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| |about 1 hour per week|about 3 hours per week|\n|--|--|--|\n|e - readers|6|4|\n|paperback books|6|4|\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| |about 1 hour per week|about 3 hours per week|\n|--|--|--|\n|e - readers|6|4|\n|paperback books|6|4|\nwhat is the probability that a randomly selected club member prefers e - readers or reads about 1 hour per week?\nsimplify any fractions.

Answer

Explanation:

Step1: Calculate total number of members

The total number of members is the sum of all values in the table. So, $6 + 6+4 + 4=20$.

Step2: Calculate number of members who prefer e - readers

The number of members who prefer e - readers is $6 + 4=10$.

Step3: Calculate number of members who read about 1 hour per week

The number of members who read about 1 hour per week is $6+6 = 12$.

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

The number of members who prefer e - readers and read about 1 hour per week is 6.

Step5: Use the formula for the probability of the union of two events

The formula for $P(A\cup B)=P(A)+P(B)-P(A\cap B)$. Here, $A$ is the event of preferring e - readers and $B$ is the event of reading about 1 hour per week. $P(A)=\frac{10}{20}$, $P(B)=\frac{12}{20}$ and $P(A\cap B)=\frac{6}{20}$. Then $P(A\cup B)=\frac{10 + 12-6}{20}=\frac{16}{20}=\frac{4}{5}$.

Answer:

$\frac{4}{5}$