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it said my answer is wrongTwo cards are selected from a standard deck of 52 cards. The first card is not replaced before the second card selected. Find the probability of selecting a two and then selectinga a king

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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