This is the same as example 2 in the link below
The probability of tossing two heads in two coins is 0.25.
HHHH, HHHT, HHTT, HTTT, TTTT. Pr(HHTT) = 6/16 = 0.375
The probability that both coins are heads is the probability of one coin landing heads multiplied by the probability of the second coin landing heads: (.5) * (.5) = .25 or (1/2) * (1/2) = 1/4
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
The probability you'd get heads is still one half.
possible
3/8
When we toss a coin getting head or tail have equal probability of 50% - that is, out of the two possible outcomes getting the specified one becomes 1/2 probability. When we toss three coins, the probability of getting all the coins showing tails is given by (1/2) * (1/2) * (1/2) equal to 1/8 or 12.5 % chance. Alikban
If both tosses are fair, the probability of that outcome is one in four.
2 out of 8
1/2
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%
If they are fair coins, the probability is 0.25
Possible outcomes are HH, HT, TH, TT; therefore the probability of HH is 1/4 or 0.25.
The probability would be once in 128 attempts. You don't have to toss seven coins simultaneously. the 7 tosses just have to be independent of one another.
It is 0.375It is 0.375It is 0.375It is 0.375
The answer to the first question is 0.5. The answer to the second is not possible to work out.