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:
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.
We will now see some solved examples:
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:
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| Fig.1.40 Probability for Number '6' |
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.
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.
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
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
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.
Solved example 1.9
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| Fig.1.41 Spin wheel |
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 |
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
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.
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
• 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
• 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
• The probability of getting an even number is 1⁄2
In the next section, we will discuss the probability related to 'drawing a card from a standard deck of cards'.
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