3bc is the answer to your problem
P(most favorable outcome) = 1 -(0.03 +0.16 -0.01) = 0.82
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"repair fails" includes the "infection and failure" case, as does "infection". By adding the probability of "repair fails" and "infection", we count the "infection and failure" case twice. So, we have to subtract the probability of "infection and failure" from the sum of "repaire fails" and "infection" in order to count each bad outcome only once.
The probability of a good outcome is the complement of the probability of a bad outcome.
Answer:
The dog will eat 3 lbs
Step-by-step explanation:
Take the amount eaten per day and multiply by the number of days
3/4 * 4 = 3
The dog will eat 3 lbs
