In the range of 1 to 10, there are five odd numbers: 1, 3, 5, 7, and 9. Since there are a total of 10 numbers, the probability of selecting an odd number is the number of odd outcomes divided by the total outcomes. This gives us a probability of 5/10, which simplifies to 1/2 or 50%.
It is 14/20 or 7/10.
10 possibilities, only three (7, 8 & 9) are greater than 6 so probability is 0.3 or 30%
To find the probability of selecting a multiple of 2 or 3 from the numbers 1 to 10, first identify the multiples: the multiples of 2 are 2, 4, 6, 8, and 10; the multiples of 3 are 3, 6, and 9. The number 6 is counted in both categories, so the unique multiples of 2 or 3 are 2, 3, 4, 6, 8, 9, and 10, totaling 7 unique numbers. Since there are 10 possible selections, the probability is 7/10 or 0.7.
There are 20 numbers in total from 1 to 20. The even numbers in this range are 2, 4, 6, 8, 10, 12, 14, 16, 18, and 20, totaling 10 even numbers. Therefore, the probability of picking an even number is the number of even numbers divided by the total numbers, which is ( \frac{10}{20} = \frac{1}{2} ). Thus, the probability of selecting an even number is 0.5 or 50%.
It is 0.4
In the range of 1 to 10, there are five odd numbers: 1, 3, 5, 7, and 9. Since there are a total of 10 numbers, the probability of selecting an odd number is the number of odd outcomes divided by the total outcomes. This gives us a probability of 5/10, which simplifies to 1/2 or 50%.
There are 10 numbers {1, 2, ..., 10} The solution set contains {4, 5, ...,10} - 7 numbers in total → probability = number of successful tries/total number of tries = 7/10 = 0.7
It is 14/20 or 7/10.
10 possibilities, only three (7, 8 & 9) are greater than 6 so probability is 0.3 or 30%
Probability = 10 is a very serious mistake since the probability of any event can never be greater than 1: so a probability of 10 is obviously a big error.
If the only numbers to pick from are 1 through 8, how can you get a factor greater than 10?
25/50 gives the probability of selecting a blue marble
To find the probability of selecting a multiple of 2 or 3 from the numbers 1 to 10, first identify the multiples: the multiples of 2 are 2, 4, 6, 8, and 10; the multiples of 3 are 3, 6, and 9. The number 6 is counted in both categories, so the unique multiples of 2 or 3 are 2, 3, 4, 6, 8, 9, and 10, totaling 7 unique numbers. Since there are 10 possible selections, the probability is 7/10 or 0.7.
There are 20 numbers in total from 1 to 20. The even numbers in this range are 2, 4, 6, 8, 10, 12, 14, 16, 18, and 20, totaling 10 even numbers. Therefore, the probability of picking an even number is the number of even numbers divided by the total numbers, which is ( \frac{10}{20} = \frac{1}{2} ). Thus, the probability of selecting an even number is 0.5 or 50%.
The numbers from 00 to 99 include a total of 100 numbers. Among these, the numbers that end in 5 are 05, 15, 25, 35, 45, 55, 65, 75, 85, and 95, which totals 10 numbers. Therefore, the probability of randomly selecting a number that ends in 5 is 10 out of 100, or 0.1 (10%).
Let P(A) = 1/10; P(A) = probability of selecting one people on a basketball team P(B) = 1/35; P(B) = probability of selecting one people on a football team P(C) = 1/10 = probability of selecting one people who plays in both teams P(D) = probability of selecting from either team. P(D) = P(A) + P(B) - P(C) P(D) = 1/10 + 1/35 - 1/10 P(D) = 1/35 or 0.0286