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Sagot :
Sure, let's break down the solution step-by-step:
### (i) Define sample space.
The sample space in probability represents all possible outcomes of a given experiment. When a normal die is thrown, it has six faces, each showing a number from 1 to 6. Thus, for a single throw of a die, the sample space (S) is:
[tex]\[ S = \{1, 2, 3, 4, 5, 6\} \][/tex]
### (ii) Find the probability of getting a prime numbered face.
First, identify which numbers on the die are prime. Prime numbers are numbers greater than 1 that have no divisors other than 1 and themselves. The prime numbers between 1 and 6 are:
[tex]\[ \{2, 3, 5\} \][/tex]
There are 3 prime numbers within the sample space of 6 possible outcomes.
The probability of an event is calculated as the number of favorable outcomes divided by the total number of possible outcomes. Therefore, the probability of rolling a prime number (P(prime)) is:
[tex]\[ P(\text{prime}) = \frac{\text{Number of prime numbers}}{\text{Total number of outcomes}} = \frac{3}{6} = 0.5 \][/tex]
### (iii) Find the probability of getting a composite numbered face.
Next, identify which numbers on the die are composite. Composite numbers are positive integers that have more than two distinct positive divisors. The composite numbers between 1 and 6 are:
[tex]\[ \{4, 6\} \][/tex]
There are 2 composite numbers within the sample space of 6 possible outcomes.
The probability of rolling a composite number (P(composite)) is:
[tex]\[ P(\text{composite}) = \frac{\text{Number of composite numbers}}{\text{Total number of outcomes}} = \frac{2}{6} = \frac{1}{3} \approx 0.3333 \][/tex]
### (iv) Write down the case under which the probability of an event is maximum.
The probability of an event is maximum when the outcome associated with the event has the highest expected count relative to the total number of possible outcomes. Since all faces of a die have an equal chance of turning up (i.e., 1/6 for each face), the event that has the highest count will have the maximum probability. In terms of equal likelihood for each face, this principle implies that among equal single-outcome events, any single outcome such as rolling a particular number (e.g., rolling a 1 or a 2) will have the highest probability possible in this uniform distribution, which is:
[tex]\[ P(\text{a specific face}) = \frac{1}{6} \approx 0.1667 \][/tex]
In summary, the maximum probability condition in a uniform probability distribution of a die is when you consider a single face among the 6 outcomes, each having a probability of 1/6 when the outcomes are equally likely.
### (i) Define sample space.
The sample space in probability represents all possible outcomes of a given experiment. When a normal die is thrown, it has six faces, each showing a number from 1 to 6. Thus, for a single throw of a die, the sample space (S) is:
[tex]\[ S = \{1, 2, 3, 4, 5, 6\} \][/tex]
### (ii) Find the probability of getting a prime numbered face.
First, identify which numbers on the die are prime. Prime numbers are numbers greater than 1 that have no divisors other than 1 and themselves. The prime numbers between 1 and 6 are:
[tex]\[ \{2, 3, 5\} \][/tex]
There are 3 prime numbers within the sample space of 6 possible outcomes.
The probability of an event is calculated as the number of favorable outcomes divided by the total number of possible outcomes. Therefore, the probability of rolling a prime number (P(prime)) is:
[tex]\[ P(\text{prime}) = \frac{\text{Number of prime numbers}}{\text{Total number of outcomes}} = \frac{3}{6} = 0.5 \][/tex]
### (iii) Find the probability of getting a composite numbered face.
Next, identify which numbers on the die are composite. Composite numbers are positive integers that have more than two distinct positive divisors. The composite numbers between 1 and 6 are:
[tex]\[ \{4, 6\} \][/tex]
There are 2 composite numbers within the sample space of 6 possible outcomes.
The probability of rolling a composite number (P(composite)) is:
[tex]\[ P(\text{composite}) = \frac{\text{Number of composite numbers}}{\text{Total number of outcomes}} = \frac{2}{6} = \frac{1}{3} \approx 0.3333 \][/tex]
### (iv) Write down the case under which the probability of an event is maximum.
The probability of an event is maximum when the outcome associated with the event has the highest expected count relative to the total number of possible outcomes. Since all faces of a die have an equal chance of turning up (i.e., 1/6 for each face), the event that has the highest count will have the maximum probability. In terms of equal likelihood for each face, this principle implies that among equal single-outcome events, any single outcome such as rolling a particular number (e.g., rolling a 1 or a 2) will have the highest probability possible in this uniform distribution, which is:
[tex]\[ P(\text{a specific face}) = \frac{1}{6} \approx 0.1667 \][/tex]
In summary, the maximum probability condition in a uniform probability distribution of a die is when you consider a single face among the 6 outcomes, each having a probability of 1/6 when the outcomes are equally likely.
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