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From a Spanish deck of 40 cards, 1 is taken:

  • If gold or cup appears, 2 coins are thrown;
  • If sword appears, 1 coin is thrown;
  • If club appears, no coins are thrown.
  1. What is the probability that a face appears?
  2. What is the probability that a gold appears and no face appears?
  3. What is the probability that 2 faces appears?

Being,

$G$: "The card is gold", $C$: "The card is club", $S$: "The card is sword", $D$: "The card is cup" and,

$F$ :"Face appears", $N$: "Number appears":

$\mathcal{P}(G)=\mathcal{P}(C)=\mathcal{P}(S)=\mathcal{P}(D)=1/4$

$\mathcal{P}(F)=\mathcal{P}(N)=1/2$

1.

This is correct?

$\mathcal{P}([(G\cap F\cap F)\cup(G\cap F\cap N)\cup (G\cap N\cap F)]$

$\cup[(D\cap F\cap F)\cup(D\cap F\cap N)\cup(D\cap N\cap F)]\cup(S\cap F))$=

$\mathcal{P}(G)\times\mathcal{P}(F)\times\mathcal{P}(F)+ \mathcal{P}(G)\times\mathcal{P}(F)\times\mathcal{P}(N)+ \mathcal{P}(G)\times\mathcal{P}(N)\times\mathcal{P}(F)+$

$\mathcal{P}(D)\times\mathcal{P}(F)\times\mathcal{P}(F)+ \mathcal{P}(D)\times\mathcal{P}(F)\times\mathcal{P}(N)+ \mathcal{P}(D)\times\mathcal{P}(N)\times\mathcal{P}(F)+ \mathcal{P}(S)\times\mathcal{P}(F)=0.5$

Could a similar reasoning be applied for the rest of the exercise? Thank you very much.

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    What's the purpose of the $\mathcal{P}(S) \times \mathcal{P}(N)$ term? What outcome do you want to account for? – 2017-02-26
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    @FabioSomenzi It was in error. – 2017-02-26
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    I see you fixed the other two small errors too. Your result is now correct. – 2017-02-26
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    @FabioSomenzi What about reasoning? Is it also correct? Thank you. – 2017-02-26
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    It's correct. It's based on decomposing the event whose probability must be computed into the union of disjoint events and then adding the probabilities of all those events. – 2017-02-26

1 Answers 1

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It is easier in my opinion for the first question to ask "What is the probability of no heads being thrown"

No heads occur when either throwing only tails when asked to throw a coin, or when not even being given the opportunity to throw any coins.

  • Clubs are drawn: no coins thrown. Occurs with probability $\frac{1}{4}$
  • Sword is drawn: the one coin comes up tails. Occurs with probability $\frac{1}{4}\cdot \frac{1}{2}=\frac{1}{8}$
  • Gold or Cup is drawn: the two coins both come up tails. Occurs with probability $\frac{1}{2}\cdot \frac{1}{2}\cdot \frac{1}{2}=\frac{1}{8}$

Thus, probability of no heads occurring is $\frac{1}{4}+\frac{1}{8}+\frac{1}{8}=\frac{1}{2}$, verifying your result.