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cat2010
A card is drawn at random from a deck of fifty-two cards (no jokers). What is the probability of drawing a diamond, a card with an even number, or a card with a number divisible by three?

(A)\(\frac{19}{26}\)

(B)\(\frac{1}{2}\)

(C)\(\frac{17}{26}\)

(D)\(\frac{45}{52}\)

(E)\(\frac{33}{52}\)


Take each case separately and ADD them.
1) Diamond: 13 cards
2) All other sets
a) Multiples of 3: 3, 6, and 9, so 3 cards in each.
a) Even number: 2, 4, 6, 8, 10, but 6 is already calculated above, so 4 cards in each.
As entire set of diamond is already taken, calculate for remaining three = (3+4)*3 = 21
Total with restrictions = 13+21 = 34

Total possible outcomes = 52

P= \(\frac{34}{52}=\frac{17}{26}\)


C


Why do we subtract the outcomes of diamonds & 6 ? there is an 'or' in between right? Is it because we're finding the total number of cases as a whole?

chetan2u

Hi

When you pick 6 of diamonds, it is just one card, so you cannot take it as three different cards, one as even card, second as diamond card and third as multiple of 3.
Thus, you take all three as one single card, which it actually is.
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cat2010
A card is drawn at random from a deck of fifty-two cards (no jokers). What is the probability of drawing a diamond, a card with an even number, or a card with a number divisible by three?

(A)\(\frac{19}{26}\)

(B)\(\frac{1}{2}\)

(C)\(\frac{17}{26}\)

(D)\(\frac{45}{52}\)

(E)\(\frac{33}{52}\)

Solution:

  • Number of diamond card \(= 13\)
  • Even number cards = 2, 4, 6, 8, and 10 of three types i.e., spade, heart, and club. Thus \(5\times 3=15\)
  • Divisible by 3 = 3 and 9 (not including 6 because its already counted) of three types i.e., spade, heart, and club. Thus, \(2\times 3=6\)
  • Total favorable cards \(=13+15+6=34\)
  • Total cards in the deck \(=52\)
  • Required probability \(=\frac{34}{52}=\frac{17}{26}\)

Hence the right answer is Option C
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Is it assumed from GMAC that we will be aware of types and number of cards in set?
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Will it be mentioned in the exam whether we are considering J,Q, and K as 11, 12, 13, or not?
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