Chapter 28.1 - Increasing number of Rolls for a Die

In the previous section we saw the results when the number of times a coin is tossed is increased. In this section we will see such experiments related to 'rolling a die'.

Let us do an experiment. We will call it experiment IIA:
1. Take a die. 
(i) Roll it once. The reading may be any one of the following:
1, 2, 3, 4, 5 or 6
(ii) What ever be the reading, note that reading on a piece of paper. 
• (i) and (ii) constitutes one cycle of our experiment.
2. Repeat the cycle 20 times. 
3. In the note book, tabulate the readings as shown in table 28.3 below 
Table 28.3

Number of times a die is thrown	Number of times these scores turn up
	1	2	3	4	5	6
20	7	1	1	3	5	3

4. Determine the following ratios:
• ^{Number of times 1 turned up}⁄_{Total number of times the die is rolled} 
• ^{Number of times 2 turned up}⁄_{Total number of times the die is rolled} 
• ^{Number of times 3 turned up}⁄_{Total number of times the die is rolled} 

• - - - 
• - - - 
• ^{Number of times 6 turned up}⁄_{Total number of times the die is rolled} 
• In our present case:
    ♦ the 1^st ratio is ⁷⁄₂₀ = 0.35
    ♦ the 2^nd ratio is ¹⁄₂₀ = 0.05
    ♦ the 3^rd ratio is ¹⁄₂₀ = 0.05
    ♦ the 4^th ratio is ³⁄₂₀ = 0.15
    ♦ the 5^th ratio is ⁵⁄₂₀ = 0.25
    ♦ the 6^th ratio is ³⁄₂₀ = 0.15
5. Once the ratios in (4) are determined, the experiment IIA is complete

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But our work is not over

■ Repeat the above experiment. We will call it experiment IIB.
• For this experiment IIB, the number of cycles in (2) must be 40
• Let the readings be as shown below:

Number of times a die is thrown	Number of times these scores turn up
	1	2	3	4	5	6
40	4	8	9	4	9	6

• In this case:
    ♦ the 1^st ratio is ⁴⁄₄₀ = 0.1
    ♦ the 2^nd ratio is ⁸⁄₄₀ = 0.2
    ♦ the 3^rd ratio is ⁹⁄₄₀ = 0.225
    ♦ the 4^th ratio is ⁴⁄₄₀ = 0.1
    ♦ the 5^th ratio is ⁹⁄₄₀ = 0.225
    ♦ the 6^th ratio is ⁶⁄₄₀ = 0.15
• When the ratios in (4) are determined, the experiment IIB is over
■ Once again repeat the experiment. We will call it experiment IIC
• For this experiment IIC, the number of cycles in (2) must be 60 
• Let the readings be as shown below:

Number of times a die is thrown	Number of times these scores turn up
	1	2	3	4	5	6
60	9	10	8	10	12	11

• In this case:
    ♦ the 1^st ratio is ⁹⁄₆₀ = 0.15
    ♦ the 2^nd ratio is ¹⁰⁄₆₀ = 0.167
    ♦ the 3^rd ratio is ⁸⁄₆₀ = 0.133
    ♦ the 4^th ratio is ¹⁰⁄₆₀ = 0.167
    ♦ the 5^th ratio is ¹²⁄₆₀ = 0.2
    ♦ the 6^th ratio is ¹¹⁄₆₀ = 0.183
• When the ratios in (4) are determined, the experiment IIC is over

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So we did the same experiment 3 times. Before proceeding further, we will discuss the importance of the six ratios:

1. We know that, the probability of obtaining 1, 2, 3, 4, 5 or 6 when rolling a die once is ¹⁄₆

• Each of the numbers 1, 2, 3, 4, 5 and 6 has the same probability ¹⁄₆.

2. But this '¹⁄₆' is a theoretical value. If ¹⁄₆ is always obtained in the real life also, we will get results such as these:

• Roll the die 6 times
    ♦ 1 will be obtained once
    ♦ 2 will be obtained once
    ♦ 3 will be obtained once
    ♦ - - -
    ♦ - - - 
    ♦ 6 will be obtained once
• Roll the die 18 times
    ♦ 1 will be obtained thrice
    ♦ 2 will be obtained thrice
    ♦ 3 will be obtained thrice
    ♦ - - -
    ♦ - - - 
    ♦ 6 will be obtained thrice

3. But we never get such exact values.
4. However, as the number of trials increase, each of the 6 ratios become closer and closer to ¹⁄₆

5. If, instead of 20,40 or 60 times, if we toss it for a 'very large number of times', then: 

• No. of times 1 is obtained = No. of times 2 is obtained = No. of times 3 is obtained = No. of times 4 is obtained = No. of times 5 is obtained = No. of times 6 is obtained. = k

• That is., 'the equality of probability while rolling a die' will become clear.
• Let the 'very large number of times' for which the die is rolled be 'n'
• Then ^k⁄_n will be equal to ¹⁄₆.

