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Why is this the answer?

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?

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P(radio) = P(red and radio) + P(blue and radio) = (0.4 * 0.7) + (0.6 * 0.8) =0.28 + 0.48 = 0.76

“Mmmmmm, something…” - H. Simpson

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.

Simplify the complicated side; don't complify the simplicated side.

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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.

Thank you for your input, although your personal opinion is inconsequential.