Five cards are drawn from a pack of 52 cards. What is the chance that

1.	Five cards are drawn from a pack of 52 cards. What is the chance that these 5 will contain: (i) just one ace (ii) at least one ace?
Answer» given: pack of 52 playing cards Formula: P(E) = \(\frac{favorable\ outcomes}{total\ possible\ outcomes}\) five cards are drawn at random, therefore total possible outcomes are ⁵²C₅ therefore n(S)=2598960 (i) let E be the event that exactly only one ace is present n(E)= ⁴C₁ ⁴⁸C₄= 778320 P(E) = \(\frac{n(E)}{n(S)}\) P(E) = \(\frac{778320}{2598960}\) = \(\frac{3243}{19829}\) (ii) let E be the event that at least one ace is present E= {1 or 2 or 3 or 4 ace(s)} n(E)= ⁴C₁ ⁴⁸C₄+ ⁴C₂ ⁴⁸C₃+ ⁴C₃ ⁴⁸C₂+ ⁴C₄ ⁴⁸C₁= 886656 P(E) = \(\frac{n(E)}{n(S)}\) P(E) = \(\frac{886656}{2598960}\) = \(\frac{18472}{54145}\)

Five cards are drawn from a pack of 52 cards. What is the chance that these 5 will contain: (i) just one ace (ii) at least one ace?

Answer»

given: pack of 52 playing cards

Formula: P(E) = \(\frac{favorable\ outcomes}{total\ possible\ outcomes}\)

five cards are drawn at random, therefore

total possible outcomes are ⁵²C₅

therefore n(S)=2598960

(i) let E be the event that exactly only one ace is present

n(E)= ⁴C₁ ⁴⁸C₄= 778320

P(E) = \(\frac{n(E)}{n(S)}\)

P(E) = \(\frac{778320}{2598960}\) = \(\frac{3243}{19829}\)

(ii) let E be the event that at least one ace is present

E= {1 or 2 or 3 or 4 ace(s)}

n(E)= ⁴C₁ ⁴⁸C₄+ ⁴C₂ ⁴⁸C₃+ ⁴C₃ ⁴⁸C₂+ ⁴C₄ ⁴⁸C₁= 886656

P(E) = \(\frac{n(E)}{n(S)}\)

P(E) = \(\frac{886656}{2598960}\) = \(\frac{18472}{54145}\)

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