Suppose you randomly choose two cards from a thoroughly shuffled deck. If you do NOT put the first card back in the deck before you draw the second, what is the probability that the first card is club and the second card is a spade?
A)13/204
B)2/51
C)1/16
D)1/17

Question

Quizplus · Accepted Answer

To solve this problem, we need to calculate the probability of two independent events happening in sequence: first drawing a club, and then drawing a spade from a standard deck of 52 cards without replacement.
A standard deck of cards has 52 cards in total, with 13 cards in each of the four suits: clubs, diamonds, hearts, and spades.
Let's calculate the probability step by step:
1. The probability of drawing a club as the first card:
There are 13 clubs in the deck, and there are 52 cards total. So the probability of drawing a club first is:
P(Club first) = Number of clubs / Total number of cards = 13/52
2. The probability of drawing a spade as the second card:
After drawing a club, there are now 51 cards left in the deck (since we do not replace the first card), and there are still 13 spades. So the probability of drawing a spade second is:
P(Spade second | Club first) = Number of spades / Remaining number of cards = 13/51
Now, to find the combined probability of both events happening in sequence (drawing a club first and then a spade), we multiply the probabilities of the two events:
P(Club first and Spade second) = P(Club first) * P(Spade second | Club first)
P(Club first and Spade second) = (13/52) * (13/51)
Now, let's do the math:
P(Club first and Spade second) = (13/52) * (13/51) = (1/4) * (13/51) = 13/204
So the correct answer is:
A) 13/204

Suppose You Randomly Choose Two Cards from a Thoroughly Shuffled