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At a phone store, the purchases for one month are recorded in the table below:

\begin{tabular}{|l|c|c|c|}
\hline & Phone I & Phone II & Phone III \\
\hline Mini & 7 & 23 & 31 \\
\hline Standard & 43 & 41 & 29 \\
\hline Maximum & 2 & 17 & 13 \\
\hline
\end{tabular}

If we choose a customer at random, what is the probability that they have purchased a standard-sized phone or a Phone II?

[tex]\[ P(\text{Standard or Phone II}) = \square \][/tex]

Give your answer in simplest form.

Sagot :

GinnyAnswer
To determine the probability that a randomly chosen customer has purchased a standard-sized phone or a Phone II, we need to follow a series of steps:

1. Calculate the total number of standard-sized phone purchases:
- Standard for Phone I: [tex]\(43\)[/tex]
- Standard for Phone II: [tex]\(41\)[/tex]
- Standard for Phone III: [tex]\(29\)[/tex]
[tex]\[ \text{Total standard purchases} = 43 + 41 + 29 = 113 \][/tex]

2. Calculate the total number of Phone II purchases:
- Mini for Phone II: [tex]\(23\)[/tex]
- Standard for Phone II: [tex]\(41\)[/tex]
- Maximum for Phone II: [tex]\(17\)[/tex]
[tex]\[ \text{Total Phone II purchases} = 23 + 41 + 17 = 81 \][/tex]

3. Calculate the total number of customers:
- Mini purchases: [tex]\(7 + 23 + 31 = 61\)[/tex]
- Standard purchases: [tex]\(43 + 41 + 29 = 113\)[/tex]
- Maximum purchases: [tex]\(2 + 17 + 13 = 32\)[/tex]
[tex]\[ \text{Total purchases} = 61 + 113 + 32 = 206 \][/tex]

4. Calculate the overlap between standard-sized phones and Phone II:
- Since we already considered the sum of standard and Phone II purchases, the overlap here (people who bought Standard sized Phone II) is:
[tex]\[ \text{Overlap} = 41 \][/tex]

5. Apply the formula for the union of two sets:
- The formula we use is [tex]\( P(A \cup B) = P(A) + P(B) - P(A \cap B) \)[/tex]
Where:
[tex]\[ P(A) = \text{Total number of standard purchases} = 113 \][/tex]
[tex]\[ P(B) = \text{Total number of Phone II purchases} = 81 \][/tex]
[tex]\[ P(A \cap B) = \text{Overlap} = 41 \][/tex]

6. Calculate the probability:
[tex]\[ \text{Probability} = \frac{P(A) + P(B) - P(A \cap B)}{\text{Total purchases}} \][/tex]
[tex]\[ \text{Probability} = \frac{113 + 81 - 41}{206} = \frac{113 + 81 - 41}{206} = \frac{153}{206} \][/tex]

7. Simplify the fraction to its simplest form:
By evaluating the fraction, we achieve:
[tex]\[ \frac{153}{206} \approx 0.7427184466019418 \][/tex]

Thus, the probability that a randomly chosen customer purchased a standard-sized phone or a Phone II is approximately [tex]\(0.7427\)[/tex], or roughly [tex]\(74.27\%\)[/tex].
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