5. We are trying to prove this using our present experiments

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• From the three experiments, we have three sets of ratios. One set from each experiment

• Each set has six ratios:
    ♦ ^{Number of times 1 turned up}⁄_{Total number of times the die is rolled} 
    ♦ ^{Number of times 2 turned up}⁄_{Total number of times the die is rolled} 
    ♦ ^{Number of times 3 turned up}⁄_{Total number of times the die is rolled} 
    ♦ - - - 
    ♦ - - - 
    ♦ ^{Number of times 6 turned up}⁄_{Total number of times the die is rolled} 
■ Let us now analyse the ratios:
1^st ratio when number of trials is 20 = 0.35
1^st ratio when number of trials is 40 = 0.1
1^st ratio when number of trials is 60 = 0.15
• We have, ¹⁄₆ = 0.1667
• As the number of trial increases, the 1^st ratio gets closer and closer to 0.1667
■ That is., as the number of trial increases, the 1^st ratio gets closer and closer to ¹⁄₆.
Consider the 2^nd ratio:
2^nd ratio when number of trials is 20 = 0.05
2^nd ratio when number of trials is 40 = 0.2
2^nd ratio when number of trials is 60 = 0.167
• We have, ¹⁄₆ = 0.1667
• As the number of trial increases, the 2^nd ratio gets closer and closer to 0.1667
■ That is., as the number of trial increases, the 2^nd ratio gets closer and closer to ¹⁄₆.
Consider the 3^rd ratio:
3^rd ratio when number of trials is 20 = 0.05
3^rd ratio when number of trials is 40 = 0.225
3^rd ratio when number of trials is 60 = 0.133
• We have, ¹⁄₆ = 0.1667
• As the number of trial increases, the 3^rd ratio gets closer and closer to 0.1667
■ That is., as the number of trial increases, the 3^rd ratio gets closer and closer to ¹⁄₆.
Consider the 4^th ratio:
4^th ratio when number of trials is 20 = 0.15
4^th ratio when number of trials is 40 = 0.1
4^th ratio when number of trials is 60 = 0.167
• We have, ¹⁄₆ = 0.1667
• As the number of trial increases, the 4^th ratio gets closer and closer to 0.1667
■ That is., as the number of trial increases, the 4^th ratio gets closer and closer to ¹⁄₆.
Consider the 5^th ratio:
5^th ratio when number of trials is 20 = 0.25
5^th ratio when number of trials is 40 = 0.225
5^th ratio when number of trials is 60 = 0.2
• We have, ¹⁄₆ = 0.1667
• As the number of trial increases, the 5^th ratio gets closer and closer to 0.1667
■ That is., as the number of trial increases, the 5^th ratio gets closer and closer to ¹⁄₆.
Consider the 6^th ratio:
6^th ratio when number of trials is 20 = 0.15
6^th ratio when number of trials is 40 = 0.15
6^th ratio when number of trials is 60 = 0.183
• We have, ¹⁄₆ = 0.1667
• As the number of trial increases, the 6^th ratio gets closer and closer to 0.1667
■ That is., as the number of trial increases, the 6^th ratio gets closer and closer to ¹⁄₆.

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So all the ratios are getting closer and closer to ¹⁄₆. When the number of trials become very large, each of the ratios will become equal to ¹⁄₆.

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• We can increase the 'number of trials' to a 'considerably large value' in another way. We will do it as a new experiment. We will call it: Experiment IIC
1. Divide the class into a number of groups. Each group must have three students
In our present case, let there be 15 groups, each with 3 students
2. Consider any one group
(i) A student in that group rolls the die once. The reading may be 1, 2, 3, 4, 5 or 6
(ii) Whatever be the reading, the other two students must check it and record it
• (i) and (ii) constitutes one cycle of our experiment.
3. Repeat the cycle 20 times. When the 20 cycles are completed, the task of this group of students is complete.
4. Now the die can be given to another group. They can do the 20 cycles.
5. After the 20 cycles, the die can be given to yet another group
6. So, when all the 15 groups have completed their tasks, the die will be rolled 15×20 = 300 times. 
• This 300 is a significantly large number. So we achieve a 'large task of rolling a die 300 times'. We achieve it by dividing the task among 15 groups.
7. If there are 15 dice, all the groups can perform the task simultaneously. This will save time. 
• There is only one condition: All the 15 dice must be identical.
If this condition is satisfied, it can be considered that a single student did all the 300 rolls.
8. When all the groups have completed their tasks, we can proceed to do the calculations. We have to combine the results in such a way that, one student did all the 300 rolls. 
9. For that, do the tabulation as shown in the table 28.4 below. 
• In the table, only 1 and 4 is given. This is to keep the size of the table small. Students may use a large sheet of paper and include all the numbers from 1 to 6. 
• The explanation about 'how this table is formed' is given below the table.

