occur is denoted as P(E)

• It is given by the ratio: ^{Number of trials in which the event happened}⁄_{Total number of trials}.

(In this chapter wherever we write 'probability', it would mean 'empirical probability'. This shortening is for convenience)

---

An example:

1. In the experiment ID, we obtained some values related to heads. They are: 0.2, 0.333, 0.356, . . . (see table 28.2 in the first section of this chapter)

2. Each of them are probability values. That is., we can write:

P(E) = 0.2

P(E) = 0.333

P(E) = 0.356

Where E is the event of 'obtaining head' in 'tossing a coin'.

3. So we find that P(E) changes when the number of trials changes.

4. Suppose someone asks us the following question:

What is the empirical probability of obtaining head when a coin is tossed once?

Ans:

If we have done 15 trials we would be able to say 0.2

If we have done 30 trials we would be able to say 0.333

If we have done 45 trials we would be able to say 0.356

---

Another example:

1. In the experiment IID:

P(E) = 0.2

P(E) = 0.125

P(E) = 0.16

Where E is the event of 'obtaining 4' in 'rolling a die'. (see table 28.4 in the second section of this chapter)

2. Suppose someone asks us the following question:

What is the empirical probability of obtaining 4 when a die is rolled once?

Ans:

If we have done 20 trials we would be able to say 0.2

If we have done 40 trials we would be able to say 0.125

If we have done 60 trials we would be able to say 0.16 

---

Now we will see some solved examples:

Solved example 28.1

A coin is tossed 1000 times. Head was obtained 455 times and tail was obtained 545 times. If E is the event of getting a head and F, the event of getting a tail, then Compute P(E) and P(F)

Solution:

1. The coin is tossed 1000 times. So total number of trials = 1000

2. In each trial there are two possible outcomes: H and T

3. Consider the probability for H.

• When we consider the probability for H, obtaining H in a trial is a favourable outcome.

• If a favourable outcome is obtained in a trial, that trial gives us an event.

• So when H is considered, 455 trials give us an event.

• In other words, when H is considered, we have 455 events

• Thus P(E) = ^{Number of trials in which the event happened}⁄_{Total number of trials} = ⁴⁵⁵⁄₁₀₀₀. = 0.455

4. Consider the probability for T.

• When we consider the probability for T, obtaining T in a trial is a favourable outcome.

• If a favourable outcome is obtained in a trial, that trial gives us an event.

• So when T is considered, 545 trials give us an event.

• In other words, when T is considered, we have 545 events

• Thus P(E) = ^{Number of trials in which the event happened}⁄_{Total number of trials} = ⁵⁴⁵⁄₁₀₀₀ = 0.545

5. Note that P(E) + P(F) = 0.455 + 0.545 = 1

This is because H and T are the only possible outcomes. If H does not occur, T will obviously occur and vice versa

Solved example 28.2

Two coins are tossed simultaneously 500 times. The results are:

HH 105 times, HT 275 times and TT 120 times

Find the probability for HH, HT and TT

Solution:

■ Let E, F and G denote the events for HH, HT and TT respectively  

1. The two coins are tossed simultaneously 500 times. So total number of trials = 500

2. In each trial there are 3 possible outcomes: HH, HT and TT

3. Consider the probability for HH.

• When we consider the probability for HH, obtaining HH in a trial is a favourable outcome.

• If a favourable outcome is obtained in a trial, that trial gives us an event.

• So when HH is considered, 105 trials give us an event.

• In other words, when HH is considered, we have 105 events

• Thus P(E) = ^{Number of trials in which the event happened}⁄_{Total number of trials} = ¹⁰⁵⁄₅₀₀. = 0.21

4. Consider the probability for HT.

• When we consider the probability for HT, obtaining HT in a trial is a favourable outcome.

• If a favourable outcome is obtained in a trial, that trial gives us an event.

• So when HT is considered, 275 trials give us an event.

• In other words, when HT is considered, we have 275 events

• Thus P(F) = ^{Number of trials in which the event happened}⁄_{Total number of trials} = ²⁷⁵⁄₅₀₀. = 0.55

5. Consider the probability for TT.

• When we consider the probability for TT, obtaining TT in a trial is a favourable outcome.

