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Sagot :
To answer this question, we need to have into account the following:
1. We have a standard deck of cards of 52 cards.
2. The first card is not replaced before the second card is selected.
3. We have four cards denominated "2" in a standard deck of cards.
4. We also have four cards denominated "king" in a standard deck of cards.
Now, the probability of selecting a two from the standard deck of cards is:
[tex]P(X=2)=\frac{4}{52}=\frac{1}{13}\Rightarrow P(X=2)=\frac{1}{13}[/tex]Since the first card is not replaced (it did not return to the deck of cards), we now have 51 cards. Now, the probability of selecting a king is:
[tex]P(X=K)=\frac{4}{51}[/tex]Then, the probability of these two events is:
[tex]P=\frac{1}{13}\cdot\frac{4}{51}\Rightarrow P=\frac{4}{663}=0.00603318250377\approx0.006[/tex]If we round to three decimals the probability is, approximately, P = 0.006.
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