The probability that a dessert sold at a certain café contains chocolate is 74%. The probability that a dessert contains both chocolate and nuts is 24%. Find the probability that a randomly chosen dessert contains nuts.
@rational
what happens when we add to things together, but they have parts in common?
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I'm not sure...
A+B = (1+2) + (2+3) ^^ ^^ the common parts get added a multiple of times so we have to adjust for that in this case: A+B = A+B-AnB
oops... I asked the question wrong... it's what is the probability that a randomly chosen chocolate dessert contains nuts
and what you just said confused me...
well i do see where i could have been clearer but im not real sure what your confusion is. total = 1+2+3 but of we just add A and B, then we are adding a part more than once.
but i misread your question and thought it asked what the probability was of only nuts
if 74% is chocolate only, then the rest must have nuts
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C = y+24 = 74 N = x+24 100 = C+N-24
I'm so confused...
you need probability that a randomly chosen chocolate dessert contains nuts, this means you need probability that a randomly chosen contains nuts given that it contains chocolate
A: the dessert contain nuts B: the dessert contain chocolate
probability that a dessert contains both chocolate and nuts is P(A and B) probability that a dessert contains chocolate is P(B)
now use the formula to get P(A|B)
since it is already known that the picked dessert is chocolate, we only look at chocolate in the venn diagram : 1) 24% in favor of nuts 2) total chocolates probability 74% take the ratio for the probability that a chocolate dessert contains nuts
ohhhh
24/74... so 32.4%?
you may also conditional probability formula for practice..
i see, i was stuck onthe question as it was first asked ...
yea, sorry, my bad..
'sok :)
its hard for these old eyes to keep up with the lag freezing my screen is all
okay.
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