Table 28.4

Group	Number of 1	Number of 4	A	B
(1)	(2)	(3)	(4)	(5)
1	3	4	3⁄20 = 0.15	4⁄20 = 0.2
2	2	1	(2+3)⁄(20+20) = 5⁄40 = 0.125	(1+4)⁄(20+20) = 5⁄40 = 0.125
3	2	5	(2+5)⁄(20+40) = 7⁄60 = 0.116	(5+5)⁄(20+40) = 10⁄60 = 0.16
_	_	_	_	_
_	_	_	_	_
15	_	_	_	_

■ Consider the row corresponding to group 1. 
• The first three columns in this row does not need any explanation because, they are the readings obtained during the experiment.
    ♦ In the 4^th column, we have a ratio ³⁄₂₀. 
    ♦ Obviously, it is the ratio: ^{Number of times 1 comes up}⁄_{Total number of times the die is rolled}
    ♦ In the 5^th column, we have a ratio ⁴⁄₂₀. 
    ♦ Obviously, it is the ratio: ^{Number of times 4 comes up}⁄_{Total number of times the die is rolled}
■ Consider the row corresponding to group 2.  
• The first three columns in this row does not need any explanation because, they are the readings obtained during the experiment.
    ♦ In the 4^th column, we have a ratio ⁵⁄₄₀.
    ♦ The numerator is (2+3). So it is the total number of 1 obtained by groups 1 and 2 together. It has the same effect of 'a single student doing 40 rolls, and obtaining 1 five times'.
    ♦ So in the denominator we have 40  
    ♦ In the 5^th column, we have a ratio ⁵⁄₄₀.
    ♦ The numerator is (1+4). So it is the total number of 4 obtained by groups 1 and 2 together. It has the same effect of 'a single student doing 40 rolls, and obtaining 4 five times'.
    ♦ So in the denominator we have 40  
■ Consider the row corresponding to group 3.  
• The first three columns in this row does not need any explanation because, they are the readings obtained during the experiment.
    ♦ In the 4^th column, we have a ratio ⁷⁄₆₀.
    ♦ The numerator is (2+5). So it is the total number of 1 obtained by groups 1, 2 and 3 together. It has the same effect of 'a single student doing 60 rolls, and obtaining 1 seven times'.
    ♦ So in the denominator we have 60  
    ♦ In the 5^th column, we have a ratio ¹⁰⁄₆₀.
    ♦ The numerator is (5+5). So it is the total number of 4 obtained by groups 1, 2 and 3 together. It has the same effect of 'a single student doing 60 rolls, and obtaining 4 ten times'.
    ♦ So in the denominator we have 60

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■ In this way, we can fill up the rows of all the 15 groups, and find all the required ratios.

■ Note: As the number of '1' and '4' in each row is added to the 'total values up to the previous row', the word cumulative can be used.

• So the heading of the 4^th column (which is written as A in the table) is:

^{Cumulative number of 1}⁄_{Total number of times the die is rolled}

• Similarly, the heading of the 5^th column (which is written as B in the table) is:

^{Cumulative number of 4}⁄_{Total number of times the die is rolled}

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Once the ratios are found out, we can analyse them:

■ Group 1:

• Number of rolls = 20
Ratio for '1' = 0.15
Ratio for '4' = 0.2

■ Groups 1 and 2:
• Number of rolls = 40
Ratio for '1' = 0.125
Ratio for '4' = 0.125

■ Groups 1, 2 and 3:

• Number of rolls = 60
Ratio for '1' = 0.116
Ratio for '4' = 0.16
■ So the ratio for '1' takes the values 0.15, 0.125, 0.116, . . .
It is getting closer and closer to ¹⁄₆.
■ The ratio for '4' takes the values 0.2, 0.125, 0.16
It is getting closer and closer to ¹⁄₆.
• If we had made the above table 28.4 on a large sheet of paper, that included numbers 2, 3, 5 and 6 also, we would get the same result for them also.  

