Question

Type the correct answer in each box. Use numerals instead of words. If necessary, use / for the fraction bar. Chris is playing a card game with his friends. The game uses only the four aces from a standard deck of cards. For each move, a player draws one card, replaces it, and then draws the second card. The probability that a player draws two cards of the same color is . The probability that a red ace is drawn first and then a black ace is

Answer 1

Answer:

4/16 or 1/4

Step-by-step explanation:

Answer 2

Let's call the aces $h, d, c, s$ for hearts, diamonds, clubs and spades. So, $h, d$ are red and [ted] c, s[/tex] are black.

Since the first card is replaced, the two picks are identical. This means that the sample space is given by all the possible couple

$(x,y)\, x, y \in \{h,d,c,s\}$

There are 16 such couples (we have four choices for the first card, and the same four choices for the second card). Now let's compute the odds in our favour to deduce the probability of winning:

If we want a player to draw two card of the same colour, the following couples are good:

$(h,h),\ (h,d),\ (d,h),\ (d,d),\ (c,c),\ (c,s),\ (s,c),\ (s,s)$

so 8 possible couples over 16. This means that the probability that a player draws two cards of the same color is 8/16 = 1/2.

Similarly, the probability of drawing a red ace first and then a black ace is represented by the following couples:

$(h,c),\ (h,s),\ (d,c),\ (d,s)$

which are 4 over the same 16 as above, thus leading to a probability of 4/16 = 1/4.