Understanding Independence in Probability: An Example

mateomy
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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
<br /> P(\frac{1}{2} * \frac{1}{2}) = \frac{1}{2} * \frac{1}{2}<br />
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.
 
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mateomy said:
The question in my book is posed as:

Find an example in which P(AB) &lt; 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
<br /> P(\frac{1}{2} * \frac{1}{2}) = \frac{1}{2} * \frac{1}{2}<br />
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.

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?
 
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That clarifies everything. Thank you.
 
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