Every day Aaron, Bebe, and Cindy eat a piece of fruit for lunch. Aaron will eat an apple, banana, peach or orange with equal likelihood. Bebe will eat an apple, banana, or mango with 50% likelihood for an apple, 30% for a banana, and 20% for a mango. Cindy will eat either an apple or banana with equal likelihood. What is the probability that they all eat the same kind of fruit for lunch today?

so you are looking for:

(A apple * B apple * C apple) + (A banana + B banana + C banana)

this is the probability that all of them will eat an apple plus the probability that all of them will eat a banana.

Let me know your answer and I'll help you further if you're wrong.

Quote:

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Originally Posted by Capuchin

so you are looking for:

(A apple * B apple * C apple) + (A banana + B banana + C banana)

this is the probability that all of them will eat an apple plus the probability that all of them will eat a banana.

Let me know your answer and i'll help you further if you're wrong.

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I'm not sure of the answer and I am having extreme difficulties trying to get an answer. Please Help?

I told you how to work it out, you just need to put in the numbers from your question!

(.50*.30*.20) + (1+1+1) = .03 +3 = 3.03
Is this the answer?

Quote:

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Originally Posted by Peachs

(.50*.30*.20) + (1+1+1) = .03 +3 = 3.03
Is this the answer?

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Obviously not. How can you have a probability greater than 1?

Reread your notes, then have another go, I've done the hard bit for you.