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

17.08

correct me if i am wrong.