A fair coin is tossed 5 times. What is the probability of obtaining exactly 4 heads.?

Consider :
Choose 4 times (₅C₄) from the 5 times of tossing to obtain 4 heads ([1/2]⁴), and another 1 time to obtain a tail (1/2).

The required probability
= ₅C₄ × (1/2)⁴ × (1/2)
= (5!/4!1!) × (1/16) × (1/2)
= 5 × (1/16) × (1/2)
= 5/32

Use binomial probability:
P(X = k) = C(n,k) * p^k * q^(n-k)

n : number of trials (5)
k : number of successes (4)
p : probability of success on each trial (1/2)
q : probability of failure on each trial (1 - p = 1/2)

P(X = 4) = C(5,4) * ½^4 * ½^1
P(X = 4) = C(5,4) * ½^5
P(X = 4) = 5 * 1/32
P(X = 4) = 5/32

Answer:
5/32

P.S. You can also think that there are 32 outcomes of the 5 tosses (2^5 = 32) and of these 5 of them (HHHHT, HHHTH, HHTHH, HTHHH, THHHH) will have an outcome with exactly 4 heads (and 1 tail).

Add your answer

(or exactly one tail :P)

Probability of getting one tail time the number of combinations of doing so...

5*(1/2)^5=5/32

5 possibilities for this:

THHHH
HTHHH
HHTHH
HHHTH
HHHHT

So the probability is ((1/2)^5)*5, or 5/32