Question

Larry reads that 1 out of 4 eggs contains salmonella bacteria, so he never uses more than 3 eggs in cooking. If eggs do or don't contain salmonella independent of each other, find the probability that at least 1 of Larry's 3 eggs contains salmonella.

Answer 1

Answer:

57.8125% or approx. 57.8%

Step-by-step explanation:

There is a 1/4, or 25%, or 0.25 chance that an egg has salmonella.

Thus, there is a 75%, or 0.75 chance that an egg DOESN'T contain salmonella.

Let's find the probability that all 3 of Larry's eggs are free from salmonella. Larry would have to hit that 75% chance 3 times in a row. The chance of that happening is:

0.75 * 0.75 * 0.75 = $0.75^3$ = 0.421875

From this, we can deduce that if there is a 0.421875 (42.1875%) chance that all eggs are safe to eat, there must be a...

1 - 0.421875 = 0.578125

...0.578125 (57.8125%) chance that 1 or more of Larry's eggs do have salmonella.

Answer: approx. 57.8% or 57.8125%