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Problem:

A gardener has twelve identical looking tulip bulbs, of which eight will produce yellow tulips and four will become pink. She randomly selects and plants five of them and gives the rest away. When the flowers start to bloom, what is the probability that exactly three of them are yellow?

I first approached this problem using combinations: $\frac{\binom{8}{3}\binom{4}{2}}{\binom{12}{5}}=\frac{14}{33}$

Then, just for the heck of it, I attempted to solve the problem using binomial probability. I defined "success" to be planting a yellow flower and "failure" to be planting a pink flower.

$P(\text{success})=\frac{2}{3}$ and $P(\text{failure})=\frac{1}{3}$

$P(\text{3 successes in 5 trials}) = \binom{5}{3} (\frac{2}{3})^{3} (\frac{1}{3})^{2} = \frac{80}{243}$

I'm guessing one of the conditions for using the binomial distribution was violated?

  1. The number of observations n is fixed.
  2. Each observation is independent.
  3. Each observation represents one of two outcomes ("success" or "failure").
  4. The probability of "success" p is the same for each outcome.

None of these seem to be violated to me. Can someone clarify please?

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    $\begingroup$ Are you sure of independence? $\endgroup$
    – amd
    Aug 15, 2017 at 23:34

1 Answer 1

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You were correct to use the hypergeometric distribution to solve the problem since the flower bulbs are selected without replacement. The binomial distribution is used when selections are made with replacement. In particular, the probability of success must be the same for each outcome, which is not the case here since selecting a yellow flower changes the probability that the next flower you select is also yellow.

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