• If a favourable outcome is obtained in a trial, that trial gives us an event.

• So when TT is considered, 120 trials give us an event.

• In other words, when TT is considered, we have 120 events

• Thus P(G) = ^{Number of trials in which the event happened}⁄_{Total number of trials} = ¹²⁰⁄₅₀₀. = 0.24

6. Note that P(E) + P(F) + P(G) = 0.21 + 0.55 + 0.24 = 1

This is because HH, HT and TT are the only possible outcomes.

Solved example 28.3
A die is thrown 1000 times with the frequencies for the outcomes 1, 2, 3, 4, 5 and 6 as given in the table below:

Outcome	1	2	3	4	5	6
Frequency	179	150	157	149	175	190

Find the probability of getting each outcome
Solution:

■ Let E₁, E₂, E₃, E₄, E₅ and E₆ denote the events for 1, 2, 3, 4, 5 and 6 respectively  

1. The die is rolled 1000 times. So total number of trials = 1000

2. In each trial there are 6 possible outcomes: 1, 2, 3, 4, 5 or 6

3. Consider the probability for 1

• When we consider the probability for 1, obtaining 1 in a trial is a favourable outcome.

• If a favourable outcome is obtained in a trial, that trial gives us an event.

• So when 1 is considered, 179 trials give us an event.

• In other words, when 1 is considered, we have 179 events

• Thus P(E₁) = ^{Number of trials in which the event happened}⁄_{Total number of trials} = ¹⁷⁹⁄₁₀₀₀ = 0.179

4. Consider the probability for 2

• When we consider the probability for 2, obtaining 2 in a trial is a favourable outcome.

• If a favourable outcome is obtained in a trial, that trial gives us an event.

• So when 2 is considered, 150 trials give us an event.

• In other words, when 2 is considered, we have 150 events
• Thus P(E₂) = ^{Number of trials in which the event happened}⁄_{Total number of trials} = ¹⁵⁰⁄₁₀₀₀ = 0.150

5. In a similar way, P(E₃) = ¹⁵⁷⁄₁₀₀₀ = 0.157
6. P(E₄) = ¹⁴⁹⁄₁₀₀₀ = 0.149
7. P(E₅) = ¹⁷⁵⁄₁₀₀₀ = 0.175
8. P(E₆) = ¹⁹⁰⁄₁₀₀₀ = 0.190
Note that P(E₁) + P(E₂) + P(E₃) + P(E₄) + P(E₅) + P(E₆)  = 1

Solved example 28.4
On one page of a telephone directory, there are 200 telephone numbers. The unit place digit of those numbers were analysed. (For example, in the number 2578682, the unit place digit is 2). It was found that 0 occurred 22 times, 1 occurred 26 times, 2 occurred 22 times and so on.. The full frequency distribution is given in the table below:

Digit	0	1	2	3	4	5	6	7	8	9
Frequency	22	26	22	22	20	10	14	28	16	20

With out looking at the page, the pencil is placed on a telephone number. What is the probability that, the digit in the unit place of that telephone number is 6?
Solution:

1. There are 200 telephone numbers. The pencil may be placed on any of those numbers. So there are 200 possible outcomes.
2. Out of the 200 possible outcomes, 14 are favourable
3. So probability = ^{Number of favourable outcomes}⁄_{Total number of possible outcomes} = ¹⁴⁄₂₀₀ = 0.07

Solved example 28.5
The record at a weather station shows that out of the past 250 consecutive days, the weather forecasts were correct on 175 days.
(i) What is the probability that on a given day it was correct
(ii) What is the probability that it was not correct on a given day?
Solution:
1. There are 250 possible outcomes
2. If the 'correct forecast' is considered, 175 outcomes are favourable
• So probability = ^{Number of favourable outcomes}⁄_{Total number of possible outcomes} = ¹⁷⁵⁄₂₅₀ = 0.7
3. No. of incorrect forecasts = 250 - 175 = 75
• If the 'in correct forecast' is considered, 75 outcomes are favourable
• So probability = ^{Number of favourable outcomes}⁄_{Total number of possible outcomes} = ⁷⁵⁄₂₅₀ = 0.3
4. Note that, the sum of two probabilities is 1

Solved example 28.6
A tyre manufacturing company kept a record of the distance covered before a tyre needed to be replaced. The table below shows the results of 1000 cases.

