regan joined a book club to spend more quality time with her cousin. at the first meeting, club members…

regan joined a book club to spend more quality time with her cousin. at the first meeting, club members recorded how many hours a week they typically read and whether they preferred e - readers or paperback books.\n\n| |about 1 hour per week|about 3 hours per week|\n|--|--|--|\n|e - readers|6|4|\n|paperback books|6|4|\n\nwhat is the probability that a randomly selected club member prefers e - readers or reads about 1 hour per week?\nsimplify any fractions.

regan joined a book club to spend more quality time with her cousin. at the first meeting, club members recorded how many hours a week they typically read and whether they preferred e - readers or paperback books.\n\n| |about 1 hour per week|about 3 hours per week|\n|--|--|--|\n|e - readers|6|4|\n|paperback books|6|4|\n\nwhat is the probability that a randomly selected club member prefers e - readers or reads about 1 hour per week?\nsimplify any fractions.

Answer

Answer:

$\frac{5}{8}$

Explanation:

Step1: Calculate total members

$6 + 4+6 + 4=20$

Step2: Calculate members who prefer e - readers

$6 + 4 = 10$

Step3: Calculate members who read about 1 hour per week

$6+6 = 12$

Step4: Calculate members who prefer e - readers and read about 1 hour per week

$6$

Step5: Use the formula for $P(A\cup B)$

$P(A\cup B)=P(A)+P(B)-P(A\cap B)$ $P(A)=\frac{10}{20}$, $P(B)=\frac{12}{20}$, $P(A\cap B)=\frac{6}{20}$ $P(A\cup B)=\frac{10 + 12-6}{20}=\frac{16}{20}=\frac{4}{5}$