4
$\begingroup$

Suppose there is $1$% of cloves with four leaves. We pick $100$ cloves. Let $X$ denote the event a clove has four leaves. What's the probability of having 4 cloves?

I am wondering which method should I use?

1st possibility: The first such clove has a probability of being picked of $\frac{1}{100}$, the second $\frac{3}{99}$ the third $\frac{2}{98}$ and the last one $\frac{1}{97}$. By multiplying these, I could get the final probability of getting $4$ such cloves. Is this correct?

2nd possibility: My second thought would have been to use the Binomial distribution, although I am unsure I can use it, as we don't repeat the experience several times, we just pick $100$ cloves and see whether it has $4$ four-leave cloves.

Any help will be appreciated.

  • 1
    This is not clear. "There is $1$ of cloves with four leaves"...out of how many? In the entire world? If $X$ is the event "a clove has four leaves" then $X$ is either "yes" or "no", it can't be $3$. – 2017-01-21
  • 0
    Yes you should mention how many cloves have 4 leaves. – 2017-01-21
  • 0
    Can you clarify your question? If not, I think it should be closed. – 2017-01-21
  • 0
    My apologies, LATEX didn't show the '%' sign. – 2017-01-21
  • 0
    And $X$ is the number of 4-leaved clovers you draw? Do you want $3$ or $4$? and do you want "exact" numbers or "at least" numbers...that is, suppose $10$ of your clovers have four leaves...is that a success or a failure? – 2017-01-21
  • 0
    X is the number of 4-leaved clovers. I want 4 of them, and only 4. 10 of them would be a failure. – 2017-01-21

2 Answers 2

2

Required probability $=\dbinom{100}{4}\left(\dfrac{1}{100}\right)^4\left(\dfrac{99}{100}\right)^{96}$

  • 0
    So you're using the Binomial distribution. Could you justify why we are allowed to use it here? I had a feeling I couldn't use it here, but have difficulties telling why or why not I could do it. – 2017-01-21
  • 0
    see https://en.wikipedia.org/wiki/Binomial_distribution. The binomial distribution is frequently used to model the number of successes in a sample of size n drawn with replacement from a population of size N – 2017-01-21
  • 0
    But isn't the Binomial distribution used when there is "no memory" of previous outcomes? – 2017-01-21
  • 0
    I have taken your question as '1% chance of success from any sample' and hence it is independent of previous outcomes. This is my view and like to see how others interpret this. – 2017-01-21
1

You can use binomial distribution.

p (% of not getting clove with leaves) = 99% = $\frac{99}{100}$

q (% of getting clove with leaves) = 1% = $\frac{1}{100}$

Probability = $\binom{100}{4} \left(\frac{1}{100} \right)^4 \left(\frac{99}{100} \right)^{96}$

  • 0
    As $\binom{100}{4}$ arranges the cases. – 2017-01-21