Distance (km)	less than 4000	4000 to 9000	9001 to 14000	more than 14000
Frequency	20	210	325	445

If you buy a tyre from this company, what is the probability that:
(i) It will need to be replaced before it has covered 4000 km?
(ii) It will last more than 9000 km?
(iii) It will need to be replaced after it has covered 4000 km, but before reaching 14000 km?
Solution:
1. The company did an experiment 1000 times. The procedure for each experiment was the same. Let us write that procedure:
Step 1: Make four segments:
• Less than 4000
• 4001 to 9000
• 9001 to 14000
• More than 14000
Step 2: This step is carried out when a vehicle comes for tyre replacement.
(a) Verify the records and find the distance travelled by the tyre. 
(b) Based on the distance, a tally mark can be placed in the appropriate segment
For example, if the distance is 5200 km, the tally mark is placed in the second segment.
If the distance is 15400 km, the tally mark is placed in the fourth segment
2. When each vehicle comes for tyre replacement, a tally mark is placed in the appropriate segment. 
When a tally mark is placed, one experiment is complete. 
So there will be 1000 tally marks when 1000 vehicles come for tyre replacement
3. When the 1000 tally marks are complete, they can be counted and the given table can be obtained.
4. If the company publish the table, future buyers can calculate different probabilities
5. If various manufacturing companies publish such a table of their own, and if the authenticity of such tables can be verified,  buyers can make a comparison between companies.
6. Let us now calculate the different probabilities of the given company:
7. Replacement before 4000 km
• Probability = ²⁰⁄₁₀₀₀ = 0.02  
8. Replacement after 9000 km
• Probability = ^(325+445)⁄₁₀₀₀ =  ⁷⁷⁰⁄₁₀₀₀ =  0.77
9. Replacement between 4000 and 14000 km
• Probability = ^(210+325)⁄₁₀₀₀ =  ⁵³⁵⁄₁₀₀₀ =  0.535

Solved example 28.7
The percentage of marks obtained by a student in the monthly unit tests are given below:

Unit test	I	II	III	IV	V
Percentage of marks obtained	69	71	73	68	74

Based on this data, find the probability that the student gets more than 70% marks in a unit test
Solution:
Each unit test can be considered as a trial. So there are 5 trials. 
When we consider 'more than 70%', there are 3 events
So probability = ^{Number of trials in which the event happened}⁄_{Total number of trials} = ³⁄₅ = 0.6

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In the next section, we will see a few more solved examples.

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Chapter 1.8 - Solved examples on Probability

In the previous section we completed the discussion on the probability related to drawing a coloured ball, tossing a coin, rolling a die, drawing a card etc., We saw some solved examples also. In this section, we will see some solved examples with a higher level of difficulty.

Solved example 1.10
Tickets marked with numbers from 1 to 25 are put in a box and mixed well. Then one ticket is drawn from the box without looking. What is the probability that the drawn ticket is a multiple of 3 or 7?

Solution:
Step 1: Write the outcomes:
Outcome 1: The drawn ticket is the one which is marked with '1' 
Outcome 2: The drawn ticket is the one which is marked with '2'
Outcome 3: The drawn ticket is the one which is marked with '3'
- - -
- - -
Outcome 25: The drawn ticket is the one which is marked with '25'

So there are 25 outcomes. We need not write them all. In fact, once we understand the basics, we need to write only this:
Step 1: There are 25 possible outcomes.

