P(x≥9) = 0.27777... ≈ 28%
Let x equal to the sum of the numbers showing on dice when you roll a pair of dice.
P(x=12) = 1/36; P(x=11) = 2/36; P(x=10) = 3/36; P(x=9) = 4/36.
So, P(x≥9) = P(x=12) + P(x=11) + P(x=10) + P(x=9) = 1/36 [1+2+3+4] = 10/36
P(x≥9) = 0.2777... ≈ 28%
An impossible event, with probability 0.
The probability level for an outcome is the probability that the outcome was at least as extreme as the one that was observed.
If it is a fair coin, the probability of getting at least one Head from 3 flips is 7/8If it is a fair coin, the probability of getting at least one Head from 3 flips is 7/8If it is a fair coin, the probability of getting at least one Head from 3 flips is 7/8If it is a fair coin, the probability of getting at least one Head from 3 flips is 7/8
This is a Binomial Probability Distribution; n=4, p=0.6. The probability of at least 1 passed is equal to the probability of 1-none passed; so x=0. The probability of x=0 (with n=4, p=0.6) is 0.0256. So, the probability of at least 1 passed is 1-0.0256 or 0.9744.
This is a binomial probability distribution. The number of trials, n, equals 30; and the probability of success is p, which is 0.1. In this problem, you want the probability of at least 5, which is the complement of at most 4. We use the complement because we can subtract from 1 that probability and we will have the solution. The related link has the binomial probability distribution table which is cumulative. Per the table, at n=30, p=0.1 and x = 4; the probability is 0.825. Therefore the probability of at least 5 is 1 - 0.825 or 0.175.
Probability not at least 1 head showing is when all 5 coins are tails: (1/2)5=1/32 Therefore probability at least 1 head is showing is 1-1/32=31/32
It is 5/18.
Probability of no heads = (0.5)^5 = 0.03125Probability of at least one head = 1 - probability of no heads = 1 - 0.03125 = 0.96875
It is 15/16.
This is easiest to solve by working out the probability that no heads show and subtracting this from 1 to give the probability that at least one head shows: Assuming unbiased coins which won't land and stay on their edge, the probability of head = probability of tail = ½ → probability no heads = probability 5 tails = ½^5 = 1/32 → probability of at least one head = 1 - 1/32 = 31/32 = 0.96875 = 96.875 % = 96 7/8 %
It depends upon how you are picking these numbers. Let's say you are rolling two dice. The probability of rolling 2 fours is 1 in 36. The probability of exactly 1 five is 10 in 36, while the probability of at least 1 five is 11 in 36. The probability of exactly 1 six is 10 in 36, while the probability of at least 1 six is 11 in 36. The probability of at least 1 five or 1 six is 19 in 36. The probability of exactly 1 five or six is 15 in 36. So no matter how you look at it, with dice rolling, the probability of 1 five or 1 six is bigger than the probability of 2 fours. However, if you are picking numbers from a hat, then the probabilities are different.
There are 36 possible combinations. Eleven of them have at least one four in it. That means it is 11 over 36, which is a 30.55% chance.
I suppose you mean, at least one of those numbers. Just calculate the probability of NOT getting any of those, and take the complement. The probability of not getting a one nor a five on a single die is 4/6 or 2/3. For two dice, the probability is 2/3 x 2/3 = 4/9. So, the probability of getting at least a one or a five with two dice is 1 - 4/9 = 5/9.I suppose you mean, at least one of those numbers. Just calculate the probability of NOT getting any of those, and take the complement. The probability of not getting a one nor a five on a single die is 4/6 or 2/3. For two dice, the probability is 2/3 x 2/3 = 4/9. So, the probability of getting at least a one or a five with two dice is 1 - 4/9 = 5/9.I suppose you mean, at least one of those numbers. Just calculate the probability of NOT getting any of those, and take the complement. The probability of not getting a one nor a five on a single die is 4/6 or 2/3. For two dice, the probability is 2/3 x 2/3 = 4/9. So, the probability of getting at least a one or a five with two dice is 1 - 4/9 = 5/9.I suppose you mean, at least one of those numbers. Just calculate the probability of NOT getting any of those, and take the complement. The probability of not getting a one nor a five on a single die is 4/6 or 2/3. For two dice, the probability is 2/3 x 2/3 = 4/9. So, the probability of getting at least a one or a five with two dice is 1 - 4/9 = 5/9.
An impossible event, with probability 0.
No matter how many coins are thrown, the possibility of having AT LEAST ONE 'head' is 50%. This changes if you specify the number of 'heads' that must be shown.
The probability level for an outcome is the probability that the outcome was at least as extreme as the one that was observed.
the electron cloud is least dense where the probability of finding an electron is LOWEST