With two flips of a coin you can get two heads, two tails, a head and a tail, or a tail and a head. There are a total of four different possible outcomes, and three of them have at least one head. That's 3 out of 4, or 3/4ths. It's also 0.75 which is the probability of getting at least one head with two flips of a coin. Note that as we use the term probability here, it is zero (no chance it can happen at all), or one (it must happen), or something in between. A probability appears in the form of a fraction or decimal, and has no units attached to it.
The probability of obtaining 4 tails when a coin is flipped 4 times is: P(4T) = (1/2)4 = 1/16 = 0.0625 Then, the probability of obtaining at least 1 head when a coin is flipped 4 times is: P(at least 1 head) = 1 - 1/16 = 15/16 = 0.9375
The probability is 0.998
The probability of flipping a fair coin four times and getting four heads is 1 in 16, or 0.0625. That is simply the probability of one head (0.5) raised to the power of 4.
The probability is 1. I have flipped a coin a lot more than 7 times.
This is easiest calculated by calculating the probability that NO SINGLE heads is obtained; this is of course the complement of the question. The probability of this is 1/2 x 1/2 x 1/2 ... 7 times, in other words, (1/2)7. The complement, the probability that at least one head is obtained, is then of course 1 - (1/2)7, or a bit over 99%.
Multiply the probability by the number of times the experiment was carried out. 0.6x10=6
It is 4*(1/2)4 = 4/16 = 1/4
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
The probability of getting an even number on at least one of the 3 rolls is 7/8.
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%
That depends how many times you flip the coin.
If a coin is flipped 4 times, the probability of getting 3 heads is: 4C3 (1/2)^3 (1/2)^1 = 4(1/8)(1/2) = 4/16 = 1/4
As the question is "what is the probability of getting at least one head" the correct way to answer this is to ask what is the probability of not getting any heads and then subtract this from 1.The probability of not getting a head in 4 flips = 0.54 (i.e. 0.5 * 0.5 * 0.5 * 0.5) = 1/16.Therefore the probability of getting at least one head is 1 - 1/16 = 15/16.
Assuming that it is a fair coin, the probability is 0.9990
you are looking for the probability of getting one tail, two,......., and six this equivalent to saying 1 - the probability of not getting any tails (or getting 6 heads = (1/2)^6). P(X>=1) = 1 - (1/2)^6=
The probability of a flipped coin landing heads or tails will always be 50% either way, no matter how many times you flip it.
It depends on whether you mean exactly one head or at least one head. There are 8 possible outcomes: TTT, TTH, THH, HHH, HHT, HTT, HTH, THT For at least one head, the probability is 7/8, for exactly one head, the probability is 3/8.
The probability of getting a sum of 2 at least once is 0.8155
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 landing on heads at least once is 1 - (1/2)100 = 1 - 7.9*10-31 which is extremely close to 1: that is, the event is virtually a certainty.
The probability that a coin flipped four consecutive times will always land on heads is 1 in 16. Since the events are sequentially unrelated, take the probability of heads in 1 try, 0.5, and raise that to the power of 4... 1 in 24 = 1 in 16
the probability of winning that is the number you get over the total number of times you play the round!!!!!!!!!!!!for example: if i flipped the spoon two times, and you were supposed to flip 18 times, then the probability of winning is 2/18, which reduces to 1/9.
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The probability of getting 11 with one throw of 2 dice is 1/6*1/6*2 = 1/18 So the probability of not getting 11 with 1 throw of the dice is 17/18. Tossing the dice 54 times, the probability of not getting 11 54 times is (17/18)54 = 0.0456... So the probability of at least 1 roll of 11 is 1 - 0.0456 = 0.954
.5 or 50% probability (if not counting draws)