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A card is drawn from a pack of 52 cards - Leaving Cert Mathematics - Question b - 2012

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A card is drawn from a pack of 52 cards. Find the probability that the card drawn is the king of clubs. A card is drawn from a pack of 52 cards. Find the probabilit... show full transcript

Worked Solution & Example Answer:A card is drawn from a pack of 52 cards - Leaving Cert Mathematics - Question b - 2012

Step 1

Find the probability that the card drawn is the king of clubs.

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Answer

To find the probability of drawing the king of clubs from a standard deck of 52 cards, we use the formula:

P(kextkingofclubs)=extNumberoffavorableoutcomesextTotalnumberofoutcomesP(k ext{ - king of clubs}) = \frac{ ext{Number of favorable outcomes}}{ ext{Total number of outcomes}}

In this case:

  • Number of favorable outcomes = 1 (there is only one king of clubs)
  • Total number of outcomes = 52 (total cards)

Thus, the probability is:

P(kextkingofclubs)=152P(k ext{ - king of clubs}) = \frac{1}{52}

Step 2

Find the probability that the card drawn is a club or a picture card.

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Answer

To find the probability of drawing either a club or a picture card, we first identify the number of clubs and picture cards:

  • Number of clubs = 13
  • Picture cards (Kings, Queens, Jacks) = 3 in each suit, total for all suits = 3 × 4 = 12

However, since the king of clubs has already been counted as a club, it is included in both categories.

Thus, the total number of unique clubs or picture cards:

Number of unique clubs or picture cards = Number of clubs + Number of picture cards - Picture cards that are clubs = 13 + 12 - 1 = 24

Now we find the probability:

P(extcluborpicturecard)=2452=613P( ext{club or picture card}) = \frac{24}{52} = \frac{6}{13}

Step 3

Find the probability that neither of them is a club or a picture card.

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Answer

To find the probability that two cards drawn are neither clubs nor picture cards, we first determine the number of such cards in a standard deck:

  • Total cards = 52
  • Clubs = 13
  • Picture cards = 12

Thus, the number of cards that are neither:

52(13+12)=2752 - (13 + 12) = 27

Now, we want the probability that both cards drawn are from these 27 cards:

For the first card: P(extfirstcardnotacluborpicture)=2752P( ext{first card not a club or picture}) = \frac{27}{52}

For the second card (after one card has been drawn, we have 26 left that are neither clubs nor picture cards): P(extsecondcardnotacluborpicture)=2651P( ext{second card not a club or picture}) = \frac{26}{51}

Therefore, the combined probability is:

P(extbothnotclubsorpicture)=2752×2651=70226520.26P( ext{both not clubs or picture}) = \frac{27}{52} \times \frac{26}{51} = \frac{702}{2652} \approx 0.26

Thus, rounding to two decimal places: P(extnotcluborpicturecard)extis0.26P( ext{not club or picture card}) ext{ is } 0.26

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