Question

tom has 2 red gummy fish and 2 green gummy fish left in his bag.
he reaches into the bag, grabs a gummy fish, and eats it.
he reaches in again and grabs another fish, and eats it.
what is the probability that tom eats 2 red gummy fish?
the table shows all possible outcomes.

Explanation:

Step1: Count total possible outcomes

There are 4 fish initially (2 red, 2 green). The total number of ways to pick 2 fish (without replacement) can be calculated by looking at the table or using permutations. From the table, we can count the non - crossed - out cells. Let's count: For the first red (1st row), the valid second picks are red 2 (second column). For red 2 (second row), the valid second pick is red 1 (first column). For green 1 (third row), and green 2 (fourth row), their cross - related. Wait, a better way: The total number of ordered pairs (since we pick one then another) is \(4\times3 = 12\) (because after picking one, 3 are left). But from the table, the non - crossed cells: Let's list them. Red1 - Red2, Red2 - Red1, Red1 - Green1, Red1 - Green2, Red2 - Green1, Red2 - Green2, Green1 - Red1, Green1 - Red2, Green1 - Green2, Green2 - Red1, Green2 - Red2, Green2 - Green1. Wait, but the crossed cells are the ones where we pick the same fish (since we can't pick the same fish twice). So total non - crossed (valid) outcomes: \(4\times3=12\) (since for each of the 4 fish, we can pick 3 others). Now, the number of favorable outcomes (picking 2 reds): Red1 - Red2 and Red2 - Red1. So that's 2 outcomes? Wait no, wait the table: Let's look at the red rows. First red (row 1): column 2 (red2) is a valid outcome (the cell has two reds). Second red (row 2): column 1 (red1) is a valid outcome (two reds). So that's 2 favorable? Wait no, wait the initial number of reds is 2, green is 2. The first pick: probability of red is \(\frac{2}{4}=\frac{1}{2}\). Then, given first is red, the second pick: there is 1 red left and 2 green left, so probability \(\frac{1}{3}\). So the combined probability is \(\frac{2}{4}\times\frac{1}{3}=\frac{1}{6}\)? Wait no, wait the table: Let's count the total number of cells. The table has 4 rows (red1, red2, green1, green2) and 4 columns (red1, red2, green1, green2). The crossed cells are the diagonal (where row and column are the same, since we can't pick the same fish twice). So total non - crossed cells: \(4\times4 - 4=12\) (since 4 diagonal cells are crossed). Now, the number of cells with two reds: row red1, column red2; row red2, column red1. So that's 2 cells. Wait, but when we pick red1 then red2, and red2 then red1, those are two different ordered outcomes. So favorable outcomes: 2. Total outcomes: 12. So probability is \(\frac{2}{12}=\frac{1}{6}\)? Wait no, wait the first pick: 2 red out of 4. Second pick: 1 red out of 3 (since one red is already eaten). So the number of ways to pick 2 reds is \(C(2,2)=\frac{2!}{2!(2 - 2)!}=1\) combination, but in ordered terms, it's 2 permutations (red1 then red2, red2 then red1). The total number of ordered ways to pick 2 fish out of 4 is \(P(4,2)=\frac{4!}{(4 - 2)!}=12\), which matches the table. The number of ordered ways to pick 2 reds out of 2 is \(P(2,2)=\frac{2!}{(2 - 2)!}=2\). So probability is \(\frac{2}{12}=\frac{1}{6}\)? Wait but let's check the table again. The cell for red1 (row) and red2 (column) has two reds, and red2 (row) and red1 (column) has two reds. So that's 2 cells. So 2 out of 12, which simplifies to \(\frac{1}{6}\)? Wait no, wait the initial count: 2 red, 2 green. The possible outcomes of picking two fish (without replacement) are: (red1, red2), (red1, green1), (red1, green2), (red2, red1), (red2, green1), (red2, green2), (green1, red1), (green1, red2), (green1, green2), (green2, red1), (green2, red2), (green2, green1). That's 12 outcomes. The number of outcomes with two reds: (red1, red2), (red2, red1) → 2 outcomes. So probability is \(\frac{2}{12}=\frac{1}{6}\)? Wait but wait, when we calculate using combinations: the number of ways to choose 2 reds out of 2 is \(C(2,2) = 1\), and the number of ways to choose 2 fish out of 4 is \(C(4,2)=\frac{4!}{2!2!}=6\). So probability is \(\frac{1}{6}\), which matches. So the table: the number of cells with two reds is 2 (ordered), and total ordered is 12, so 2/12 = 1/6. Or using combinations, 1/6. So the probability is \(\frac{1}{6}\).

Step1: Calculate total possible outcomes

Total number of fish: \(2 + 2=4\). The number of ways to pick 2 fish (without replacement) is \(P(4,2)=\frac{4!}{(4 - 2)!}=4\times3 = 12\) (ordered outcomes) or \(C(4,2)=\frac{4!}{2!2!}=6\) (unordered). But from the table, we use ordered (since first and second pick are distinct).

Step2: Calculate favorable outcomes

Number of red fish: 2. Number of ways to pick 2 red fish (ordered) is \(P(2,2)=\frac{2!}{(2 - 2)!}=2\times1 = 2\) (red1 then red2, red2 then red1).

Step3: Calculate probability

Probability is \(\frac{\text{favorable outcomes}}{\text{total outcomes}}=\frac{2}{12}=\frac{1}{6}\).

Answer:

\(\frac{1}{6}\)