There are 8 permutations of three coins. Of these, 3 of them have two heads, so the probability of tossing two heads on three coins is 3 in 8, or 0.375.
However, you said, "at least", so that includes the case of three heads, so the probability of throwing at least two heads is 4 in 8, or 0.5.
T T T
T T H
T H T
T H H *
H T T
H T H *
H H T *
H H H *
Probability of no heads = (0.5)^5 = 0.03125Probability of at least one head = 1 - probability of no heads = 1 - 0.03125 = 0.96875
The probability of tossing 6 heads in 6 dice is 1 in 26, or 1 in 64, or 0.015625. THe probability of doing that at least once in six trials, then, is 6 in 26, or 6 in 64, or 3 in 32, or 0.09375.
3 coins can land in 8 different ways. Only one of these ways is all tails. So the probability of rolling at least one heads is 7/8 = 87.5% .
The probability is 5/16.
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.
Probability of no heads = (0.5)^5 = 0.03125Probability of at least one head = 1 - probability of no heads = 1 - 0.03125 = 0.96875
The probability of tossing 6 heads in 6 dice is 1 in 26, or 1 in 64, or 0.015625. THe probability of doing that at least once in six trials, then, is 6 in 26, or 6 in 64, or 3 in 32, or 0.09375.
3 coins can land in 8 different ways. Only one of these ways is all tails. So the probability of rolling at least one heads is 7/8 = 87.5% .
The probability of tossing heads on all of the first six tosses of a fair coin is 0.56, or 0.015625. The probability of tossing heads on at least one of the first six tosses of a fair coin is 1 - 0.56, or 0.984375.
The probability is 5/16.
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.
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 is 0.3125
you toss 3 coins what is the probability that you get exactly 2 heads given that you get at least one head?
It is 15/16.
The probability that you will toss five heads in six coin tosses given that at least one is a head is the same as the probability of tossing four heads in five coin tosses1. There are 32 permutations of five coins. Five of them have four heads2. This is a probability of 5 in 32, or 0.15625. ----------------------------------------------------------------------------------- 1Simplify the problem. It asked about five heads but said that at least one was a head. That is redundant, and can be ignored. 2This problem was solved by simple inspection. If there are four heads in five coins, this means that there is one tail in five coins. That fact simplifies the calculation to five permutations exactly.
The probability of getting at least 1 tails is (1 - probability of getting all heads) The probability of getting all heads (no tails) is ½ x ½ x ½ x ½ x ½ x ½ x ½ x ½ = 1/256 = 0.00390625 so the probability of getting at least ONE tails is 1-0.30390625 = 0.99609375 = 255/256