Imagine performing a coin-flip experiment in which the coin is "loaded'' in such a way that the probability of landing heads is twice greater than tails. After tossing the coin 10 times, what is the probability of observing two heads?
I can set it up like:
P(H).. probability of heads landing
P(T).. probability of tails landing.
P(H)= 2/3
P(T)= 1/3
It looks as if the heads can land twice in any order of the 10 times.
I'm just confused as to what to formula to use. :(
Good explanation would be helpful, thanks!
10 points to the best answer, of course.
I can set it up like:
P(H).. probability of heads landing
P(T).. probability of tails landing.
P(H)= 2/3
P(T)= 1/3
It looks as if the heads can land twice in any order of the 10 times.
I'm just confused as to what to formula to use. :(
Good explanation would be helpful, thanks!
10 points to the best answer, of course.
-
Two steps.
First step, how many ways can you select exactly 2 heads out of 10 tosses (Answer: C(10,2) as the order doesn't matter).
Second step, what is the probability of each of these patterns. (Answer: Each of 8 tails has probability 1/3, and each of the 2 heads has probability of 2/3. So the total probability for each pattern with 8 tails and 2 heads is (1/3)^8 * (2/3)^2
Putting these together, total probability is:
C(10,2) * (1/3)^8 * (2/3)^2
First step, how many ways can you select exactly 2 heads out of 10 tosses (Answer: C(10,2) as the order doesn't matter).
Second step, what is the probability of each of these patterns. (Answer: Each of 8 tails has probability 1/3, and each of the 2 heads has probability of 2/3. So the total probability for each pattern with 8 tails and 2 heads is (1/3)^8 * (2/3)^2
Putting these together, total probability is:
C(10,2) * (1/3)^8 * (2/3)^2