I have 3 trees.

  • These particular trees are dioecious (male or female).
  • I don't know the gender of any of the trees.
  • The chance of a tree being male or female is 50/50.
  • I need at least one male and one female for successful pollination to occur.

What are the chances that I have at least one male and one female tree?

  • 1
    $\begingroup$ Another Hint: Binomial Distribution $\endgroup$ – Graham Kemp Apr 30 '18 at 22:48

HINT: Let $M$ denotes male tree, $F$ — female tree. The probability is $$ 1-P(MMM)-P(FFF). $$ But, from independency, $P(MMM)=P(M)P(M)P(M)=(1/2)^3$.

BTW: What are female trees? Have they a hollow? ;-)

  • 2
    $\begingroup$ Some species are dioecious; each tree having either male or female flowers. Eg: Willow, Juniper, Teak... $\endgroup$ – Graham Kemp Apr 30 '18 at 22:57
  • $\begingroup$ @GrahamKemp Thank you. I knew such other plants, but not trees. $\endgroup$ – Przemysław Scherwentke Apr 30 '18 at 23:02
  • $\begingroup$ @Wilson Please see my extended hint. Can you finish now? $\endgroup$ – Przemysław Scherwentke May 1 '18 at 16:49
  • $\begingroup$ @Wilson Oh, sorry! Misprint. Corrected. So $1-(1/8)-(1/8)=3/4$. $\endgroup$ – Przemysław Scherwentke May 1 '18 at 17:09

This is a binomial distribution with $n=3, p=\frac12$. Let $X$ shows the number of male trees, then: $$P(X=1)+P(X=2)={3 \choose 1}\left(\frac12\right)^1\left(\frac12\right)^{3-1}+{3 \choose 2}\left(\frac12\right)^2\left(\frac12\right)^{3-2}=\\ 3\cdot \frac12\cdot \frac14+3\cdot \frac14\cdot \frac12=3\cdot \frac14=\frac34.$$


Here is my layman's interpretation of @PrzemysławScherwentke's answer:

1 - (1/2 * 1/2 * 1/2) - (1/2 * 1/2 * 1/2) = .75


Here are all of the possible combinations that I can come up with:


There are a total of 8 possible combinations. Of the 8, only 6 of them have at least one male and one female.

Therefore, I think the probability of getting at least one male and one female is 6 / 8 = 75%.


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