Answered

A standard deck of cards has total cards divided evenly into suits — there are clubs, diamonds, hearts, and spades. Clubs and spades are black cards, while diamonds and hearts are red cards. Kenny and Aja are playing a game that involves drawing cards from a standard deck to start the game. They draw these cards with replacement, so they replace the first card and shuffle before drawing the second card. If both cards are the same color (both red or both black), then Kenny goes first. Otherwise, Aja goes first. Is this a fair way to decide who goes first? Why or why not? Choose 1 answer:



Answer :

Xaioo

Answer:

[tex][/tex] This isn't a fair way to decide who goes first.

The probability of drawing two cards of the same color (both red or both black) in a row is not equal to the probability of drawing one red card and one black card. Since the players are drawing cards with replacement, the probabilities are independent for each draw.

When drawing one card, the probability of drawing a red card is [tex]\frac{26}{52}[/tex] (since half the deck is red) and the probability of drawing a black card is [tex]\frac{26}{52}[/tex].

For the second draw, the probability of drawing another red card is [tex]\frac{26}{52}[/tex], and the probability of drawing another black card is [tex]\frac{26}{52}[/tex].

So, the probability of drawing two cards of the same color (both red or both black) in a row is ([tex]\frac{26}{52}[/tex]) [tex] \times[/tex] ([tex]\frac{26}{52}[/tex]) + ([tex]\frac{26}{52}[/tex]) [tex] \times[/tex] ([tex]\frac{26}{52}[/tex]) = 0.5 [tex] \times[/tex] 0.5 + 0.5 [tex] \times[/tex] 0.5 = 0.5.

Therefore, the chances of Kenny going first using this method is 0.5, making it an unfair way to decide.

Other Questions