15
255/256 (complement formula)
Ignoring the very small possibility that a coin tossed will land on its edge and stay there, I'll assume each coin can land either Head or Tails. The reverse of the condition "at least one Head" is "no Heads" Thus the number of ways of at least one Head is the same as the total number of ways less the number of way of no Heads. There are 2 × 2 × 2 = 2³ = 8 possible ways the three coins can land. Of these there is only 1 way which shows no Heads: TTT Thus there are 8 - 1 = 7 ways that at least one Head shows You could list all the possible ways that at least one head shows (which are: HTT, THT, TTH, HHT, HTH, THH, HHH - 7 of them), but the "subtractive" method above is much quicker and easier and less prone to error in the listing.
Assuming the coins are fair, two-sided coins, and landing on their sides is not an option, there are four possible outcomes if you consider coin a having a head and coin b having a tail being a different instance from coin a being a tail and coin be having a head. Here they are; Coin A | Coin B Heads | Tails Heads | Heads Tails....| Heads Tails....| Tails
A trick coin that you buy at a magic shop, I reckon.
>>> 1:7 (or, if you like probability, 87.5%)I disagree. There are four possible combinations of three tosses (where order does not matter):HHHHHTHTTTTTThree of these combinations will show at least one head - only by throwing three tails will you not throw at least one head.Thus, the probability of throwing at least one head in three flips is 75%.
Is possible.
255/256 (complement formula)
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
Let put the question in an other way : obtaining at least one head is the contrary of obtaining two tails at the same time. The probability to obtain one tail with first coin is 1/2, the probability to obtain one tail with the other is also 1/2, so the probability to obtain one tail on each coin is 1/2x1/2 = 1/4 Thus the probability to obtain at least one head is 1-1/4=3/4
you toss 3 coins what is the probability that you get exactly 2 heads given that you get at least one head?
50%
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 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 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.
The probability of 2 coins both landing on heads or both landing on tails is 1/2 because there are 4 possible outcomes. Head, head. Head, tails. Tails, tails. Tails, heads. Tails, heads is different from heads, tails for reasons I am unsure of.
It is 15/16.
If you throw a single fair coin multiple times, the probability of getting NO head is:For 1 throw: 1/2 For 2 throws: 1/2 squared = 1/4 For 3 throws: 1/2 cubed = 1/8 etc. The probability of getting AT LEAST ONE head is the complement; for example, for 3 throws, it would be 1 minus 1/8.