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Q:
In the grocery store, there are 2 kinds of egg cartons (12 eggs each) .
$80\%$ of the cartons have exactly $1$ broken egg. In the other $20\%$ there are exactly $3$ broken eggs. You want to buy a carton that include exactly one broken egg, but you can't examine each egg (because you have no time ) you only check $3$ eggs for each carton.
If you find one or more broken eggs in the $3$ that you examine, you return the carton back, else you take it.

How many cartons do you need to return until you will find the carton you want?

My go:
This sounds like a Geometric distribution because you search and search until you find something (and then stop) so basically $X \sim G(p)$ The chance of finding the right carton is $0.8 \cdot \frac{1}{12} + 0.2 \cdot \frac{9}{12} \cdot \frac{8}{11} \cdot \frac{7}{10} \approx 0.143$
And so the mean is $E[X] = \frac{1}{p} = \frac{1}{0.143} \approx 6.9915 \approx 7$ so $7$ cartons, so you will return $6$ Am I right?


Another question:

Same exact story as described above.
You would like to take $4$ cartons, what is the probability you will need more than $7$ cartons in order to get $4$ that satisfy your rules? (that you examine 3 random eggs and if you find $\geq$ 1 eggs you return it and search another one) ?

My go:

I did not quite understood how to approach it using distributions, it sounds like HyperGeometric however I am not sure! The tip for the question is:

Define $X_3$ to be the number of cartons you return until you get to $4$ cartons

Thank you for helping!

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  • $\begingroup$ "If you find one or more broken eggs in the 3 ..." Don´t you find always at least one broken egg in a carton? $\endgroup$ May 20 '20 at 17:32
  • $\begingroup$ @callculus you examine 3 eggs per carton , you can find 3 eggs that are completely fine $\endgroup$
    – CSch of x
    May 20 '20 at 18:47
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Let $X$ be a random variable indicating the number of broken eggs in a randomly selected carton, so $X = 1$ if it has one broken egg, and $X = 3$ if it has three. Then $$\frac{X - 1}{2} \sim \operatorname{Bernoulli}(p = 0.2).$$ Let $$Y \mid X \sim \operatorname{Hypergeometric}(N = 12, n = X, m = 3), \\ \Pr[Y = y \mid X = x] = \frac{\binom{x}{y} \binom{12-x}{3-y}}{\binom{12}{3}}$$ represent the random number of broken eggs observed when a sample of $m = 3$ eggs are taken without replacement. Then we want to compute the unconditional probability of acceptance $p$, i.e., $$\begin{align*} p = \Pr[Y = 0] \\ &= \Pr[Y = 0 \mid X = 1]\Pr[X = 1] + \Pr[Y = 0 \mid X = 3]\Pr[X = 3] \\ &= \frac{\binom{1}{0}\binom{11}{3}}{\binom{12}{3}} (0.8) + \frac{\binom{3}{0}\binom{9}{3}}{\binom{12}{3}} (0.2) \\ &= \frac{3}{4} (0.8) + \frac{21}{55} (0.2) \\ &= \frac{186}{275}. \end{align*}$$ Therefore, the number of rejected cartons $R$ until the first accepted carton is a geometric random variable with success probability $p$; i.e. $$\Pr[R = r] = (1-p)^r p, \quad r \in \{0, 1, 2, \ldots \}.$$

For the second part of the question, this is simply a negative binomial probability. Let $T$ represent the number of cartons you need to try until you get four that you want. Then $$T \sim \operatorname{NegativeBinomial}(k = 4, p = \tfrac{186}{275}), \\ \Pr[T = t] = \binom{t-1}{k-1} (1-p)^{t-k} p^k, \quad t \in \{k, k + 1, k + 2, \ldots\}.$$ The desired probability is $$\Pr[T \ge 7].$$


Additional exercises for the reader:

In the first part of the question, we were interested in how many cartons on average it would take to accept the first one. What is the probability that the carton we accepted has only one broken egg?

In the second part of the question, we were interested in the probability that we would need to try at least $7$ cartons before accepting $4$. Once we accept $4$ cartons, what is the posterior probability distribution for the number of cartons we accepted with only one broken egg?

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  • $\begingroup$ Thank you for your answer! just a little thingy I noticed: it is $\frac{186}{275}$ =) thank you!! I will take a shot at the bonus question you added $\endgroup$
    – CSch of x
    May 20 '20 at 19:50
  • $\begingroup$ @Remember1312 Indeed you are correct; I made a calculation error which has now been corrected. $\endgroup$
    – heropup
    May 20 '20 at 22:31

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