toss a fair coin 8 times. What is the probability of each event?

The Binomial distribution has the probability

function P(x=r)= nCr p^r (1-p)^(n-r)

r=0,1,2,.....,n

where nCr = n! / r! (n-r)!

1)

n=8 (number of tosses)

p=1/2 (probability of heads in a single toss)

r= 3

P(r=3) = 8C3 (1/2)^3 (1/2)^5 = 0.218750

2)

P( at least 1 head) = 1-P(no heads)

=1-8C0 (1/2)^0(1/2)^8 = 1-(0.5)^8 = 0.996094

3)

P( more than 8 heads)=0 (impossible event)

without writing the pattern area, P(E) = 4C3 * (a million/2)^3 * (a million - a million/2)^a million (binomial distribution (in case you arent acquainted with it, it says that the prospect of success is given by using nCx * p^x * (a million - p)^(n - x) the place n is the form of trials (for that reason 4), p is the prospect of success (for that reason a million/2), and x is the form of successes (for that reason 3)) = a million/4 P(F) = a million/2 * a million/2 because of the fact there's a a million/2 probability that the 1st toss is heads and that the 2nd toss is heads. w/e occurs afterwards would not influence that danger of this P(E or F) = P(E) + P(F) - P(E and F) = a million/4 + a million/4 - P(E and F) P(E and F) = P(F) * P(E|F) P(E|F) = danger of having precisely one head = 2C1 * (a million/2)^a million * (a million/2)^a million (binomial distribution) = a million/2 so.. P(E and F) = a million/4 * a million/2 = a million/8 P(E or F) = a million/4 + a million/4 - a million/8 = 3/8 All this math exchange into probable pointless and writing out each and all of the feasible combos could be greater undemanding yet it incredibly is basically because of the fact there have been 4 coin tosses. Had the form of coin tosses been greater advantageous, writing out the feasible combos could take lots longer.

Watch out! You are in danger of falling into The Gambler's Fallacy!

With a coin toss, the probability (if the coin is perfectly balanced) is 50/50. If you toss it 100 times, the probability of each toss is 50/50, NO MATTER WHAT HAPPENED ALL THE OTHER TIMES.

In other words, if you toss 72 heads in a row, what is the probability of getting heads on the 73rd toss? It is STILL 50/50.

Each toss is completely independent of any other toss. The coin does not remember the result of the last 72 tosses. The universe does not somehow keep count and make sure that every 100 tosses results in 50 heads.

Now, the probability of getting 72 heads in a row is another matter entirely, because you are looking at a different event - 72 in a row. That's completely different from each individual coin toss.

Now, look at your question. It's asking about the probability of three different events, rather like the 72 heads in a row. You need to calculate this based on an expected distribution of 4 heads.

Of course, the probability for #3 is ZERO, for obvious reasons!

If a fair coin is tossed 4 times, what is the probability of NOT obtaining exactly 3 heads?

1 = 1/2*1/2*1*2*1*1*1*1

= 1/8

2 = 1/2*1*1*1*1*1*1*1*1

= 1/2

3 = 0