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The research department of a board game company surveyed some shoppers at a local mall. The team asked shoppers how often they read the newspaper, among other questions. Then the team played a memory card game with the shoppers. They recorded how many cards each shopper remembered. 1-3 cards 4-6 cards 0 times a week 8 9 1-2 times a week 15 10 What is the probability that a randomly selected shopper reads the newspaper 1-2 times a week or remembered 1-3 cards? Simplify any fractions.

Respuesta :

Step 1: Write out the formula

[tex]P(\text{A or B) = P(A) + P(B) - P(A}\cap B)[/tex][tex]\begin{gathered} \text{where } \\ A\text{ and B are events} \end{gathered}[/tex]

Step 2: Write out the given values and substitute them into the formula

Let A be the event of "shopper reads the newspaper 1-2 times.

and

Let B be the event of "shopper remembered 1-3 cards".

n(A) = 15 + 10 = 25

n(B) = 8 + 15 = 23

n(U) = 8 + 15 + 9 + 10= 42

[tex]n(A\cap B)=15[/tex]

Therefore,

[tex]\begin{gathered} P(A)=\frac{25}{42} \\ P(B)=\frac{23}{42} \\ P(A\cap B)=\frac{15}{42} \end{gathered}[/tex]

Hence,

[tex]P(\text{A or B) = }\frac{25}{42}+\frac{23}{42}-\frac{15}{42}=\frac{11}{14}[/tex]

Thus the probability that a randomly selected shopper reads the newspaper 1-2 times a week or remembered 1-3 cards is 11/14

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