The probability that both coins are heads is the probability of one coin landing heads multiplied by the probability of the second coin landing heads: (.5) * (.5) = .25 or (1/2) * (1/2) = 1/4
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The answer depends on how many times the coin is tossed. The probability is zero if the coin is tossed only once! Making some assumptions and rewording your question as "If I toss a fair coin twice, what is the probability it comes up heads both times" then the probability of it being heads on any given toss is 0.5, and the probability of it being heads on both tosses is 0.5 x 0.5 = 0.25. If you toss it three times and want to know what the probability of it being heads exactly twice is, then the calculation is more complicated, but it comes out to 0.375.
The sample space is HH, HT, TH, HH. Since the HH combination can occur once out of four times, the probability that if a coin is flipped twice the probability that both will be heads is 1/4 or 0.25.
The probability of getting all heads or all tails in 5 flips of a coin is 1 in 16.The probability of getting a head or a tail on the first flip is 1 in 1. The probability of each of the following coins matching the first coin is 1 in 2. Simply multiply the five probabilities (1 in 1) (1 in 2) (1 in 2) (1 in 2) (1 in 2) and you get 1 in 16.It is true that the probability of getting all heads is 1 in 32, and the probability of getting all tails is also 1 in 32. Since the question asked the probability of both cases (all heads or all tails), the answer is 1 in 16.
The probability is 25%. The probability of flipping a coin once and getting heads is 50%. In your example, you get heads twice -- over the course of 2 flips. So there are two 50% probabilities that you need to combine to get the probability for getting two heads in two flips. So turn 50% into a decimal --> 0.5 Multiply the two 50% probabilities together --> 0.5 x 0.5 = 0.25. Therefore, 0.25 or 25% is the probability of flipping a coin twice and getting heads both times.
You have 12 options (2 x 6) that are all equally likely; the only favorable options is (coin 1, die 2), so that makes a probability of 1/12.