Knowing the results in advance, there is a 7/10 chance that the first toss is heads.
In this case "seven out of ten" is quite literal. The first toss is one of the 10, and 7 of them came up heads, so 7/10 is the chance that this particular one is one of the heads.
7/10
The answer depends on how many coins were tossed.
Assuming that it is a fair coin, the probability is 0.9990
To find the probability of getting at least 2 heads in 5 tosses of a fair coin, we can use the binomial probability formula. First, we calculate the probabilities of getting 0 and 1 head, then subtract these from 1. The probability of getting at least 2 heads is 1 minus the sum of the probabilities of getting 0 heads (1/32) and 1 head (5/32). Thus, the probability of getting at least 2 heads is (1 - (1/32 + 5/32) = 26/32 = 13/16) or 0.8125.
1/6
It is 0.3125
1/2
7/10
There is a 50% chance that it will land on heads each toss. You need to clarify the question: do you mean what is the probability that it will land on heads at least once, exactly once, all five times?
well, it will have 6 times of the greater chance.
Prob(Exactly 9 Heads) = 0.0269 Prob(At least 2 Heads) = 0.0327 Prob(At most 8 Heads) = 0.9673
Prob(at least 4 heads) = Prob(4 heads) + Prob(5 heads) = 5/32 + 1/32 = 6/32 = 3/16
you have 63 chances out of 64. i once witnessed a coin being tossed seven times and giving up 7 consecutive heads. we never tried it an eighth time, 7 heads and you had to go to the bar.
The probability of tossing heads on all of the first six tosses of a fair coin is 0.56, or 0.015625. The probability of tossing heads on at least one of the first six tosses of a fair coin is 1 - 0.56, or 0.984375.
The answer depends on how many coins were tossed.
It is 0.75
Assuming that it is a fair coin, the probability is 0.9990