If 10 coins are tossed, you could get 4 heads and 6 tails, 3 heads and 7 tails, 2 heads and 1 tail, 0 heads and 10 tails all giving fewer heads than tails.
Using the binomial distribution ,
P(4 heads) = 10C4 (.5)^4 (.5)^6 = 0.205078.
P(3 heads) = 10C3 (.5)^3 (.5)^7 = 0.117188
P(2 heads) = 10C2 (.5)^2 (.5)^8 = 0.043945
P(1 heads) = 10C1 (.5)^1 (.5)^9 = 0.009766
P(0 heads) =(.5)^10 = 0.000977
Adding all of these probabilities, we have P(fewer heads than tails)= 0.376953
The probability is 1/2 if the coin is flipped only twice. As the number of flips increases, the probability approaches 1.
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 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.
No, the possible results of two coin flips are; HH HT TH TT The probability of getting exactly one tail is 1/2 or 50%.
Since each event is independent (heads in one coin does not affect the probability of the other two coin flips), the multiplication rule applies: 1/2 x 1/2 x 1/2 = 1/8 or 0.125. So we can say the probability is 12.5%.
To find the probability of getting heads on the first two flips and tails on the third flip when flipping three fair coins, we multiply the probabilities of each individual event. The probability of getting heads on a flip is 1/2, so for the first two flips, it is (1/2) * (1/2) = 1/4. The probability of getting tails on the third flip is also 1/2. Therefore, the overall probability is (1/4) * (1/2) = 1/8.
The probability of HTTTT is (1/2)5 = 1/32 The correct flips can also be achieved with THTTT, TTHTT, TTTHT, & TTTTH. Hence, total probability is 5/32.
The probability is 1/2 if the coin is flipped only twice. As the number of flips increases, the probability approaches 1.
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
1/2 apex It does not matter what each prior flip's result was. Each flip has a probability of 0.5 heads or tails. Coins do not have "memory".
The probability of getting only one tails is (1/2)7. With seven permutations of which flip is the tails, this gives a probability of: P(six heads in seven flips) = 7*(1/2)7 = 7/128
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.
No, the possible results of two coin flips are; HH HT TH TT The probability of getting exactly one tail is 1/2 or 50%.
50%
It is 0.3125
It is 0.999531889
1 in 4