^{11}) = 0.3781. The 12 out front means that there are 12 ways this configuration can happen.

^{12}= 0.4186, and there is only one way for this to happen.

solve by complement I think. Its independent probability I think as well

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Hi Julia,

For each egg, there is a probability p = 0.07 that the egg is broken and a probability q = 1 - p = 0.93 that the egg is not broken.

The probability that at least two eggs are broken = 1 - (probability that all eggs except 1 are in tact and the probability that all eggs are fine).

The probability that all eggs are in tact except 1 = 12*(0.07)*(0.93^{11}) = 0.3781. The 12 out front means that there are 12 ways this configuration can happen.

The probability that all eggs are good = 0.93^{12} = 0.4186, and there is only one way for this to happen.

Therefore, your answer is P = 1 - 0.3781 - 0.4186 = 0.2033

If you did not calculate this answer this way, you would have a lot of complicated calculations of how many ways to have 2 broken eggs times the probability of this plus how many ways to have 3 broken eggs times the probability of that + ... + how many ways to have 10 broken eggs times the probability of that. It is far easier to get the same answer via the complement method.

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