Question

a teacher offers gift cards as a reward for classroom participation. the teacher places the gift cards from four different stores into a bag and mixes them well. a student gets to select two gift cards at random (one at a time and without replacement). each outcome in the sample space for the random selection of two gift cards is equally likely. what is the probability of each outcome in the sample space?

$\frac{1}{16}$

$\frac{1}{6}$

$\frac{1}{4}$

$\frac{1}{2}$

Explanation:

Step1: Calculate the number of ways to select the first gift card

There are 4 gift cards. The number of ways to select the first gift card is 4.

Step2: Calculate the number of ways to select the second gift card

After selecting the first gift card (without replacement), there are 3 gift cards left. So the number of ways to select the second gift card is 3.

Step3: Calculate the total number of outcomes in the sample space

By the multiplication principle, the total number of outcomes (selecting two gift cards one - at - a - time without replacement) is \(n = 4\times3=12\)

Step4: Calculate the probability of each outcome

Since each outcome is equally likely, and the sum of probabilities of all outcomes in a sample space is 1. Let the probability of each outcome be \(P\). If there are \(n = 12\) equally - likely outcomes, then \(P=\frac{1}{12}\) (Wait, there is a mistake above. Let's use the combination formula. The number of ways to choose 2 cards out of 4 (where order matters, since we are picking one - at - a - time) is \(n=A_{4}^2=\frac{4!}{(4 - 2)!}=\frac{4!}{2!}=4\times3 = 12\). Another way: Let the four cards be \(A\), \(B\), \(C\), \(D\). The possible ordered pairs \((x,y)\) (first card \(x\), second card \(y\), \(x
eq y\)) are: \((A,B)\),\((A,C)\),\((A,D)\),\((B,A)\),\((B,C)\),\((B,D)\),\((C,A)\),\((C,B)\),\((C,D)\),\((D,A)\),\((D,B)\),\((D,C)\) - 12 outcomes.
If we use the formula \(P=\frac{1}{n}\), where \(n\) is the number of equally - likely outcomes.

Answer:

\(\frac{1}{12}\) (But since there is no \(\frac{1}{12}\) in the options, maybe the problem is about combinations (where order does not matter). The number of combinations of choosing 2 cards out of 4 is \(C_{4}^2=\frac{4!}{2!(4 - 2)!}=\frac{4\times3}{2\times 1}=6\). If we assume that the problem was misphrased (and order does not matter, i.e., selecting two cards at once rather than one - at - a - time). Then, using the formula \(P=\frac{1}{n}\), where \(n = C_{4}^2=6\). So the probability of each outcome (in the case of combination, when we consider unordered pairs) is \(\frac{1}{6}\))

So the answer is \(\frac{1}{6}\) (assuming combination, i.e., unordered selection of two cards)