Why is this the answer?

A

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A parking lot has 100 red and blue cars in it.
  • 40% of the cars are red.
  • 70% of the red cars have radios.
  • 80% of the blue cars have radios.
What is the probability of selecting a car at random that is either red or has a radio?
I calculated it as:
.40 + .70 - (.40 X .70) = .82
I got this wrong and here is the explanation:
P(red or radio) = P(red) + P(radio) – P(red and radio) = 0.40 + 0.76 – 0.28 = 0.88 or 88%.
Where did the .76 come from?
 
P(radio) = P(red and radio) + P(blue and radio) = (0.4 * 0.7) + (0.6 * 0.8) =0.28 + 0.48 = 0.76
 
Perhaps it’s easier to see this way:
P(red or radio) = P(red) + P(blue with radio) = P(red) + P(radio|blue)P(blue) = 0.40 + 0.80(0.60) = 0.40 + 0.48 = 0.88.
 
Only independent events could be multiplied. Here, these given symbols indicate the events are dependent. Hence, the 0.88 seems correct.
 
Red Cars = 40
Red Cars with Radio = 28
Blue Cars = 60
Blue Cars with Radio = 48
Cars with Radio = 48 + 28 = 76
P(red or radio) is P(red) + P(radio) – P(red and radio) = 0.40 + 0.76 – 0.28 = 0.88 or 88%
 
This is a really easy question btw. You shouldn’t even need a formula to obtain the correct answer.
 
Topperharley wrote:
This is a really easy question btw. You shouldn’t even need a formula to obtain the correct answer.
Click to expand...
Thank you for your input, although your personal opinion is inconsequential.
 
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