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# Probability problem

Sat 25 Nov, 2017 09:01 am
We have 12 eggs. 8 of these eggs are boiled and 4 are not boiled. George chooses 2 of these eggs for his breakfast.
a.) What is the probability that the 2 chosen eggs are boiled?
b.) If George broke the 1 of the 2 eggs and it is boiled, what is the probability that the other egg is boiled?
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Type: Question • Score: 1 • Views: 862 • Replies: 5

fresco

2
Sat 25 Nov, 2017 01:47 pm
@clevergamer,
I suggest
(a) 56/132
(b) 7/11
but am open to correction.
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clevergamer

1
Sun 26 Nov, 2017 08:22 am
@fresco
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engineer

3
Sun 26 Nov, 2017 08:48 am
@clevergamer,
a) There are 66 combinations of 2 eggs, 12x11/2. There are 28 combinations of two eggs where both are boiled. 8x7/2. 28/66.

b) Two ways to do this depending on your geek level. First you could say that if the first egg is boiled, there are 11 eggs left, seven of which are boiled, therefore 7/11. The second way is Bayes Theorem. The probability of both boiled is 28/66, the probability of the first being boiled is 8/12, so find 28/66 divided by 8/12 = 7/11
clevergamer

1
Sun 26 Nov, 2017 09:56 am
@engineer,
Thank you! I think that's the right solution.
fresco

2
Sun 26 Nov, 2017 10:20 am
@clevergamer,
An ordinary tree diagram works with two branches for first egg, each with two branches for the second egg, probabilities being found by multiplication along the branches.
Pedagogically, it would be a poor question because the answer to b is required for the answer to a.
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