The following exercise refers to choosing two cards from a thoroughly shuffled deck. Assume that the...

The following exercise refers to choosing two cards from a thoroughly shuffled deck. Assume that the deck is shuffled after a card is returned to the deck.

If you do not put the first card back in the deck before you draw the next, what is the probability that the first card is a 2 and the second card is a ace? (Enter your probability as a fraction.

Expert Solution

Number of cards in the deck = 52

Number of 2's in the deck = 4

Number of aces in the deck = 4

A: Event of first drawn card is 2

B : Event of second drawn card is Ace

Probability that the first card is a 2 and the second card is a ace = P(A and B)

P(A and B) = P(A) P(B|A)

P(A) = Probability of first drawn card is 2 = Number of 2' in deck / Total number of cards = 4/52

P(B|A) = Probability of second drawn card is Ace given that first drawn card is 2

Given the first drawn cards is 2 : Number of cards left in deck (you do not put the first card back in the deck before you draw the next) = 52-1=51

Number of aces left in deck (as first drawn card is a '2'; all aces in deck are intact) = 4

P(B|A) = Probability of second drawn card is Ace given that first drawn card is 2

= Number of aces left in deck /Number of cards left in deck = 4/51

P(A and B) = P(A) P(B|A) = (4/52) x (4/51) = (1/13) x (4/51) = 4/663

probability that the first card is a 2 and the second card is a ace = 4/663

orchestra answered 1 year ago

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