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.
It is 15/16.
1/4
Probability of no heads = (0.5)^5 = 0.03125Probability of at least one head = 1 - probability of no heads = 1 - 0.03125 = 0.96875
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
The probability that exactly one will land heads up is 0.15625
It is 15/16.
1/4
Probability of no heads = (0.5)^5 = 0.03125Probability of at least one head = 1 - probability of no heads = 1 - 0.03125 = 0.96875
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
The probability that exactly one will land heads up is 0.15625
6/16 or 3/86/16 or 3/86/16 or 3/86/16 or 3/8
The probability of tossing two coins that are different is 1 in 2, or 0.5.The probability of tossing something on the first coin is 1. The probability of not matching that on the second coin is 0.5. Multiply 1 and 0.5 together, and you get 0.5.
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.
the probability is 0.03125 or 3.123%
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 %
25%
7/8