It is 15/16.
Probability of no heads = (0.5)^5 = 0.03125Probability of at least one head = 1 - probability of no heads = 1 - 0.03125 = 0.96875
7/8
Probability not at least 1 head showing is when all 5 coins are tails: (1/2)5=1/32 Therefore probability at least 1 head is showing is 1-1/32=31/32
1/4
It is 15/16.
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%
Probability of no heads = (0.5)^5 = 0.03125Probability of at least one head = 1 - probability of no heads = 1 - 0.03125 = 0.96875
7/8
Probability not at least 1 head showing is when all 5 coins are tails: (1/2)5=1/32 Therefore probability at least 1 head is showing is 1-1/32=31/32
1/4
No matter how many coins are thrown, the possibility of having AT LEAST ONE 'head' is 50%. This changes if you specify the number of 'heads' that must be shown.
The probability is 0.998
This is easiest to solve by working out the probability that no heads show and subtracting this from 1 to give the probability that at least one head shows: Assuming unbiased coins which won't land and stay on their edge, the probability of head = probability of tail = ½ → probability no heads = probability 5 tails = ½^5 = 1/32 → probability of at least one head = 1 - 1/32 = 31/32 = 0.96875 = 96.875 % = 96 7/8 %
Assuming a two-sided coin, and that you make the the toss, the probability of tossing a head or a tail is 100%. The probability of tossing a head is 50%. The probability of tossing a tail is 50%.
There are 2 coins, 1 and 2. Each has two possibilities, H or T. The possibilities are: 1H, 2H 1H, 2T 1T, 2H 1T, 2T Each possibility has an equal chance of happening. The chance of tossing at least one head is 3/4.
Is possible.