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Sagot :
Let's solve each part of the question step-by-step.
(i) The probability of getting 4 on a throw of a die:
When throwing a fair six-sided die, each face (number) has an equal probability of landing face up. The six sides are numbered 1, 2, 3, 4, 5, and 6.
The probability [tex]\( P \)[/tex] of any specific number, such as 4, appearing on a throw of the die is given by:
[tex]\[ P(4) = \frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}} \][/tex]
Here, there is only one favorable outcome (the die shows the number 4), and there are six possible outcomes (the numbers 1 through 6), thus:
[tex]\[ P(4) = \frac{1}{6} \][/tex]
This simplifies to approximately 0.1667 (rounded to four decimal places).
(ii) The probability of getting either 2 or 3 on a throw of a die:
To find the probability of getting either 2 or 3, we again count the number of favorable outcomes and the total number of possible outcomes.
The favorable outcomes are 2 and 3, so:
[tex]\[ P(2 \text{ or } 3) = \frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}} \][/tex]
There are two favorable outcomes (2 and 3), and the total number of possible outcomes remains six:
[tex]\[ P(2 \text{ or } 3) = \frac{2}{6} \][/tex]
Simplifying this fraction gives:
[tex]\[ P(2 \text{ or } 3) = \frac{1}{3} \][/tex]
This simplifies to approximately 0.3333 (rounded to four decimal places).
(iii) The probability of getting a number greater than 2 on a throw of a die:
To find the probability of rolling a number greater than 2, we consider the numbers greater than 2: 3, 4, 5, and 6.
Thus, the probability [tex]\( P \)[/tex] is given by:
[tex]\[ P(\text{greater than 2}) = \frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}} \][/tex]
There are four favorable outcomes (3, 4, 5, and 6), so:
[tex]\[ P(\text{greater than 2}) = \frac{4}{6} \][/tex]
Simplifying this fraction gives:
[tex]\[ P(\text{greater than 2}) = \frac{2}{3} \][/tex]
This simplifies to approximately 0.6667 (rounded to four decimal places).
The mean of Shivansh's marks in 5 subjects is 86:
The mean (average) of a set of numbers is calculated as the sum of the numbers divided by the count of the numbers.
If the mean of Shivansh's marks in 5 subjects is 86, it indicates that if you sum up Shivansh's marks in all 5 subjects and divide by 5, you get 86:
[tex]\[ \text{Mean of marks} = 86 \][/tex]
This means that the total sum of Shivansh's marks in 5 subjects is:
[tex]\[ \text{Total sum of marks} = 86 \times 5 = 430 \][/tex]
Thus, the mean remains 86.
In summary, the detailed solutions are:
(i) The probability of getting 4 is approximately 0.1667.
(ii) The probability of getting 2 or 3 is approximately 0.3333.
(iii) The probability of getting a number greater than 2 is approximately 0.6667.
The mean of Shivansh's marks in 5 subjects is 86.
(i) The probability of getting 4 on a throw of a die:
When throwing a fair six-sided die, each face (number) has an equal probability of landing face up. The six sides are numbered 1, 2, 3, 4, 5, and 6.
The probability [tex]\( P \)[/tex] of any specific number, such as 4, appearing on a throw of the die is given by:
[tex]\[ P(4) = \frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}} \][/tex]
Here, there is only one favorable outcome (the die shows the number 4), and there are six possible outcomes (the numbers 1 through 6), thus:
[tex]\[ P(4) = \frac{1}{6} \][/tex]
This simplifies to approximately 0.1667 (rounded to four decimal places).
(ii) The probability of getting either 2 or 3 on a throw of a die:
To find the probability of getting either 2 or 3, we again count the number of favorable outcomes and the total number of possible outcomes.
The favorable outcomes are 2 and 3, so:
[tex]\[ P(2 \text{ or } 3) = \frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}} \][/tex]
There are two favorable outcomes (2 and 3), and the total number of possible outcomes remains six:
[tex]\[ P(2 \text{ or } 3) = \frac{2}{6} \][/tex]
Simplifying this fraction gives:
[tex]\[ P(2 \text{ or } 3) = \frac{1}{3} \][/tex]
This simplifies to approximately 0.3333 (rounded to four decimal places).
(iii) The probability of getting a number greater than 2 on a throw of a die:
To find the probability of rolling a number greater than 2, we consider the numbers greater than 2: 3, 4, 5, and 6.
Thus, the probability [tex]\( P \)[/tex] is given by:
[tex]\[ P(\text{greater than 2}) = \frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}} \][/tex]
There are four favorable outcomes (3, 4, 5, and 6), so:
[tex]\[ P(\text{greater than 2}) = \frac{4}{6} \][/tex]
Simplifying this fraction gives:
[tex]\[ P(\text{greater than 2}) = \frac{2}{3} \][/tex]
This simplifies to approximately 0.6667 (rounded to four decimal places).
The mean of Shivansh's marks in 5 subjects is 86:
The mean (average) of a set of numbers is calculated as the sum of the numbers divided by the count of the numbers.
If the mean of Shivansh's marks in 5 subjects is 86, it indicates that if you sum up Shivansh's marks in all 5 subjects and divide by 5, you get 86:
[tex]\[ \text{Mean of marks} = 86 \][/tex]
This means that the total sum of Shivansh's marks in 5 subjects is:
[tex]\[ \text{Total sum of marks} = 86 \times 5 = 430 \][/tex]
Thus, the mean remains 86.
In summary, the detailed solutions are:
(i) The probability of getting 4 is approximately 0.1667.
(ii) The probability of getting 2 or 3 is approximately 0.3333.
(iii) The probability of getting a number greater than 2 is approximately 0.6667.
The mean of Shivansh's marks in 5 subjects is 86.
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