The following is probably simple, but somehow I don't see where my logic takes a wrong turn.
Let's assume that Sarah has lunch (L) today with 0.9. If (and only if) she has lunch, she will independently choose Avocado (A) with 0.9, Beef (B) with 0.9 and/or Celery (C) with 0.5 (multiple choices allowed). To me, this looks like follows:
$P (A \mid L) = 0.9$
$P (B \mid L) = 0.9$
$P (C \mid L) = 0.5$
The (a priori?) likelihood of her having lunch and the respective dish today would therefore be:
$P (A\cap L)= 0.9 \cdot 0.9 = 0.81$
$P (B\cap L)= 0.9 \cdot 0.9 = 0.81$
$P (C\cap L)= 0.9 \cdot 0.5 = 0.45$
Based on these results, how high is the probability that she will not eat any of these dishes today?
$(1 – 0.81) * (1 – 0.81) * (1 – 0.45) = 0.019855$
The opposite scenario, meaning that she will eat at least one of these dishes, could be:
$1 – 0.019855 = 0.980145$
Which is most likely wrong, as the probability of her having at least one of the dishes is higher than the probability of her having lunch (0.9).
I'd be glad if anyone could point me to the mistake in this calculation. Thanks in advance for any hint!
Best Answer
What is independent is her choice of the dishes given that she eats lunch, and not the general probability of getting each dish.
The formula is
$P = 0.1 + 0.9 \cdot((1 - 0.9)\cdot(1 - 0.9)\cdot( 1 - 0.5))$
This is the probability of not getting lunch, plus the probability of getting lunch and not choosing any of the three dishes.