Question #6626cac3 - SAT Suite Question Bank

ID: 6626cac3

Modded SAT Question Bank by Abdullah Mallik

	Phone	Email
Dinner dance	55%	80%
Football game	20%	10%
Picnic	20%	5%
Pool party	5%	5%
Total	100%	100%

An alumni association survey asked each high school graduate to select the one activity he or she preferred for the association’s next event. Some of the people responded by phone, and the others responded by email. The table above shows the distribution of preferred activity, in percent, for each response type used. For the survey, the number of email responses was twice the number of phone responses. If a person who preferred a picnic is selected at random, what is the probability that the person responded by email?

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Correct Answer:

Rationale

The correct answer is $one third$ . It’s given that the number of email responses is twice the number of phone responses. Therefore, if the number of phone responses is p, then the number of email responses is $2 p$ . The table shows that 20% of people who responded by phone preferred a picnic. It follows that the expression $0 point 2 0 p$ represents the number of these people. The table also shows that 5% of the people who responded by email preferred a picnic. The expression $0 point 0 5 times 2 p$ , or $0 point 1 p$ , represents the number of these people. Therefore, a total of $0 point 2 0 p plus 0 point 1 p$ , or $0 point 3 p$ people preferred a picnic. Thus, the probability of selecting at random a person who responded by email from the people who preferred a picnic is $the fraction 0 point 1 p over 0 point 3 p$ , or $one third$ . Note that 1/3, .3333, and 0.333 are examples of ways to enter a correct answer.

Question Difficulty: Hard