Question

7. A dog watches as a human places 4 cups upside down in a row on a table. The human places a dog treat under one of the cups as the dog watches, and then the human quickly rearranges the cups. The dog then chooses one of the cups; if the chosen cup has the treat underneath it, the dog gets the treat. (In case the setup for this problem is unclear - and also just for fun - here's a video: https://www.youtube.com/watch?v=c9lbKeBe9CI) Let's assume our dog is not as smart as the one in the video, and only chooses at random. He gets to play 4 times. Let X denote the total number of treats our dog, guessing at random, gets to eat over the 4 times he plays the game. Assume each game is independent (i.e. he doesn't get any better at the game) (a) Create the pmf of X (round to 4 decimal places). Hint: The probability the dog chooses correctly by guessing is 0.25 and the probability of choosing wrong is 0.75. Assuming each game is independent of the others, find how many ways you can get X to be 0, 1, 2, 3, and 4 b) What is the probability that our dog gets less than 2 treats? (c) Find the probability that our dog gets to eat at least 3 treats d) What is the expected number of treats our dog gets to eat?

Answer 1