+2236 Here’s the math portion of the problem:
\(P(B)=0.3 \tiny \text{ Read as “The probability that a child (from the sample set) likes (B)rownies" }\\ P(C)=0.5 \tiny \text{ The probability that a child likes (C)upcakes}\\ P(B\cap C)=0.15 \tiny \text{ (The intersection of these probabilities )}\\ \\ P(B|C)=\dfrac{P(B\cap C)}{P(C)}=\dfrac{0.15}{0.5}=0.3 \tiny \text { Read as “The probability that a child likes brownies, given that s/he likes cupcakes.}\\ P(C|B)=\dfrac{P(B\cap C)}{P(B)}=\dfrac{0.15}{0.3}=0.5 \tiny \text { The probability that a child likes cupcakes, given that s/he likes brownies.}\\ \small \text {Note that }P(B|C)=P(B) \small \text { and }P(C|B)=P(C)\\ \)
With this, you can logically determine which statement is true.
Post your response on here, clearly explaining your logic.
If you do not respond, I will not answer any more of your questions.
Feeding the bears is causing an educational and social disturbance on this forum. You are one of the panhandling bears on this forum. Your response will help to mitigate the concerns. If you continue to respond (to other questions), you will become a participant in your own education and you will then be a domesticated bear. Domesticated bears are well tolerated here, and they learn as they are fed. Doing this will greatly augment and enhance your education, and you will actually learn the mathematics, instead of just passing the class.
GA
