Probability And Statistics

The question asks for the geometric probability of a randomly placed dot falling within the shaded region of a circle. The shaded region is the major sector, and the unshaded region is the minor sector with a central angle of 120°.

Geometric probability is calculated as the ratio of the favorable area to the total area.
Probability = Area of the favorable regionTotal Area Since the area of a sector is directly proportional to its central angle, we can also use the ratio of the angles.
Probability = Central angle of the favorable regionTotal angle in a circle The total angle in a circle is 360°.
The central angle of the unshaded minor sector is given as 120°.
The shaded region is the major sector. Its central angle is the total angle minus the angle of the minor sector.
Angle of shaded region = Total angle - Angle of unshaded region
Angle of shaded region = 360^ - 120^ = 240^ This is the favorable angle.
Now, we can calculate the probability:
P(dot in shaded region) = Angle of shaded regionTotal angle in a circle P(dot in shaded region) = (240^)/(360^) To simplify the fraction, we can divide both the numerator and the denominator by their greatest common divisor, which is 120.
P(dot in shaded region) = (240 ÷ 120)/(360 ÷ 120) = (2)/(3) The probability that the dot is in the shaded region is (2)/(3).

We are drawing one number from each of two boxes and finding the product. We need to calculate the probability that this product is a prime number.
Box 1: 1, 3, 5, 7
Box 2: 2, 4, 6
P(Event) = Number of favorable outcomesTotal number of possible outcomes We need to list all possible pairs and their products to find the favorable and total outcomes.

Total number of possible outcomes:
There are 4 choices from Box 1 and 3 choices from Box 2.
Total number of pairs = 4 × 3 = 12.
The possible pairs (Box1, Box2) are:
(1,2), (1,4), (1,6)
(3,2), (3,4), (3,6)
(5,2), (5,4), (5,6)
(7,2), (7,4), (7,6)
Number of favorable outcomes:
We need to find the product for each pair and check if it is a prime number. (A prime number is a natural number greater than 1 that has no positive divisors other than 1 and itself).
Products:
1 × 2 = 2 (Prime)
1 × 4 = 4 (Not prime)
1 × 6 = 6 (Not prime)
3 × 2 = 6 (Not prime)
3 × 4 = 12 (Not prime)
3 × 6 = 18 (Not prime)
5 × 2 = 10 (Not prime)
5 × 4 = 20 (Not prime)
5 × 6 = 30 (Not prime)
7 × 2 = 14 (Not prime)
7 × 4 = 28 (Not prime)
7 × 6 = 42 (Not prime)
The only product that is a prime number is 2, which comes from the pair (1, 2).
So, there is only 1 favorable outcome.

Calculating the probability:
P(product is prime) = (1)/(12) The probability that the product is a prime number is (1)/(12).

Using the same setup as the previous question, we now need to find the probability that the product of the two numbers drawn is a multiple of 7.
Box 1: 1, 3, 5, 7
Box 2: 2, 4, 6
P(Event) = Number of favorable outcomesTotal number of possible outcomes Total number of possible outcomes:
As before, the total number of possible pairs is 4 × 3 = 12.

Number of favorable outcomes:
For the product of two numbers to be a multiple of 7, at least one of the numbers must be a multiple of 7.
In Box 1, the number 7 is present.
In Box 2, there are no multiples of 7.
Therefore, a favorable outcome occurs only when we draw the number 7 from Box 1. The number drawn from Box 2 can be any of its three numbers (2, 4, or 6).
The favorable pairs are:
(7, 2) → Product = 14 (Multiple of 7)
(7, 4) → Product = 28 (Multiple of 7)
(7, 6) → Product = 42 (Multiple of 7)
There are 3 favorable outcomes.

Calculating the probability:
P(product is a multiple of 7) = Favorable outcomesTotal outcomes = (3)/(12) Simplifying the fraction:
P(product is a multiple of 7) = (1)/(4) The probability that the product is a multiple of 7 is (1)/(4).

We need to find the probability of randomly choosing a three-digit number where all three digits are identical (e.g., 111, 222).

The probability of an event is calculated as:
P(Event) = Number of favorable outcomesTotal number of possible outcomes Total number of possible outcomes:
Three-digit numbers start from 100 and end at 999.
Total count = (Last number - First number) + 1
Total count = (999 - 100) + 1 = 899 + 1 = 900.
So, there are 900 possible three-digit numbers.

Number of favorable outcomes:
We are looking for numbers where all three digits are the same.
These numbers are: 111, 222, 333, 444, 555, 666, 777, 888, 999.
The number 000 is not a three-digit number.
So, there are 9 favorable outcomes.

Calculating the probability:
P(all digits are the same) = Favorable outcomesTotal outcomes = (9)/(900) Simplifying the fraction:
P(all digits are the same) = (1)/(100) The probability that all three digits are the same is (1)/(100).

We need to find the probability that a randomly chosen three-digit number is a multiple of 5 and is also less than 200.

The probability of an event is calculated as:
P(Event) = Number of favorable outcomesTotal number of possible outcomes Total number of possible outcomes:
As calculated in the previous part, the total number of three-digit numbers is 900.

Number of favorable outcomes:
We need to count the three-digit numbers that are multiples of 5 and are less than 200.
The three-digit numbers start from 100. So we are looking for multiples of 5 in the range [100, 199].
The multiples of 5 in this range are: 100, 105, 110, ..., 195.
This forms an arithmetic progression with first term a = 100, last term aₙ = 195, and common difference d = 5.
We can find the number of terms (n) using the formula aₙ = a + (n-1)d.
195 = 100 + (n-1)5 95 = (n-1)5 n-1 = (95)/(5) = 19 n = 19 + 1 = 20 So, there are 20 favorable outcomes.

Calculating the probability:
P(multiple of 5 and < 200) = Favorable outcomesTotal outcomes = (20)/(900) Simplifying the fraction:
P(multiple of 5 and < 200) = (2)/(90) = (1)/(45) The probability is (1)/(45).

We are given a set of 7 data points (weights of students) and asked to find the median of this data.

The median is the middle value in a dataset that has been arranged in ascending or descending order.
1. Arrange the data in order.
2. If the number of observations (n) is odd, the median is the ((n+1)/(2))-th term.
3. If n is even, the median is the average of the ((n)/(2))-th and ((n)/(2)+1)-th terms.

The given weights are: 44, 42, 40, 47, 41, 38, 45.
First, we arrange these numbers in ascending order:
38, 40, 41, 42, 44, 45, 47
The number of students (observations) is n = 7, which is an odd number.
The position of the median is given by ((n+1)/(2))-th term.
Position = ((7+1)/(2)) = (8)/(2) = 4-th term.
Now we find the 4th term in the ordered list:
1st term = 38
2nd term = 40
3rd term = 41
4th term = 42
5th term = 44
6th term = 45
7th term = 47
The 4th term is 42. Therefore, the median weight is 42.