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Sagot :
Answer:
The answer is "0.00842".
Step-by-step explanation:
Due to these cards, a conventional 52-card set is dealt one at a time.
While the first two cards are spades, then it would be expected that we could find the next three cards tents, too.
Let A draw 2 spade cards out of 52 cards. Let B become the occasion to draw 3 spade cards the remaining 50 end cards [tex](A \cap B)[/tex] was its case where a 52 card spade is chosen [tex](2+3)=5[/tex].
The number of plots drawn first from 13 regular 52-cerd decks equals number of,[tex]n(A)=\binom{13}{2}[/tex] probability of event [tex]A, PA =\frac{\binom{13}{2}}{\binom{52}{2}}[/tex]
Furthermore, the multitude of possibilities we can pull from of the remaining 11 spades of the previous 3 cards is 50-card decks standard, [tex]n (B) = \binom{11}{3}[/tex] then, the probability of event, [tex]B, P(B)=\frac{\binom{11}{3}}{\binom{50}{3}}[/tex]
The chances of five cards being drawn (three spades and two spades),
[tex]P(B \cap A)=\frac{\binom{13}{5}}{\binom{52}{5}}[/tex]
Then there is the chance that the next three cards will be picked if the first two are both pads, [tex]P(\frac{B}{A})[/tex]
[tex]\to P(\frac{B}{A}) \\\\ =\frac{P(B\cap A )}{P(A)}\\\\=\frac{\frac{\frac{13}{5}}{\frac{52}{2}}}{\frac{\frac{13}{2}}{\frac{52}{2}}}\\\\= 0.00842[/tex]
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