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What is the probability of getting 10 different numbers out of 1 to 20 numbers?
Wiki User
∙ 14y ago
1/1 - 1/20 - 1/19 - 1/18 - 1/17- 1/16 -1/15 -1/14 - 1/13 - 1/12 = about 45.2 percent
Wiki User
∙ 14y ago
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10 coins are tossed once .The probalilty of getting exactly eight heads is?
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.
What is the probability of getting a factor of 10?
1 out of 2
What is the probality of not getting 1 out of 2 dice?
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.
What is the probability of rolling a 4 on a 10-sided dice?
If there is a different number on each face of the die and no numbers a repeated it is a 1 in 10 chance.
What are the chances of for the probability 2 to 5?
The chances of picking a number between 2 to 5 is 4/10 if the numbers to be picked from are 1-10. However, if the numbers to be picked from are 1-100, then the probability drops to 4/100.
What is the probability of getting two prime numbers 1-10?
The probability of getting two prime numbers when two numbers are selected at random and without replacement, from 1 to 10 is 2/15.
10 coins are tossed once .The probalilty of getting exactly eight heads is?
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.
What is the probability to choose a prime number from numbers 1 to 50?
The probability is 3 out of 10.
Theoretical probability of rolling a factor of 10 on a number cube?
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.
What is the probability of getting a factor of 10?
1 out of 2
What is the probability of choosing a divisor of 48 numbers 1-50?
The probability is 10/50 = 1/5.
What the probability of getting not a yellow sweet if there 3 yellow sweets and 10 blue sweets?
Probability of getting not a yellow sweet if there 3 yellow sweets and 10 blue sweets is 10/13.
What is the probality of not getting 1 out of 2 dice?
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.
What is the probability of getting less than 10 when you roll a dice?
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
A number is chosen at random from the first 10 whole number What is the probability that it is not exactly divisible by 3?
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%.
What is the probability of rolling a 4 on a 10-sided dice?
If there is a different number on each face of the die and no numbers a repeated it is a 1 in 10 chance.
What is the probability of 1 to 10 to get a number greater then 3?
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