■ In this experiment we made 300 trials. This '300' is significantly large but not even close to 'a very large number n'. 
• A very large number is like 'infinity'
• At such a very large number, all the ratios will become equal to ¹⁄₆.

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In the next section, we will see the above procedure when 'two coins are tossed simultaneously'.

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Chapter 1.9 - Examples on Probability

In the previous section we saw some solved examples on probability. In this section, we will see some more examples.
Solved example 1.13
A jar contains 3 red balls, 5 black balls and 4 white balls. A ball is drawn at random from the jar. What is the probability that the ball drawn is (i) White (ii) Red (iii) Black (iv) Not red?


Solution:
As there are more than one ball of a kind, let us name them.
• The 3 red balls will be R1, R2 and R3
• The 5 black balls will be B1, B2, B3, B4 and B5
• The 4 white balls will be W1, W2, W3 and W4

Step 1: Write the possible outcomes:
Outcome 1: The drawn ball is R1
Outcome 2: The drawn ball is R2
Outcome 3: The drawn ball is R3
Outcome 4: The drawn ball is B1
Outcome 5: The drawn ball is B2
Outcome 6: The drawn ball is B3
Outcome 7: The drawn ball is B4
Outcome 8: The drawn ball is B5
Outcome 9: The drawn ball is W1
Outcome 10: The drawn ball is W2
Outcome 11: The drawn ball is W3
Outcome 12: The drawn ball is W4

So there are 12 outcomes. We need not write them all. In fact, once we understand the basics, we need to write only this:
Step 1: There are 12 possible outcomes.
This result can be used for all the 4 cases of the question. So for each of the cases, only Step 2 (which is analysis of each outcome) need to be done.
(i) Drawn ball is white:
There are 4 outcomes (outcomes 9, 10, 11 and 12. This is equal to the number of white balls) that are favourable for this event. So the probability = ⁴⁄₁₂ = ¹⁄₃ = 33.33% 
(ii) Drawn ball is red:
There are 3 outcomes (outcomes 1, 2, and 3. This is equal to the number of red balls) that are favourable for this event. So the probability = ³⁄₁₂ = ¹⁄₄ = 25% 
(iii) Drawn ball is black:
There are 5 outcomes (outcomes 4, 5, 6, 7 and 8. This is equal to the number of black balls) that are favourable for this event. So the probability = ⁵⁄₁₂ = 41.67%
(iii) Drawn ball is not red:
There are 9 outcomes (outcomes 4 to 12. This is equal to the total number of balls which are not red) that are favourable for this event. So the probability = ⁹⁄₁₂ = ³⁄₄ = 75%

Solved example 1.14
Two dice are rolled simultaneously. Find the probability of getting 
(i) an even number as the sum
(ii) a total of atleast 11
(iii)  a doublet [a doublet is the situation in which both the dice show the same number on the top face. See the fifth definition given here]
(iv)  a doublet of even number 
(v)  getting a sum divisible by 5
(vi)  getting a multiple of 3 as the sum
(vii) getting a multiple of 2 on one die and a multiple of 3 on the other die
(viii) getting sum ≤ 3 


Solution:
When two dice are rolled simultaneously, the possible outcomes can be written in the form of a table:

The above 36 are the only possible outcomes. As the problem involves some cases related to 'sum', it is also tabulated for each outcome. The table can be used for all the 8 cases in the question.

(i) an even number as the sum
Analysing each outcome, we find that 18 outcomes (Outcomes 1, 3, 5, 8, 10, 12, 13, 15, 17, 20, 22, 24, 25, 27, 29, 32, 34, 36) satisfies this condition. So the probability of getting an even number as the sum = ¹⁸⁄₃₆ = ¹⁄₂ = 50%
(ii) a total of at least 11
This condition means that, the sum should not be less than 11. It should be 11 or greater than 11. Three outcomes (outcomes 30, 35 and 36) satisfies this condition. So the probability = ³⁄₃₆ = ¹⁄₁₂ = 8.33%
(iii) a doublet
6 outcomes (outcomes 1, 8, 15, 22, 29 and 36) satisfies this condition. So the probability = ⁶⁄₃₆ = ¹⁄₆ = 8.33%
(iv) a doublet of even numbers
3 outcomes (outcomes 8, 22, and 36) satisfies this condition. So the probability = ³⁄₃₆ = ¹⁄₁₂ = 16.67%
(v) a sum divisible by 5
7 outcomes (outcomes 4, 9, 14, 19, 24, 29 and 34) satisfies this condition. So the probability = ⁷⁄₃₆ = 19.44%
(vi) a multiple of 3 as the sum
12 outcomes (outcomes 2, 5, 7, 10, 15, 18, 20, 23, 25, 28, 33 and 36) satisfies this condition. So the probability = ¹²⁄₃₆ = ¹⁄₃ = 33.33%
(vii) a multiple of 2 on one die and a multiple of 3 on the other die
11 outcomes (outcomes 8, 12, 14, 16, 18, 21, 24, 32, 33, 34, and 36) satisfies this condition. So the probability = ¹¹⁄₃₆ = ¹⁄₃ = 30.55%
(viii) getting sum ≤ 3 
3 outcomes (outcomes 1, 2, and 7) satisfies this condition. So the probability = ³⁄₃₆ = ¹⁄₁₂ = 8.33%

