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A tree has 20 fruits. 15 of which have no seeds and the rest do have seeds. A bird eats 5 of these fruits picked at random.

a) If i pick one fruit from whats left on the tree whats the probability it has seeds

b) Given that this one fruit i pick has seeds whats the probability that the bird had consumed at least one with seeds.

What i have so far:

So before the bird eats anything the probability of a fruit with no seeds is 15/20=0.75 and the probability of fruit with seeds is 5/20=0.25.

For part a is the probability that the fruit i pick(i.e. 6th fruit picked) has seeds is given by (15/20)* (14/19)* (13/18)* (12/17) *(11/16) *(5/15)=0.065

Am i on the right track with part a? As for part b im not sure where to begin.

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  • 2
    $\begingroup$ For part a, it doesn't really matter that the bird ate any fruit. The probability that the sixth fruit has seeds is the same as the probability that the first fruit has seeds. This is easier to confirm with calculations if the bird only eats one fruit, but the principle is the same. $\endgroup$ – Arthur Mar 15 '18 at 8:25
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    $\begingroup$ For part b, this is similar to asking whether with $19$ fruit, $4$ of which have seeds, the bird eats at least one with seeds (i.e. the bird does not eat $5$ seedless fruit) $\endgroup$ – Henry Mar 15 '18 at 8:34
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a)

The probability that you calculated is the probability that the bird only picked fruits without seeds and you picked one fruit with seeds. That is not the correct answer to the question.

The correct answer is $\frac5{20}$. Think of it like this: all fruits are placed randomly in a row. Now what is the probability that on spot number $6$ (corresponding to your pick after the $5$ picks of the bird) a fruit is placed that has seeds? It will not differ from the probability of any other spot to achieve a fruit that has seeds.

b)

Let $E$ denote the event that the bird consumed none with seeds and let $F$ denote the probability that you picked a fruit with seeds. Then: $$P(E\mid F)P(F)=P(E\cap F)$$where $P(F)=\frac5{20}$ (the probability calculated at a) and $P(E\cap F)=\frac{15}{20}\frac{14}{19}\frac{13}{18}\frac{12}{17}\frac{11}{16}\frac{5}{15}$ (the probability that you calculated in your answer).

This allows you to find $P(E\mid F)$ and also $P(E^{\complement}\mid F)=1-P(E\mid F)$

Note that $P(E^{\complement}\mid F)$ is the probability that you are looking for.

For a more elegant route see the comment of Henry.

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$(a)$

Notice that this is a hypergeometric random variable. Using the same variables as Matti

  • $P_0 = \frac{{5\choose0}{15\choose5}}{20\choose5}$
  • $P_1 = \frac{{5\choose1}{15\choose4}}{20\choose5}$
  • $P_2 = \frac{{5\choose2}{15\choose3}}{20\choose5}$
  • $P_3 = \frac{{5\choose3}{15\choose2}}{20\choose5}$
  • $P_4 = \frac{{5\choose4}{15\choose1}}{20\choose5}$
  • $P_5 = \frac{{5\choose5}{15\choose0}}{20\choose5}$

These calculations can be made in R statistical software:

> dhyper(0:5, m=5, n=15, k=5, log = FALSE)
[1] 1.936920e-01 4.402090e-01 2.934727e-01 6.772446e-02 4.837461e-03 6.449948e-05

Now given that $P_0,P_1,...,P_5$ happen the probabilities then of selecting a fruit with seeds are $\frac{5}{15},\frac{4}{15},\frac{3}{15},\frac{2}{15},\frac{1}{15}$, and $\frac{0}{15}$, respectively.

Then if we pick one fruit the probability that it has seeds is $$P_0\cdot\frac{5}{15}+P_1\cdot\frac{4}{15}+P_2\cdot\frac{3}{15}+P_3\cdot\frac{2}{15}+P_4\cdot\frac{1}{15}+P_5\cdot\frac{0}{15}=\frac{1}{4}$$

as Arthur alluded to. This can be calculated in R using

> sum(dhyper(0:5, m=5, n=15, k=5, log = FALSE)*(c(5/15,4/15,3/15,2/15,1/15,0)))
[1] 0.25

$(b)$

Let $P_6$ denote the event of getting a fruit with seeds. We wish to find the probability that $P_0$ did not occur. We have $$\begin{align*} P(P_0^C \mid P_6) &=\frac{P(P_0^C \cap P_6)}{P(P_6)}\\\\ &=\frac{P_1\cdot\frac{4}{15}+P_2\cdot\frac{3}{15}+P_3\cdot\frac{2}{15}+P_4\cdot\frac{1}{15}+P_5\cdot\frac{0}{15}}{P_0\cdot\frac{5}{15}+P_1\cdot\frac{4}{15}+P_2\cdot\frac{3}{15}+P_3\cdot\frac{2}{15}+P_4\cdot\frac{1}{15}+P_5\cdot\frac{0}{15}=\frac{1}{4}}\\\\ &=\frac{\frac{1}{4}-P_0\cdot\frac{5}{15}}{\frac{1}{4}}\\\\ &\approx 0.742 \end{align*}$$

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    $\begingroup$ Nice confirmation of intuition. $\endgroup$ – drhab Mar 15 '18 at 9:10
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Since there bird eats only five fruits, it's relatively easy to consider all possible cases. Let's write

  • $P_0 = P(\text{bird eats zero fruits with seeds}) = \frac{15}{20} \frac{14}{19} \frac{13}{18} \frac{12}{17} \frac{11}{16} = \frac{1001}{15504}$
  • $P_1 = P(\text{bird eats exactly one fruit with seeds}) =$ $5\choose 1$$\frac{5}{20} \frac{14}{19} \frac{13}{18}\frac{12}{17} \frac{11}{16} = \frac{5005}{15504}$
  • $P_2 = P(\text{bird eats exactly two fruits with seeds}) =$ $5\choose 2$$\frac{5}{20} \frac{4}{19} \frac{13}{18}\frac{12}{17} \frac{11}{16} = \frac{715}{3876}$
  • $P_3 = P(\text{bird eats exactly three fruits with seeds}) =$ $5\choose 3$$\frac{5}{20} \frac{4}{19} \frac{3}{18}\frac{12}{17} \frac{11}{16} = \frac{55}{1292}$
  • $P_4 = P(\text{bird eats exactly four fruits with seeds}) =$ $5\choose 4$$\frac{5}{20} \frac{4}{19} \frac{3}{18}\frac{2}{17} \frac{11}{16} = \frac{55}{15504}$
  • $P_5 = P(\text{bird eats exactly five fruits with seeds}) =$ $5\choose 5$$\frac{5}{20} \frac{4}{19} \frac{3}{18}\frac{2}{17} \frac{1}{16} = \frac{1}{15504}$

So for part a, you can consider the five cases separately.

  • If the bird ate zero fruits with seeds, there are $5$ out of $15$ fruits with seeds left. So you have to multiply $\frac{1001}{15504}\frac{5}{15}$
  • If the bird ate one fruit with seeds, there are $4$ out of $15$ fruits left with seeds. So you have to multiply $\frac{5005}{15504} \frac{4}{15}$

etc ...

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  • $\begingroup$ e.g. $P_1=\binom51\frac5{20}\frac{15}{19}\frac{14}{18}\frac{13}{17}\frac{12}{16}$ and there are more mistakes of that sort. $\endgroup$ – drhab Mar 15 '18 at 9:08

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