Question

The probability that a dessert sold at a certain cafe contains chocolate is 86%. The probability that a dessert containing chocolate also contains nuts is 30%. Find the probability that a dessert chosen at random contains nuts given that it contains chocolate. Round to the nearest tenth of a percent.

Answer 1

N= total desserts
Choc N1= chocolate desserts= .86N
Choc desserts containing nuts= N2= .30 x .86 N as it will be thirty % if the choclate desserts
so probabilty= (choc+nuts desserts )/total desserts = (.30* .86*N)/N= .258= 25.8%

Answer 2

is that the answer??

Answer 3

i think so!

Answer 4

\[
\Pr(A|B) = \frac{\Pr(A\cap B)}{\Pr(B)}
\]

Answer 5

So let \(A\) be the event it contains nuts and let \(B\) be the event it's chocolate.

Answer 6

"The probability that a dessert sold at a certain cafe contains chocolate is 86%"
This means: \(\Pr(B)=0.86\)

"The probability that a dessert containing chocolate also contains nuts is 30%. "
This means \(\Pr(A\cap B)=0.3\)

Answer 7

"Find the probability that a dessert chosen at random contains nuts given that it contains chocolate."
This means: What is \(\Pr(A|B)\)

Answer 8

Or rather, the probability of event A (has nuts) if B (has chocolate) is true. It's conditional probability.

Answer 9

The probability that a dessert sold at a certain cafe contains chocolate is 86%.

The probability that a dessert containing chocolate also contains nuts is 30%.

Find the probability that a dessert chosen at random contains nuts given that it contains chocolate

P(nuts given chocolate) = .30/.86 = .349 or 34.9% (rounded to the nearest tenth

Answer 10

ur anwser would be 349 or 34.9%

Answer 11

your all nerds ha naw jp thanks though

Answer 12

lol ight brah