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A rose breeder is trying to cross yellow and red roses. He finds, on average, that $1$ in every $4$ of his test seeds produce a yellow rose. If he plants $12$ seeds, determine the prob that:

  1. Exactly $2$ roses are purely yellow.
  2. Less than $3$ of the $12$ are purely yellow.

My attempt

  1. $n=2 p=0.25$, ${12\choose2}= 66$ so $66 * .25^{12} * .25^2$. [Different answer to the solution paper, which doesnt give the working.]
  2. Since it is less than $3$, I would combine the prob of ($2$ of $12$) and ($1$ of $12$)?
mathochist
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StuM
  • 45

2 Answers2

1

The Probability Mass Function for a Binomial Distribution is: $\mathsf P(X=k) = {^n{\sf C}_k}\, p^k\, (1-p)^{n-k}$

Where $p^k\, (1-p)^{n-k}$ is the probability of $k$ yellow roses and $n-k$ not yellow roses in a certain arrangement, and ${^{n}{\sf C}_{k}}$ is the count of distinct permutations of that arrangement.

Here, $n=12, p=0.25, k=2$ so $$\mathsf P(X=2)={^{12}{\sf C}_2}\, 0.25^{2}\, \color{red}{\underline{\color{black}{0.75^{10}}}} = \frac{1948617}{8388608}$$

Now find $\mathsf P(X<3) = \mathsf P(X=0)+\mathsf P(X=1)+\mathsf P(X=2)$

Graham Kemp
  • 129,094
1

For 1. you are almost correct. Instead of $0.25^{12}$ you should write $(1-0.25)^{12-2}$.

For 2. you should add (instead of combine) these probabilities, but do not forget that you also need to add the probability that there are $0$ (not only $1$ or $2$) yellow roses.

Jimmy R.
  • 35,868