Solved example 1.15
One card is drawn from a pack of 52 cards, each card being equally likely to be drawn. Find the probability that the card drawn is: Either red card or king

Solution:
Step 1: We have 52 possible outcomes.
Step 2: Let us analyse each outcome:
If the drawn card is red, then we have an event. If the drawn card is a king, then also, we have an event. So 
• we can write 'favourable' towards 26 red cards. 
• we can write 'favourable' towards 4 king cards.


But 2 red kings are already marked 'favourable' when we mark the 26 red cards. So the total number of favourable outcomes are 26 + 2 = 28. Thus the probability = ²⁸⁄₅₂ = ⁷⁄₁₃ = 53.84%

Solved example 1.16
The king, queen and jack of clubs are removed from a standard pack of 52 cards. The pack is well shuffled, and one card is drawn. Find the probability that the drawn card is (i) a heart (ii) a king (iii) an '8' of spades (iv) a club

Solution:
Step 1: 3 cards are removed from the pack, which leaves 52 -3 = 49 cards. So there are 49 possible outcomes. This result can be used for all the 4 cases in the question.
(i) a heart
There are 13 hearts in the pack. Each of them will give a favourable outcome. So the probability = ¹³⁄₄₉
(ii) a king
There are 3 kings left in the pack. Each of them will give a favourable outcome. So the probability = ³⁄₄₉
(iii) an '8' of spades  
This is an unique card. There is only one favourable outcome. So the probability = ¹⁄₄₉
(iv) a club
There are 10 clubs left in the pack. Each of them will give a favourable outcome. So the probability = ¹⁰⁄₄₉ 

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We have now obtained a basic idea about the topic of probability. We have also seen some solved examples. We now know to calculate 'the probability of a certain event to occur'. The probability that we calculate is a 'theoretical' value. Let us now see how it is related to the 'real world events'.

Let us take the example of drawing a ball from the jar containing 2 red balls and 5 blue balls. (Details here) We found out that, the probability of getting a red ball is ²⁄₇ and that for a blue ball is ⁵⁄₇. So the blue ball has greater probability. 

Suppose there is a game of bet based on drawing a ball from this jar. The game is between two players A and B. It is played as follows:

• Player A chooses a color: red or blue

• The balls are well mixed and player A draws a ball without looking. If the drawn ball is of his chosen color, he wins. 

• If the drawn ball is not of his chosen color, then he looses. The winner will then be B

• Drawing a ball by player A completes the game

So we know how the game is played. If A ask us for advice before the starting of the game, we would certainly advise to select blue. Because it has greater probability.

But there is no guarantee that the drawn ball will be blue. As the balls are mixed well, red ball also has the chance for being drawn, though it's probability is less. It may so happen that A chooses blue, and the ball he draws is red. Then, A looses the game.

Let the game be played again. This time also A chooses blue. The balls are mixed well and one ball is drawn. The drawn ball can be red or blue. The player has no control over the outcome. If the ball is red, A has lost again.

So even though blue has a high probability of ⁵⁄₇, player A may loose the game. Then how do we relate theory to practice? The answer is that, when the number of trials increases, the practical values approaches the theoretical values. This can be illustrated based on the above game:

In the above example, one game is complete when a ball is drawn. The winner is decided immediately when a ball is drawn. Instead of this, let the winner be chosen based on 5 results. That is., the ball is drawn once. If it is blue, player A wins. If it is red, he looses. This result is noted down. This completes one cycle. The ball is placed back and mixed well. A ball is drawn again. The result is noted down. In this way, a total of 5 cycles are completed, and 5 cycles completes one game. The winner is decided based on the results of all the 5 cycles. A possible result is tabulated below:

Based on the results in the above table, we can say that Player A has lost the game. Even though blue has a higher probability, it was drawn only 2 times, while red was drawn 3 times. [It may be noted that this is only a possibility]

Let the game consists of 10 cycles. The results are tabulated below.

