.125
It is (1/2)3 = 1/8
P(4T1H)) = 5C4 ∙ (0.5)5 = 5/32 = 0.15625 ≈ 15.6% where 5C4 = 5!/[3!∙(5-3)!] = 5
When tossing 4 coins, there are a total of (2^4 = 16) possible outcomes. The combinations for getting exactly 2 heads (and consequently 2 tails) can be calculated using the binomial coefficient, which gives us (\binom{4}{2} = 6) ways. Therefore, the probability of getting exactly 2 heads is ( \frac{6}{16} = \frac{3}{8} ). Since getting 2 heads and 2 tails are the same event, the total probability of getting either 2 heads or 2 tails is also (\frac{3}{8}).
Pr(3 Heads) = 0.125 Pr(2 Heads and 1 Tail) = 0.375 Pr(1 Head and 2 Tails) = 0.375 Pr(3 Tails) = 0.125
3 coins can land in 8 different ways. Only one of these ways is all tails. So the probability of rolling at least one heads is 7/8 = 87.5% .
1/4
3 out of 6
It is (1/2)3 = 1/8
Each toss has 2 possible results.A series of 3 tosses has (2 x 2 x 2) = 8 possible overall scenarios.Three scenarios have exactly two tails: TTH, THT, and HTT.So the probability is 3/8 = 37.5% .
The outcomes are: heads, tails, tails or tails, heads, tails or tails, tails, heads. You can see that there are 3 possible outcomes with exactly 1 head.
It is 3/8.
P(4T1H)) = 5C4 ∙ (0.5)5 = 5/32 = 0.15625 ≈ 15.6% where 5C4 = 5!/[3!∙(5-3)!] = 5
The probability is 3/8 = 0.375
4C2(1/2)4 = 6/16 = 3/8
By tossing two coins the possible outcomes are:H & HH & TT & HT & TThus the probability of getting exactly 1 head is 2 out 4 or 50%. If the question was what is the probability of getting at least 1 head then the probability is 3 out of 4 or 75%
Pr(3 Heads) = 0.125 Pr(2 Heads and 1 Tail) = 0.375 Pr(1 Head and 2 Tails) = 0.375 Pr(3 Tails) = 0.125
3 coins can land in 8 different ways. Only one of these ways is all tails. So the probability of rolling at least one heads is 7/8 = 87.5% .