In Ms. Q's deck of cards, every card is one of four colors (red, green, blue, and yellow), and is labeled with one of seven numbers ($1,$ $2,$ $3,$ $4,$ $5,$ $6,$ and $7$). Among all the cards of each color, there is exactly one card labeled with each number. The cards in Ms. Q's deck are shown below.

Question

[asy]
unitsize(0.5 cm);
pen cbred=RGB(224,109,128);
pen cbblue=RGB(136,204,238);
pen cbgreen=RGB(72,189,107);
pen cbyellow=RGB(236,217,120);

real cardwidth = 1, cardheight = 1.5, spacewidth = 0.5, spaceheight = 0.5;
int i, j;
path card = (0,0)--(cardwidth,0)--(cardwidth,cardheight)--(0,cardheight)--cycle;

for (i = 0; i <= 6; ++i) {
filldraw(shift((i*cardwidth + i*spacewidth,0))*card,cbred);
label("$" + string(i + 1) + "$", (i*spacewidth + (i + 0.5)*cardwidth, 0.5*cardheight));
filldraw(shift((i*cardwidth + i*spacewidth,-cardheight - spaceheight))*card,cbgreen);
label("$" + string(i + 1) + "$", (i*spacewidth + (i + 0.5)*cardwidth, 0.5*cardheight - cardheight - spaceheight));
filldraw(shift((i*cardwidth + i*spacewidth,-2*cardheight - 2*spaceheight))*card,cbblue);
label("$" + string(i + 1) + "$", (i*spacewidth + (i + 0.5)*cardwidth, 0.5*cardheight - 2*cardheight - 2*spaceheight));
filldraw(shift((i*cardwidth + i*spacewidth,-3*cardheight - 3*spaceheight))*card,cbyellow);
label("$" + string(i + 1) + "$", (i*spacewidth + (i + 0.5)*cardwidth, 0.5*cardheight - 3*cardheight - 3*spaceheight));
}
[/asy]

Professor Grok draws two cards from Ms. Q's deck at random without replacement. What is the probability that the first card Grok draws is labeled with an even number, and the second card Grok draws is labeled with a multiple of $3?$

Answer 1

To find the probability that the first card drawn has an even number and the second card drawn has a multiple of 3, we first need to determine how many cards belong to each category.

The cards are labeled with the numbers 1 through 7. The even numbers in this range are:

• 2
• 4
• 6

Thus, there are 3 even numbers.

The multiples of 3 in the range from 1 to 7 are:

• 3
• 6

This means there are 2 multiples of 3.

Next, we calculate the total number of cards in the deck. Since there are 4 colors and each color has 7 numbered cards, the total number of cards is: \[ 4 \times 7 = 28. \]

Now we'll determine how many successful outcomes there are for our specific drawing scenario.

1. Choosing the first card: The first card must be one of the 3 cards labeled with an even number. After this card is drawn, there will still be a total of 27 cards left in the deck.

2. Choosing the second card: Now, we need to count how many cards labeled with multiples of 3 remain. There are initially 2 multiples of 3 (3 and 6), and we need to consider if the first card that was drawn had a multiple of 3 as a label:

   • Case 1: The first card drawn is an even number that is not a multiple of 3 (i.e., 2 or 4). Then both multiples of 3 (3 and 6) are available for drawing. Thus, there are 2 suitable cards left.
   • Case 2: The first card drawn is the even number 6, which is also a multiple of 3. In this case, only 1 multiple of 3 (the number 3) remains to be drawn.

Let's compute the total number of favorable scenarios:

• Case 1: First card is 2 or 4 (2 outcomes).

  • Probability of picking an even number (2 or 4) first: \( \frac{2}{28} = \frac{1}{14} \).
  • Then, second card can be either 3 or 6: \( 2 \) favorable outcomes remain from \( 27 \) total cards.
  • Probability of picking a multiple of 3 second: \( \frac{2}{27} \).

  The combined probability for this case: \[ \frac{1}{14} \times \frac{2}{27} = \frac{2}{378} = \frac{1}{189}. \]

• Case 2: First card is 6 (1 outcome).

  • Probability of picking 6 first: \( \frac{1}{28} \).
  • Then, only 3 remains as a multiple of 3: \( 1 \) favorable outcome remains from \( 27 \) total cards.
  • Probability of picking a multiple of 3 second: \( \frac{1}{27} \).

  The combined probability for this case: \[ \frac{1}{28} \times \frac{1}{27} = \frac{1}{756}. \]

Adding the probabilities from both cases gives: \[ \frac{1}{189} + \frac{1}{756}. \] To add these fractions, we need a common denominator. The least common multiple of 189 and 756 is 756. Thus: \[ \frac{1}{189} = \frac{4}{756}. \] Now we can combine: \[ \frac{4}{756} + \frac{1}{756} = \frac{5}{756}. \]

Therefore, the final probability that the first card drawn is labeled with an even number and the second card drawn is labeled with a multiple of 3 is: \[ \boxed{\frac{5}{756}}. \]