This time we see that blue is drawn 7 times while red is drawn only 3 times. So player A is the winner. [Again, this is also only one of the many possibilities]

While playing such games, and doing experiments like tossing coins, rolling dice etc., it is seen that, when the number of cycles increases, the actual results come closer and closer to theoretical results. So, if it is decided to base the winner of the above game on the results of a large number of cycles, Player A will surely be the winner.

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Let us take the example of tossing a coin. There are 2 players A and B. The coin is tossed once. If it is heads A wins. If it is tails, then B wins. Both A and B have equal chance of winning. Because the probability for both Heads and Tails is ¹⁄₂. But there can be only one winner. The winner is decided immediately after tossing the coin once.

Instead of this, let there be a large number of cycles. Each cycle consists of tossing the coin once. The winner in each cycle is noted down. After completing all the cycles, the player who has obtained the largest number of wins will be declared the final winner. The tabulation can be done as shown below:

If the number of cycles in the above table is large, we will find that the wins obtained by player A and player B will almost be the same.

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The relation between theoretical and practical results is that, when the number of cycles increases, the actual results become closer and closer to theoretical results.

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We have completed the present discussion on the basics of Probability. The next level discussion is given in chapter 28. In the next chapter, we will discuss about Graphs.

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Chapter 1.6 - Probability related to rolling an unbiased die

In the previous section we saw the experiments of drawing a ball and tossing a coin. Now we will consider the experiment of rolling an unbiased die. 

We know that a die is used in many games. When a player rolls the die, he may be wishing to get a '6' or any particular number from 1 to 6. It depends on the situation he is in. What ever is the wish, the die can land with any one of the numbers (from 1 to 6) on it's top face. The player who rolls the die has no control over it. Let us try to derive the probability for obtaining a '6'. So getting a 6 is the event in this problem. We will write the steps as usual. The possible outcomes are:
Outcome 1: The die lands with 1 on upper face.
Outcome 2: The die lands with 2 on upper face.
Outcome 3: The die lands with 3 on upper face.
Outcome 4: The die lands with 4 on upper face.
Outcome 5: The die lands with 5 on upper face.
Outcome 6: The die lands with 6 on upper face.
● No outcomes other than the above 6 can possibly occur. So we say that the maximum number of outcomes possible is equal to 6.
● We want to present the probability for the ‘getting a 6’. If we get 6, we will call it an event. Let us examine each outcome:

Outcome 1: The die lands on 1 → Not a favourable outcome → We don’t have an event
Outcome 1: The die lands on 2 → Not a favourable outcome → We don’t have an event
Outcome 1: The die lands on 3 → Not a favourable outcome → We don’t have an event
Outcome 1: The die lands on 4 → Not a favourable outcome → We don’t have an event
Outcome 1: The die lands on 5 → Not a favourable outcome → We don’t have an event
Outcome 1: The die lands on 6 → A favourable outcome → We have an event

So, out of the 6 possible outcomes, 1 is favourable, and gives us an event. So,the probability for the occurrence of the event (which is ‘getting a 6’) is 1 in 6. In mathematical form, we write: The probability for the event of getting a '6' = 1/6. In the form of a pie chart, it can be presented as shown in the fig. below:

Fig.1.40 Probability for Number '6'

Just like 6, all the other numbers 1, 2, 3, 4 and 5 are also marked only once on the die. So all these numbers also have a probability of ¹⁄₆.

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We will now see some solved examples:

Solved example 1.6

List the outcomes of the experiment shown below in fig.1.41, and write the probabilities.

Fig.1.41 Spin wheel

Solution:
The fig. shows a Spin wheel with a pointer at the centre. The pointer (shown in yellow colour) is stationary, while the wheel can spin about the centre. The wheel is divided into 3 equal sectors: Red, Blue and Green. As the sectors are equal, each sector has equal chance to have the pointer inside it when the spinning stops.

Let us do the experiment once. That is., the wheel is made to spin once.
Step 1: Write the possible outcomes:
Outcome 1: The wheel comes to rest with the pointer inside red sector.
Outcome 2: The wheel comes to rest with the pointer inside blue sector.
Outcome 3: The wheel comes to rest with the pointer inside green sector.
● No outcomes other than the above 3 can possibly occur. So we say that the maximum number of outcomes possible is equal to 3.

