I assume you mean what's the chance of at least two heads showing when three fair coins are tossed.
There are 8 possible outcomes as each coin can either be head or tails.
For 3 heads, all 3 coins must show a head → 1 success
For 2 heads, one coin will be a Tail; each coin could be a tail in turn → 3 successes
→ Pr = (1+3)/8 = 4/8 = 1/2
If you are wanting the probability that the first TWO specific coins are heads and the last, third, coin is either, then:
Pr(head) = 1/2
→ Pr(1st 2 heads, 3rd anything) = 1/2 × 1/2 × 1 = 1/4
The chances of heads on first flip is 1/2 and on second flip 1/2 so chances of 2 heads on 2 flips is 1/2 x 1/2 = 1/4 (one in four). Chance of third flip either heads or tails is 1/1 so for the 3 flips it is 1/2 x 1/2 x 1/1 = 1 in 4 still
The probability is 0.25
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%.
It is approx 0.2461
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 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
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%.
The probability of getting four heads when flipping four identical coins at once is 1/16 or 0.0625. This is because each coin flip has a 1/2 probability of landing on heads, and since all four coin flips are independent events, we multiply the probabilities together: (1/2)^4 = 1/16.
Two flips can have four possible results:T - TT - HH - TH - HThe question defines 'success' as either H-T or T-H.Two successes out of four possibilities = probability of 2/4 = 1/2 = 50% .
50%
It is 0.3125
It is 0.999531889