The sample space is 23 or 8; which can be listed out as: HHH, HHT, HTH, HTT, THH, THT, TTH, TTT. There are 2 of the 8 that have exactly 2 heads; so the probability of exactly two coins landing on heads is 2/8 or 1/4.
The probability is 3/8 = 0.375
The probability is 90/216 = 5/12
Since a coin has two sides and it was tossed 5 times, there are 32 possible combinations of results. The probability of getting heads three times in 5 tries is 10/32. This is 5/16.
it is a fair chance so 1/2 :P
The probability is 0.5The probability is 0.5The probability is 0.5The probability is 0.5
The probability is 3/8 = 0.375
The probability is 90/216 = 5/12
3/8ths
3 out of 8
The answer depends on how many times the coin is tossed. The probability is zero if the coin is tossed only once! Making some assumptions and rewording your question as "If I toss a fair coin twice, what is the probability it comes up heads both times" then the probability of it being heads on any given toss is 0.5, and the probability of it being heads on both tosses is 0.5 x 0.5 = 0.25. If you toss it three times and want to know what the probability of it being heads exactly twice is, then the calculation is more complicated, but it comes out to 0.375.
6/16 = 3/8
3/8 * * * * * That is the probability of getting EXACTLY 1H. The prob of getting one (or more) head is 7/8
HHHH, HHHT, HHTT, HTTT, TTTT. Pr(HHTT) = 6/16 = 0.375
Since a coin has two sides and it was tossed 5 times, there are 32 possible combinations of results. The probability of getting heads three times in 5 tries is 10/32. This is 5/16.
The probability is 1/16.
The required probability is 1/8.
1/4 if they are tossed only once.