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im working on the following problem:


Problem 3.

The probability that a seed of a certain sort will become a plant is 0.8, while the probability that a seed from another sort will become a plant is 0.9.

a) If you sow 5 seeds of each sort (10 in total), what is the probability that all seeds will become plants? You can assume independence between plants. Why is this not a binomial situation?


The problem is a part of an assignment in course in mathematical statistics.

For me it seems pretty obvious that each plant should have its binomial distribution, and to find the probability that all the seeds of each sort becomes a plant is the product of the individual probabilities that all of sort 1 and sort 2 become plants due to indepence.

What makes me so confused is the last question, why this should not be a binomial situation.

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Binomial distribution means that you conduct a sequence of Bernoulli trials (i.e experiments with two possible outcomes) which all have the SAME probability of success. In your 10 trials, you have two different probabilities of success. But you are right with the probability: $P = 0.8^5 \cdot 0.9^5$.

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  • $\begingroup$ Im sorry but i still dont understand that question. I have tried to find a solution for over a week, and i have asked other places and be given the same explanation. Do you think that taken togheter, they should be a trinomial distribution? $\endgroup$ – user390653 Nov 18 '16 at 18:54
  • $\begingroup$ Solution for what? You do already have the solution. And, as stated, it is NOT a binomial distribution just because it does not meet the definition of it. It is neither a multinomial distribution, where you could have more than 2 possible outcomes in each Bernoulli trial. But again, these probabilities must be the same in all trials. $\endgroup$ – UweM. Nov 21 '16 at 8:35

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