• Background Information:

    I am studying Discrete Math regarding finite probability. I understand If |S| = n, a ∈ S, and A ⊆ S, then Pr({a}) = Pr(a) = 1/n (the probability that "a" occurs). Thus, Pr(A) = |A|/ n "the probability that "A" occurs".

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  • Original Question:

    100 tickets, numbered 1,2,3,…, 100, are sold to 100 different people for a drawing. Four different prizes are awarded, including a grand prize (a trip to Tahiti). Find the probability that ticket 47 wins a prize while ticket 73 does not.

  • My thinking:

    We know that n = 100, so the sample space S = n. Considering that A is the number of events, then A = 1 because ticket 47 is considered only one ticket from the original 100 tickets, and ticket 73 is still A = 1 because there is only one ticket 73 from 100 tickets. Therefore, both 47 and 73 tickets have the same probability of happening, so n = 100, then S = {1,2, ..., 100}, then Pr(A) = |A|/n = 1/100.

    Am I right? If not, could you please clarify the question for me?

  • $\begingroup$ I think you meant the sample space $S=\{1,2,3,\ldots,n\}$. $\endgroup$ – The Phenotype Dec 21 '17 at 23:21
  • $\begingroup$ Yes sorry I made a small mistake, I just fixed it. The equation is Pr(A) = |A|/n where n is the number of values in the sample space. $\endgroup$ – Kourosh Dec 21 '17 at 23:24

Since OP asks for comments and clarifications for his/her approach which involves sample space, I think it's better to first start with a discussion in , before moving on to practical calculations. In this way, the theoretical side, which brings about a rigourous understanding of the problem, can be learnt better.

Theoretical settings

The sample space $S$ represents the set of all a priori possible outcomes (of the lucky draw). Since four different prizes are randomly attributed to a ticket number, $S$ should be the set of all possible permutations of four distinct elements drawn from $\{1\,\dots,100\}$. i.e. $$S=\{(i,j,k,l)\in\{1,\dots,100\}^4\mid i,j,k,l \text{ pairwise distinct}\}.$$ $|S|=100P4=100\times99\times98\times97$.

N.B.: According to Foundations of the Theory of Probability by A. N. Kolmogorov (Chap I, section 1), we shall call elements in $S$ elementary events.

The event $A$ is "ticket $47$ wins a prize while ticket $73$ does not". Therefore, $$\small{A=\{(i,j,k,l)\in S\mid(i=47 \lor j=47 \lor k=47 \lor l=47) \land (i \neq 73 \land j \neq 73 \land k \neq 73 \land l \neq 73)\}}$$

The probability of the event $A$ is $\Bbb{P}(A)=|A|/|S|$. In words, the set $A$ has one $47$ in exactly one of its components ($i,j,k$ or $l$), and none of the components of $A$ contain $73$.


To calculate this probability, we first include $47$ as a lucky number. Next, pick three other lucky numbers out of $\{1,\dots,100\}\setminus\{47,73\}$ without considering their order: there are $98C3$ possible combinations. To finish, we calculate the number of possible permutations by multiplying the previous number by $4!$, followed by a division by $|S|=100P4$.

To sum up, $$\Bbb{P}(A)=\frac{|A|}{|S|}=\frac{98C3\times4!}{100P4}=\frac{98\times97\times96\times4}{100\times99\times98\times97}=\frac{32}{825}.$$


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