One hundred A&M parents completed a survey regarding the music they listen to. O
ID: 3272622 • Letter: O
Question
One hundred A&M parents completed a survey regarding the music they listen to. Out of these 100 parents, 80 parents reported that they listen to music on Compact Discs (CDs), 60 parents reported that they listen to music on the radio, and 50 parents reported that they listen to music on both CD's and the radio. What is … (a) the probability that a randomly selected parent listens to music on CDs or on the radio or both? (b) the probability that a randomly selected parent listens to music on neither CDs nor the radio? (c) the probability that a randomly selected parent listens to the radio, but does not listen to CDs?
please use the proper formulas and step by step on how you go the numbers and what equations you used. I dont want just the raw answer
Explanation / Answer
P(listen to music on Compact Discs) = 80/100 = 0.8
P(listen to music on the radio) = 60/100 = 0.6
P(listen to music on both CDs and radio) = 50/100 = 0.5
a) P(listens to music on CDs or on the radio or both) = P(listen to music on Compact Discs) + P(listen to music on the radio) - P(listen to music on both CDs and radio)
= 0.8 + 0.6 - 0.5
= 0.9
b) P(listens to music on neither CDs nor the radio) = 1 - P(listens to music on CDs or on the radio or both)
= 1 - 0.9
= 0.1
c) P(listens to the radio, but does not listen to CDs) = P(listen to music on the radio) - P(listen to music on both CDs and radio)
= 0.6 - 0.5
= 0.1
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