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Problem 1. $10$ identical bulbs, of which $6$ will be yellow. He randomly plants $6$ of the bulbs. Prob that exactly $4$ are yellow.

Isn't this binomial prob?

I did as follows $6C4 (.6)^4 (.4)^2$ to get $0.311$. Answer is $\frac 37$.

Problem 2. Bag of bulbs of which $40$% produce red tulips. Plant $15$ of them. Prob that $6$ will be red? I did similar: $15C6 (.4)^6 (.6)^9$ to get $20.7$% which I think is correct.

Are these the same type of problems, and why isn't Problem 1 working?

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    You can't pick the same bulb twice to be planted. Once one yellow bulb has been selected, the probability of picking another yellow bulb has decreased, it doesn't remain a constant 60%. The second question the probability is assumed to be *close* to 20.7% but the true value will depend on exactly how many bulbs are in the bag. We make the simplification in the second question to *assume* that there there is a *large enough* number of bulbs that by taking one it doesn't change the probabilities in a noticable way. – 2017-02-25
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    So, in problem 2 the draws are assumed to be close to independent, whereas in problem 1 not the case. correct? – 2017-02-25

1 Answers 1

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The first question does not follow binomial distribution as each draw is not identical.

The first part can be computed as follows:

$$\frac{\begin{pmatrix} 6 \\ 4\end{pmatrix}\begin{pmatrix} 4 \\ 2\end{pmatrix}}{\begin{pmatrix} 10 \\ 6\end{pmatrix}}=\frac37$$

where the numerator is the number of ways to have exactly $4$ yellows and the denominator is the total possible way of outcome when we choose $6$ bulbs.

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    Thank you for taking the time to answer. – 2017-02-25