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  • Four cards are successively drawn without replacement from a deck of 52 playing cards. What is the probability that all the four cards are kings?

Four cards are successively drawn without replacement from a deck of 52 playing cards. What is the probability that all the four cards are kings?

Answers (1)

Let E1, E2, E3 and E4 be the events that the first, second, third and fourth card is king respectively.

As we know, there are 4 kings,

\mathrm{P}\left(\mathrm{E}_{1}\right)=\frac{4}{52}

when 1 king is taken out, there are 3 kings and total 51 cards left.
Therefore, probability of drawing a king when one king has been taken out is:

\mathrm{So}, \mathrm{P}\left(\mathrm{E}_{2} \mid \mathrm{E}_{1}\right)=\frac{3}{51}$
When 2 kings are taken out, there are 2 kings and 50 cards left. Therefore, probability of drawing a king when two kings have been taken out is:
P\left(E_{3} \mid E_{1} \cap E_{2}\right)=\frac{2}{50}$
When 3 kings are taken out, 1 king and 49 cards are left.
Therefore, probability of drawing a king when three kings have been taken out is:
\mathrm{P}\left[\mathrm{E}_{4} \mid\left(\mathrm{E}_{1} \cap \mathrm{E}_{2} \cap \mathrm{E}_{3} \cap \mathrm{E}_{4}\right)\right]=\frac{1}{49}$
Probability that all 4 cards are king is:
\\\left.\mathrm{P}\left(\mathrm{E}_{1} \cap \mathrm{E}_{2} \cap \mathrm{E}_{3} \cap \mathrm{E}_{4}\right)=\mathrm{p}\left(\mathrm{E}_{1}\right) . \mathrm{P}\left(\mathrm{E}_{2} \mid \mathrm{E}_{1}\right) . \mathrm{P}\left(\mathrm{E}_{3} \mid \mathrm{E}_{1} \cap \mathrm{E} 2\right) . \mathrm{P}\left[\mathrm{E}_{4}\right]\left(\mathrm{E}_{1} \mathrm{NE}_{2} \cap \mathrm{E}_{3} \mathrm{NE}_{4}\right)\right]$ $=\frac{4}{52} \times \frac{3}{51} \times \frac{2}{50} \times \frac{1}{49}=\frac{24}{52 \times 51 \times 50 \times 49}$ \\$=\frac{1}{13 \times 14 \times 25 \times 49}$ \\$=\frac{1}{270725}$

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infoexpert22

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