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Answered 2011-05-14 23:53:24

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 H H *


H T H *

H H T *

H H H *

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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.

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 %

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.

you toss 3 coins what is the probability that you get exactly 2 heads given that you get at least one head?

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

The probability of tossing a coin 9 times and getting at least one tail is: P(9 times, at least 1 tail) = 1 - P(9 heads) = 1 - (0.50)9 = 0.9980... ≈ 99.8%

By tossing two coins the possible outcomes are:H & HH & TT & HT & TThus the probability of getting exactly 1 head is 2 out 4 or 50%. If the question was what is the probability of getting at least 1 head then the probability is 3 out of 4 or 75%

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

The probability of getting at least one tail in a flip of six coins is the same as the probability of not getting all heads, which is 1 - (0.56), or 0.984375.

The empirical probability can only be determined by carrying out the experiment a very large number of times. Otherwise it would be the theoretical probability.

The requirement that one coin is a head is superfluous and does not matter. The simplified question is "what is the probability of obtaining exactly six heads in seven flips of a coin?"... There are 128 permutations (27) of seven coins, or seven flips of one coin. Of these, there are seven permutations where there are exactly six heads, i.e. where there is only one tail. The probability, then, of tossing six heads in seven coin tosses is 7 in 128, or 0.0546875.

Possibilities: HHH, HHT, HTH, THH, HTT, THT, TTH, TTT. There are 3 chances out of 8 that there will be two heads and one more that there will be AT LEAST two heads.

There are 2 coins, 1 and 2. Each has two possibilities, H or T. The possibilities are: 1H, 2H 1H, 2T 1T, 2H 1T, 2T Each possibility has an equal chance of happening. The chance of tossing at least one head is 3/4.

There are four outcomes possible (not considering order)HHHHHTHTTTTTOnly in two of the cases are there two or more headsThe probability is 0.5

Number of possible outcomes of one coin = 2Number of possible outcomes of six coins = 2 x 2 x 2 x 2 x 2 x 2 = 64Number of possible outcomes with six heads = 1Probability of six heads = 1/64Probability of not six heads = at least one tails = 63/64 = 98.4375%

This is one of those cases where it is probably easier to think what is the probability of not doing it, then subtracting that from 1 to get the probability of doing it. To not get at least one head and one tail, you would have to get all heads or all tails. To get all heads, the probability is (1/2)5. To get all tails is the same probability; so double it to get the probability of either of those. 2(1/2)5=1/16. Subtract the 1/16 from 1 to get 15/16. Answer: 15/16

The correct answer is 31/32 or 0.96875 because .50/2/2/2/2= 3.125%

Pr(At least one head in three tosses) = 1 - Pr(No heads in three tosses) = 1 - Pr(Three tails in three tosses) = 1 - (1/2)*(1/2)*(1/2) = 1 - 1/8 = 7/8 or 0.875 or 87.5%

The chance for each toss is 1/2. The chance of all three flips being tails is 1/2 * 1/2 * 1/2 = 1/8. So the probability of at least one head is 7/8.

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