Probability, flipping a rigged coin problem

[danielmcg90]

A coin is loaded so that the probabilities of heads is .52 and the probability of tails is .48.
If the coin is tossed 3 times what is probabilities that

a) all heads

b) 2 tails and a head in that order

I figured out part a). So i just thought of it with an and statement. ( I want heads and heads and heads) since the previous flip has no bearing on the next flip i can just multiply the quantities,

so P(all heads) = .52(.52)(.52) = .1406 which is the right answer

For part b) i am confused some how i have to implement that order is going to matter so I can just multiply them and that's it.
But still each flip of the coin has no bearing on the other flips so..
Can someone explain? the answer in the book is .1198

 

[Dick]

Why do you think you can't just multiply them like in the first case? You have three events occurring in a definite order with a definite probability for each.