A normal pack of 52 cards comprises four suits – Spades, Clubs, Diamonds and Hearts − of 13 cards each, the Spades and Clubs being black and the other two red. Further, each suit of 13 cards comprises nine numbered cards − numbered from 2 to 10 − and four honour cards − Ace, King, Queen and Jack. Now, it is known that a card is missing from a normal pack of 52 cards. If from that pack, two black cards can be selected in 300 ways and two numbered cards can be selected in 630 ways, which of the following could be the missing card?Spade AceDiamond kingClub kingMore than one of the above
Question
A normal pack of 52 cards comprises four suits – Spades, Clubs, Diamonds and Hearts − of 13 cards each, the Spades and Clubs being black and the other two red. Further, each suit of 13 cards comprises nine numbered cards − numbered from 2 to 10 − and four honour cards − Ace, King, Queen and Jack. Now, it is known that a card is missing from a normal pack of 52 cards. If from that pack, two black cards can be selected in 300 ways and two numbered cards can be selected in 630 ways, which of the following could be the missing card?Spade AceDiamond kingClub kingMore than one of the above
Solution
To solve this problem, we need to understand the composition of a deck of cards and the concept of combinations in probability.
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A normal deck of 52 cards has 26 black cards (13 Spades and 13 Clubs) and 36 numbered cards (9 in each suit: Spades, Clubs, Diamonds, Hearts).
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The problem states that a card is missing, so there are 51 cards in the deck.
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It also states that two black cards can be selected in 300 ways. The number of ways to select 2 black cards from a set of n black cards is given by the combination formula nC2 = n*(n-1)/2.
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If we set this equal to 300, we can solve for n: n*(n-1)/2 = 300. Solving this quadratic equation gives n = 25 or n = -24. Since n can't be negative, we have 25 black cards in the deck.
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This means one black card is missing from the deck (since there should be 26).
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Similarly, the problem states that two numbered cards can be selected in 630 ways. Using the combination formula again, we can set n*(n-1)/2 = 630 and solve for n. This gives n = 36 or n = -35. Since n can't be negative, we have 36 numbered cards in the deck.
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This means no numbered card is missing from the deck.
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Therefore, the missing card must be a black card that is not numbered. This means it must be an Ace, King, Queen, or Jack of Spades or Clubs.
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Looking at the options given, the only card that fits this description is the Spade Ace. So, the Spade Ace could be the missing card.
Similar Questions
A standard deck of playing cards has 52 cards divided equally into four suits: hearts , diamonds , clubs , and spades . Within each suit, there are 13 cards: one ace, the numbers from 2 to 10, a jack, queen, and king.Suppose you are dealt a hand of 3 cards from a standard deck. In how many ways can this be done?How many of the sets will contain at least 1 ace among the 3 cards?What is the probability of at least 1 ace among the 3 cards you are dealt?
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There are four suits in a deck of 52 playing cards: hearts, diamonds, clubs, and spades. Each suit has 13 cards: A, 2, 3, 4, 5, 6, 7, 8, 9, 10, J, Q, K. Suppose you draw two cards randomly without replacement. What is the probability that the first card is a heart and the second card is a king? (The result of directly drawing the King of Hearts on the first draw is also acceptable; therefore, this problem has two answers, and either is correct.
a classic deck of cards is made up of 52 cards, 26 are black, 26 are red. Each color is split into two suits of 13 cards each (clubs and spades are black and hearts and diamonds are red). Each suit is split into 13 individual cards (Ace, 2-10, Jack, Queen, and King). Leave answers as reduced fractions. If a card is randomly selected, what is the probability of drawing a(n) 9 of one suit?
If 5 cards are drawn from a pack of 52 well-shuffled cards, find the probability of (a) 4 ace, (b) 4 aces and 1 is a king, (c) 3 are tens and 2 are jacks, (d) a nine, ten, jack, queen, king is obtained in any order, (e) 3 are of any one suit and 2 are of another, (f) at least one ace is obtained.
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