Probability with Combinations (Coin Toss) Toss a fair coin six times. What is probability of :


*

*a- all heads

*b- one head

*c- two heads

*d- an even number of heads

*e- at least four heads.


I have the answers so I ask if you could explain how you should derive rather than just giving me the answers please.
I know for a, you just multiple 1/2 6 times.
But for the rest I need your help please.
 A: Do you know Pascal's triangle?  The line 1 6 15 20 15 6 1 (just below the bottom one shown in the article) is the one you want.  For two heads, you can choose the first head in 6 ways, then the second in 5, but you have counted each possibility twice (two different orders).  Thus there are 6*5/2=15 ways to get two heads and the probability is $\frac{15}{2^6}$.  For e you just add the chance of 4, 5, and 6.
A: There are $2^6$ possible outcomes, one for each sequences of six heads-or-tails.
a. How many sequences of six tosses have all heads? One. Since there is one good outcome and $2^6$ possible outcomes, then the probability is
$$\frac{1}{2^6}.$$
b. How many sequences of six tosses have exactly one head? Six (pick which toss is heads). So the probability of getting exactly one head is...
c. How many possible sequences have exactly two heads? Pick the two tosses that are heads; then you get...
d. An even number of heads means either no heads, two heads, four heads, or six heads. There is one sequence with no heads, one sequences with six heads, then xxx sequences with two heads, and yyy sequences with four heads. So the total number of "good" outcomes is $1+1+$xxx$+$yyy. So the probability is...
e.  At least four heads... that means either no tails, one tail, or two tails. You already figured each of those in (a), (b), and (c) (well, for heads, but you can see that it's the same answer if you want to know the probability of "exactly one head" and the probability of "exactly one tail", right?). So...
