Usually, it takes 2-5 flips until a head appears.
2/6=1/3
A fair coin means that the probability of a head = probability of a tail = 1/2 So you would expect half the tosses to be heads, ie 1/2 x 75 = 371/2 heads. ...oooOOOooo... Having 1/2 a head doesn't seem possible, but when the question asks about expectation, it is saying: if you repeated the experiment lots of times, how often, on average, would the required result appear. So the expectation of heads when a fair coin is tossed 75 times is asking: if a fair coin was repeatedly tossed 75 times, what would be the (mean) average number of heads achieved? As more and more trials are done and the (mean) average of the number of heads got is taken, it will get closer and closer to 371/2 37 or 38 times. (Obviously, you can't have half of a time.) You will either get one or the other, and a fair coin means that either is just as likely. So, it should split evenly down the middle.
The probability of rolling a number greater than 1 is 5/6.
mean = 5, variance = 5
n(S)=8 let A be the event that more than one tail appears n(A)=4 so,P(A)=4\8=0.5
A coin is tossed until ahead appears. What is the expectation of the number of tosses required?
The probability is 90/216 = 5/12
The number of combinations - not to be confused with the number of permutations - is 2*21 = 42.
7878
48
There are 72 permutations of two dice and one coin.
The answer in the back of the book says that it is: TT, TH, H1, H2 H3, H4, H5, H6= 8 possibilities. But I don't understand what that means or how the answer was found. Help?
A homophone for "tossed" is "tost."
The number of sequences is 27 or 128.
tossed
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 suffix for tossed is ed