- #1
mateomy
- 307
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The question in my book is posed as:
Find an example in which [itex]P(AB) < P(A)P(B)[/itex].
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
[tex]
P(\frac{1}{2} * \frac{1}{2}) = \frac{1}{2} * \frac{1}{2}
[/tex]
because the probabilities for both are [itex]\frac{1}{2}[/itex]. So they are equivalent at [itex]\frac{1}{4}[/itex]. Clearly I'm incorrect, can someone point out why?
Thanks.
Find an example in which [itex]P(AB) < P(A)P(B)[/itex].
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
[tex]
P(\frac{1}{2} * \frac{1}{2}) = \frac{1}{2} * \frac{1}{2}
[/tex]
because the probabilities for both are [itex]\frac{1}{2}[/itex]. So they are equivalent at [itex]\frac{1}{4}[/itex]. Clearly I'm incorrect, can someone point out why?
Thanks.