# Independence (Probability)

1. Oct 23, 2013

### mateomy

The question in my book is posed as:

Find an example in which $P(AB) < P(A)P(B)$.
The answer is given as a coin toss letting A be the event of obtaining heads and B being the even of obtaining tails. I'm confused as to how that's working.

My understanding is no doubt 'off', but I thought it was
$$P(\frac{1}{2} * \frac{1}{2}) = \frac{1}{2} * \frac{1}{2}$$
because the probabilities for both are $\frac{1}{2}$. So they are equivalent at $\frac{1}{4}$. Clearly I'm incorrect, can someone point out why?

Thanks.

2. Oct 23, 2013

### Dick

AB is shorthand for the cases where A is true AND B is true. Wouldn't it be true that the probability of getting heads AND tails is 0?

3. Oct 23, 2013

### mateomy

That clarifies everything. Thank you.