1/1 - 1/20 - 1/19 - 1/18 - 1/17- 1/16 -1/15 -1/14 - 1/13 - 1/12 = about 45.2 percent
The probability of getting exactly eight heads when tossing 10 coins once can be found using the binomial probability formula. Assuming a fair coin, the probability of getting a heads is 1/2. Plugging in the numbers, the probability of getting exactly eight heads is (10 choose 8) * (1/2)^8 * (1/2)^2 = 45/1024, which is approximately 0.04395.
1 out of 2
I'm assuming your question is the same as this: "If 2 dice are rolled, what is the probability of not getting 1 on either die?" To answer this question, we need to look at what IS possible. If I'm 2 rolling normal, fair dice, then I have equal probability of getting each of the numbers 1-6 on either die. If I'm trying to NOT get 1, then I want to get any of the numbers 2-6 on both dice. This gives me 10 desired outcomes (5 numbers * 2 dice) out of 12 possible outcomes (6 numbers * 2 dice), so the probability is 10/12, which simplifies to 5/6.
If there is a different number on each face of the die and no numbers a repeated it is a 1 in 10 chance.
7/10
The probability of getting two prime numbers when two numbers are selected at random and without replacement, from 1 to 10 is 2/15.
The probability of getting exactly eight heads when tossing 10 coins once can be found using the binomial probability formula. Assuming a fair coin, the probability of getting a heads is 1/2. Plugging in the numbers, the probability of getting exactly eight heads is (10 choose 8) * (1/2)^8 * (1/2)^2 = 45/1024, which is approximately 0.04395.
The probability is 3 out of 10.
The factors of 10 are the numbers that divide 10 evenly: 1, 2, 5 and 10. To answer your question, you have to figure out what the probability of rolling one of these numbers is on a number cube.
1 out of 2
Probability of getting not a yellow sweet if there 3 yellow sweets and 10 blue sweets is 10/13.
The probability is 10/50 = 1/5.
I'm assuming your question is the same as this: "If 2 dice are rolled, what is the probability of not getting 1 on either die?" To answer this question, we need to look at what IS possible. If I'm 2 rolling normal, fair dice, then I have equal probability of getting each of the numbers 1-6 on either die. If I'm trying to NOT get 1, then I want to get any of the numbers 2-6 on both dice. This gives me 10 desired outcomes (5 numbers * 2 dice) out of 12 possible outcomes (6 numbers * 2 dice), so the probability is 10/12, which simplifies to 5/6.
A die normally has six sides with the numbers 1 to 6 on them, so any roll will be less than 10 and thus the probability of getting less than 10 with a die is 1. With two normal dice, the sum of the digits on the dice added together ranges from 2 to 12 and the probability of getting less than 10 is the same as 1 minus the probability of getting 10 or more. There are 36 ways the two dice can fall and 10 can be achieved in 3 ways (4&6, 5&5, 6&4), 11 can be achieved in 2 ways (5&6, 6&5) and 12 in 1 way (6&6). Thus the probability of getting 10 or more with 2 dice is (3+2+1)/36 = 6/36 = 1/6 So the probability of getting less than 10 is 1-1/6 = 5/6
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%.
If there is a different number on each face of the die and no numbers a repeated it is a 1 in 10 chance.
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