Step 2: In the question, probability of a specific colour is not asked. But we can see that each colour has a probability of ¹⁄₃. That is:
• The probability that 'the wheel comes to rest with the pointer inside red sector' is ¹⁄₃
• The probability that 'the wheel comes to rest with the pointer inside blue sector' is ¹⁄₃
• The probability that 'the wheel comes to rest with the pointer inside green sector' is ¹⁄₃

Solved example 1.7
The jar shown in fig.1.42 contains five balls of different colours: Green, Yellow, White, Red, and Blue. The balls are identical in all respects except for the colours. They are well mixed, and one ball is drawn with out looking. List the outcomes of the experiment and write the probabilities.

Fig.1.42

Solution:
Let us do the experiment once. That is., drawing of the ball is done once.
Step 1: Write the possible outcomes:
Outcome 1: The drawn ball is green
Outcome 2: The drawn ball is yellow
Outcome 3: The drawn ball is white
Outcome 4: The drawn ball is red
Outcome 5: The drawn ball is blue
● No outcomes other than the above 5 can possibly occur. So we say that the maximum number of outcomes possible is equal to 5.

Step 2: In the question, probability of a specific colour is not asked. But we can see that each colour has a probability of ¹⁄₅. That is:
• The probability that 'the drawn ball is green' is ¹⁄₅
• The probability that 'the drawn ball is yellow' is ¹⁄₅
• The probability that 'the drawn ball is white' is ¹⁄₅
• The probability that 'the drawn ball is red' is ¹⁄₅
• The probability that 'the drawn ball is blue' is ¹⁄₅

Solved example 1.8
In the experiment of a spin wheel shown in fig.1.43(a), Find the probability of 
(i) Getting a green sector
(ii) Not getting a green sector

Fig.1.43 Spin wheel with more than one Red and Green sectors

Solution:
In this problem, two probabilities are asked. We will do them separately as two parts.
Part (i): Probability for getting a green sector. So for this, if we get a green sector, we have an event. We will do the experiment once. That is., spinning of the wheel will be done once.

Step 1: Write the possible outcomes: [There are more than one sector with the same colours. 3 reds and 4 greens. So the sectors are named as in fig.1.43(b)]

Outcome 1: The wheel comes to rest with the pointer inside R1.
Outcome 2: The wheel comes to rest with the pointer inside R2.
Outcome 3: The wheel comes to rest with the pointer inside R3.
Outcome 4: The wheel comes to rest with the pointer inside G1.
Outcome 5: The wheel comes to rest with the pointer inside G2.
Outcome 5: The wheel comes to rest with the pointer inside G3.
Outcome 7: The wheel comes to rest with the pointer inside G4.
Outcome 8: The wheel comes to rest with the pointer inside G5.

● No outcomes other than the above 8 can possibly occur. So we say that the maximum number of outcomes possible is equal to 8.

Step 2: Analyse each outcome to see if it is a favourable outcome (favourable if the wheel comes to rest with the pointer inside a green sector) to give us an event:

Outcome 1: sector R1 → Not a favourable outcome → We don’t have an event
Outcome 2: sector R2 → Not a favourable outcome → We don’t have an event
Outcome 3: sector R3 → Not a favourable outcome → We don’t have an event
Outcome 4: sector G1 → A favourable outcome → We have an event
Outcome 5: sector G2 → A favourable outcome → We have an event
Outcome 6: sector G3 → A favourable outcome → We have an event
Outcome 7: sector G4 → A favourable outcome → We have an event
Outcome 8: sector G5 → A favourable outcome → We have an event

So, out of the 8 possible outcomes, 5 are favourable, and gives us an event. So,the probability for the occurrence of the event = ⁵⁄₈. That is:
• The probability of getting a green sector is ⁵⁄₈

Part (ii): Probability of 'Not getting a green sector'. In this problem, we have an event when the wheel comes to rest with the pointer inside any sector whose colour is not green. As this is a separate question from part (i), we will consider that the experiment is done again one more time, specifically to study the possibilities for part (ii).

