Consider a biased coin where P ({H}) = 1/3 and P ({T}) = 2/3. We toss this biased coin 3 times and we observe the sequence of the heads and the tails. a) Let A be the event that heads appear only in the third toss. Find P (A). b) Let B be the event that heads appear in the third toss. Find P (B). c) Let C be the event that heads appear two times. Find P (C).

a) 4/27

b) 1/3

c) 6/27

Step-by-step explanation:

HI!

a)

The event A has only one member:

T T H

Therefore:

P (A) = P (T) P (T) P (H) = (2/3) (2/3) (1/3) = 4/27

b)

For event B, we are only interested in the last toss

x x H

Since the first two tosses are not important for event B they do not contribute to P (B)

P (B) = 1 * 1 * P (H) = 1/3

c)

The following are elements of C:

H H T

H T H

T H H

Therefore:

P (C) = P (HHT) + P (HTH) + P (THH)

We can easily see that the probability of the three members is the sam:

P (HHT) = (1/3) (1/3) (2/3) = 2/27

P (HTH) = (1/3) (2/3) (1/3) = 2/27

P (THH) = (2/3) (1/3) (2/3) = 2/27

Therefore:

P (C) = 3 P (HHT) = 6/27