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NISA
2 months ago
12

RANDOMLY CHOOSING TWO POTENTIAL CLIENTS

Mathematics
1 answer:
Svet_ta [12.7K]2 months ago
6 0
Let Jacob, Carol, Geraldo, Meg, Earvin, Dora, Adam, and Sally be denoted as J, C, G, M, E, D, A, and S respectively. In part IV, we need to identify the pairs of potential clients that could potentially be selected. The sample space consists of all possible outcomes, therefore we create a set of all valid pairs, listed as follows: {(J, C), (J, G), (J, M), (J, E), (J, D), (J, A), (J, S), (C, G), (C, M), (C, E), (C, D), (C, A), (C, S), (G, M), (G, E), (G, D), (G, A), (G, S), (M, E), (M, D), (M, A), (M, S), (E, D), (E, A), (E, S), (D, A), (D, S), (A, S)}. We can verify the number of elements in the sample space, n(S) is 1+2+3+4+5+6+7=28. This gives us the answer to the first question: What is the count of pairs of potential clients that can be randomly selected from the pool of eight candidates? (Answer: 28.) II) What is the chance of a certain pair being chosen? The chance of picking a specific pair is 1/28, as there’s just one way to select a particular pair out of the 28 possible options. III) What is the probability that the selected pair consists of Jacob and Meg or Geraldo and Sally? The probability of selecting (J, M) or (G, S) is 2 out of 28, which equates to 1/14. Answers: I) 28 II) 1/28 ≈ 0.0357 III) 1/14 ≈ 0.0714 IV) {(J, C), (J, G), (J, M), (J, E), (J, D), (J, A), (J, S), (C, G), (C, M), (C, E), (C, D), (C, A), (C, S), (G, M), (G, E), (G, D), (G, A), (G, S), (M, E), (M, D), (M, A), (M, S), (E, D), (E, A), (E, S), (D, A), (D, S), (A, S).}
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