Step 1: Write the possible outcomes
The possible outcomes are the same as those for part (i). So we need not write them again.
Step 2: Analyse each outcome to see if it is a favourable outcome (favourable if the wheel comes to rest with the pointer inside a non-green sector) to give us an event:

Outcome 1: sector R1 → A favourable outcome → We have an event
Outcome 2: sector R2 → A favourable outcome → We have an event
Outcome 3: sector R3 → A favourable outcome → We have an event
Outcome 4: sector G1 → Not a favourable outcome → We don’t have an event
Outcome 5: sector G2 → Not a favourable outcome → We don’t have an event
Outcome 6: sector G3 → Not a favourable outcome → We don’t have an event
Outcome 7: sector G4 → Not a favourable outcome → We don’t have an event
Outcome 8: sector G5 → Not a favourable outcome → We don’t have an event

So, out of the 8 possible outcomes, 3 are favourable, and gives us an event. So,the probability for the occurrence of the event = ³⁄₈. That is:
• The probability of getting a non-green sector is 3/8

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In the above problem we can see that [⁵⁄₈ = 1 - ³⁄₈]. So we can write:The probability of getting a green sector = 1 - The probability of getting a non-green sector.

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Solved example 1.9

When a die is rolled, find the probability of:

(i) Getting a number greater than 5

(ii) Getting a number less than 5

(iii) Getting an even number

Solution:

Part (i): Probability for getting a number greater than 5. For this, if we get '6', we have an event. We will do the experiment once. That is., rolling the die will be done once.

Step 1: Write the possible outcomes:

Outcome 1: The die lands with 1 on upper face.

Outcome 2: The die lands with 2 on upper face.

Outcome 3: The die lands with 3 on upper face.

Outcome 4: The die lands with 4 on upper face.

Outcome 5: The die lands with 5 on upper face.

Outcome 6: The die lands with 6 on upper face.

● No outcomes other than the above 6 can possibly occur. So we say that the maximum number of outcomes possible is equal to 6.

Step 2: Analyse each outcome to see if it is a favourable outcome (favourable if the die lands with number '6' on the upper face) to give us an event:

Outcome 1: Number is 1 → Not a favourable outcome → We don’t have an event

Outcome 2: Number is 2 → Not a favourable outcome → We don’t have an event

Outcome 3: Number is 3 → Not a favourable outcome → We don’t have an event

Outcome 4: Number is 4 → Not a favourable outcome → We don’t have an event

Outcome 5: Number is 5 → Not a favourable outcome → We don’t have an event

Outcome 6: Number is 6 → A favourable outcome → We have an event

So, out of the 6 possible outcomes, 1 is favourable, and gives us an event. So,the probability for the occurrence of the event = ¹⁄₆. That is:
• The probability of getting number greater than 5 is ¹⁄₆

Part (ii): Probability for getting a number less than 5. For this, if we get any of the numbers: '1', '2', '3' or '4', we have an event. We will do the experiment once. That is., rolling the die will be done once.

Step 1: Write the possible outcomes:

The possible outcomes are the same as those for part (i). So we need not write them again.

Step 2: Analyse each outcome to see if it is a favourable outcome (favourable if the die lands with any of the numbers 1, 2, 3, or 4 on the upper face) to give us an event:

Outcome 1: Number is 1 → A favourable outcome → We have an event

Outcome 2: Number is 2 → A favourable outcome → We have an event

Outcome 3: Number is 3 → A favourable outcome → We have an event

Outcome 4: Number is 4 → A favourable outcome → We have an event

Outcome 5: Number is 5 → Not a favourable outcome → We don’t have an event

Outcome 6: Number is 6 → Not a favourable outcome → We don’t have an event

So, out of the 6 possible outcomes, 4 are favourable, and gives us an event. So,the probability for the occurrence of the event = ⁴⁄₆ = ²⁄₃. That is:
• The probability of getting number less than 5 is ²⁄₃

Part (iii): Probability for getting an even number. For this, if we get any of the numbers: '2', '4', or '6', we have an event. We will do the experiment once. That is., rolling the die will be done once.

Step 1: Write the possible outcomes:

The possible outcomes are the same as those for part (i). So we need not write them again.

Step 2: Analyse each outcome to see if it is a favourable outcome (favourable if the die lands with any of the numbers 2, 4, or 6 on the upper face) to give us an event:

Outcome 1: Number is 1 → Not a favourable outcome → We don’t have an event

Outcome 2: Number is 2 → A favourable outcome → We have an event

Outcome 3: Number is 3 → Not a favourable outcome → We don’t have an event

Outcome 4: Number is 4 → A favourable outcome → We have an event

Outcome 5: Number is 5 → Not a favourable outcome → We don’t have an event

Outcome 6: Number is 6 → A favourable outcome → We have an event

So, out of the 6 possible outcomes, 3 are favourable, and gives us an event. So,the probability for the occurrence of the event = ³⁄₆ = ¹⁄₂. That is:
• The probability of getting an even number is ¹⁄₂

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In the next section, we will discuss the probability related to 'drawing a card from a standard deck of cards'.

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