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Q: If a number is chosen at random from the numbers 1 to 20 inclusive what is the probability that an even number will be picked?

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From 75 to 100 (inclusive), there are 26 numbers, and 13 of them are odd.The probability of picking an odd number is 13/26 = 50%.

3 out of 4

90%

50 50 odd or even same probability

Using the formula n!/r!(n-r)! where n is the number of possible numbers and r is the number of numbers chosen, there are 13983816 combinations of six numbers between 1 and 49 inclusive.

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There are 8 out of 20 numbers that are prime, so 8/20, or 2/5.

From 75 to 100 (inclusive), there are 26 numbers, and 13 of them are odd.The probability of picking an odd number is 13/26 = 50%.

15/49ExplanationThere are 15 prime numbers between 1 and 49 (2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47). If you randomly choose one natural number from the 49 numbers between 1 and 49 inclusive, there is a 15/49 probability that it will be prime.

3 out of 4

90%

1 out of 20 this is because there are 20 numbers in total, and there is only one 7 in there. (Assuming that there is the same probability for each number to be chosen, and that 17 is excluded as an affirmative outcome)

50%

1/3

There are 12 composite (and 8 primes) in the first twenty whole numbers. So the probability of randomly choosing a non-prime is 12/20 or 60%.

Assuming no repetition is allowed, there are 8582777280 ways in which you can pick any of the numbers from 1 to 99 inclusive with 5 numbers. This is given by the formula n! / (n - r)! where n is the number of numbers you have to choose from, and r is the number of numbers chosen.

the first 10 whole numbers are numbers 1 to 10 and in those numbers only 3 numbers are divisible by 3 in which 3, 6 and 9 therefore the probability of from those figures that the numbers won't be divisible by 3 is 7/10 or 70%.

A probability must needs be a number between 0 and 1 (often expressed as 0% and 100%), inclusive.