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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?

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Solution

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 a total of 51 cards left.

Therefore, the 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, the 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, the 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}
The 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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