Question

a. in exploration 1(a), find the theoretical probability that the sum of the two numbers rolled is 7. then compare your answer with the experimental probability you found in exploration 2(a).
b. in exploration 1(b), find the theoretical probability that the sum of the two numbers selected is 7. then compare your answer with the experimental probability you found in exploration 2(b).
c. compare the probabilities you obtained in parts (a) and (b).

Explanation:

Step1: Assume Exploration 1(a) is two dice rolls

When rolling two standard six - sided dice, the total number of possible outcomes is $n(S)=6\times6 = 36$. The pairs of numbers that sum to 7 are: (1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1). So the number of favorable outcomes $n(A)=6$. The theoretical probability $P(A)=\frac{n(A)}{n(S)}=\frac{6}{36}=\frac{1}{6}\approx0.1667$. (We would then compare this with the experimental probability from Exploration 2(a), but since that data is not given, we focus on the theoretical part here.)

Step2: Assume Exploration 1(b) is two number selections (e.g., without replacement)

Suppose we are selecting two numbers from a set, say from 1 - 6 without replacement. The total number of ways to select 2 numbers from 6 is $n(S)=\binom{6}{2}=\frac{6!}{2!(6 - 2)!}=\frac{6\times5}{2\times1}=15$. The pairs that sum to 7 are (1, 6), (2, 5), (3, 4). So the number of favorable outcomes $n(B) = 3$. The theoretical probability $P(B)=\frac{n(B)}{n(S)}=\frac{3}{15}=\frac{1}{5}=0.2$. (Again, comparison with experimental probability from Exploration 2(b) is based on given data, here we do the theoretical part.)

Step3: Compare probabilities from (a) and (b)

The probability from part (a) is $\frac{1}{6}\approx0.1667$ and from part (b) is $\frac{1}{5} = 0.2$. So the probability in part (b) is greater than that in part (a) because in part (a) we are dealing with independent events (dice rolls, with replacement) and in part (b) we are dealing with dependent events (selection without replacement) which changes the sample space and the number of favorable outcomes in a way that increases the probability of getting a sum of 7 in this case.

Answer:

Part (a) Theoretical Probability: $\boldsymbol{\frac{1}{6}}$ (Comparison with experimental depends on data)

Part (b) Theoretical Probability: $\boldsymbol{\frac{1}{5}}$ (Comparison with experimental depends on data)

Part (c) Probability in (b) > Probability in (a) (due to replacement vs non - replacement)