Step 2: Analyse each outcome. According to the question, the card should be a multiple of 3 OR 7. So if the card drawn is either a multiple of 3 OR a multiple of 7, we can write 'favourable' towards that outcome. And we will have an event. Let us see on which all out comes, we can write 'favourable':

Ticket marked with '1' → 1 is not a multiple of 3 or 7 → Not favourable
Ticket marked with '2' → 2 is not a multiple of 3 or 7 → Not favourable
Ticket marked with '3' → 3 is a multiple 3 → Favourable
Ticket marked with '4' → 4 is not a multiple of 3 or 7 → Not favourable
Ticket marked with '5' → 5 is not a multiple of 3 or 7 → Not favourable
Ticket marked with '6' → 6 is a multiple 3 → Favourable
Ticket marked with '7' → 7 is a multiple 7 → Favourable
- - -
- - -
Ticket marked with '25' → 25 is not a multiple of 3 or 7 → Not favourable

In this step 2 also, it is not necessary to write the 25 lines. We need to write only this:
Step 2: Analysing each outcome, we find that, 3, 6, 7, 9, 12, 14, 15, 18, 21, and 24 satisfy the required conditions. So there are a total of 10 favourable outcomes.


Thus the required probability = ¹⁰⁄₂₅ = ²⁄₅ = 40%. So there is a 40% probability that we will get a ticket which is either a multiple of 3 or a multiple of 7.

Solved example 1.11
Two unbiased coins are tossed simultaneously. Find the probability of getting (i) Two heads (ii) One tail (iii) One head (iv) at most one head (v) at least one head (vi) no head

Solution:
Step 1: Write the outcomes:

When two coins are tossed simultaneously, the possible outcomes can be written in the form of a table:

Outcome	First coin	Second coin
Outcome 1	Heads	Heads
Outcome 2	Heads	Tails
Outcome 3	Tails	Tails
Outcome 4	Tails	Heads

The above 4 are the only possible outcomes. This table can be used for all 6 cases of the question.
(i) Two heads: Analysing each outcome, we find that only one outcome (Outcome 1) satisfies this condition. So the probability of getting 2 heads = ¹⁄₄ = 25%
(ii) One tail: Two outcomes (Outcome 2 and 4) satisfies this condition. So the probability = ²⁄₄ = ¹⁄₂ = 50%
(iii) One head: Two outcomes (Outcome 2 and 4) satisfies this condition. So the probability = ²⁄₄ = ¹⁄₂ = 50%
(iv) At most one head: This condition means that when two coins are tossed simultaneously, there must be only one head. There must not be more than one head.
Two outcomes (Outcome 2 and 4) satisfies this condition. So the probability = ²⁄₄ = ¹⁄₂ = 50%
(v) At least one head: This condition means that, when two coins are tossed simultaneously, there must be atleast one head. So even if there are two heads, the condition will be satisfied. Three outcomes (outcomes 1, 2 and 4) satisfies this condition. So the probability = ³⁄₄ = 75%
(vi) No head: This condition means that when two coins are tossed simultaneously, there must not be any head. Only one outcome (Outcome 3) satisfies this condition. So the probability = ¹⁄₄ = 25% 

Solved example 1.12
Three unbiased coins are tossed simultaneously. Find the probability of getting (i) All heads (ii) two heads (iii) One head (iv) at least two heads

Solution:
Step 1: Write the outcomes:
When three coins are tossed simultaneously, the possible outcomes can be written in the form of a table:

The above 8 are the only possible outcomes. This table can be used for all 4 cases of the question.
(i) All heads: Analysing each outcome, we find that only one outcome (Outcome 1) satisfies this condition. So we can write: The probability of getting all heads = ¹⁄₈ = 12.5%
(ii) Two heads: Three outcomes (outcomes 2, 6 and 8) satisfies this condition. So the probability = ³⁄₈ = 37.5%
(iii) One head: Three outcomes (Outcome 3, 5 and 7) satisfies this condition. So the probability = ³⁄₈ = 37.5%
(iv) At least two heads: This condition means that when three coins are tossed simultaneously, there must be atleast two heads. So even if there are 3 heads, the condition will be satisfied.
Four outcomes (Outcome 1, 2, 6 and 8) satisfies this condition. So the probability = ⁴⁄₈ = ¹⁄₂ = 50%

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In the next section we will see some more solved examples

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