At a grocery store, eggs come in cartons that hold 12 eggs. Experience indicates that 78.5% of cartons have no broken eggs, 19.2% have one broken egg, 2.2% have two broken eggs, 0.1% have three broken eggs, and the percentage of cartons with four or more broken eggs is negligible. An egg selected at random from a carton is found to be broken. What is the probability that this egg is the only broken one in the carton?

The probability that an egg is the only broken one in the carton, given that it is broken, is determined as the percentage of cartons with exactly one broken egg, divided by the percentage of cartons with broken eggs.

19.2% of cartons have exactly one broken egg.

The percentage of cartons with broken eggs is:

[tex]B = 19.2\%+2.2\%+0.1\%\\B=21.5\%[/tex]

Therefore, the probability that this egg is the only broken one in the carton is:

[tex]P = \frac{19.2\%}{21.5\%}=0.8930 =89.30\%[/tex]

There is a 0.8930 or 89.30% that this is the only broken egg.

AFOKE88 AFOKE88 · Answer 2 · 2021-11-25T15:42:27+01:00

The probability that this egg is the only broken one in the carton is;

P(only one egg broken) = 0.8930 = 89.3%

We are given;

% of cartons that have no broken eggs = 78.5%

% of cartons that have one broken eggs = 19.2%

% of cartons that have two broken eggs = 2.2%

% of cartons that have three broken eggs = 0.1%

Total percentage of cartons that have broken eggs is;

T = 19.2% + 2.2% + 0.1%

T = 21.5%

Since the percentage that has only one broken egg is 19.2%, then probability that only one egg is broken in the selection is;

P(only one egg broken) = 19.2%/21.5%

P(only one egg broken) = 0.8930 = 89.3%

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