A major electronics manufacturer has determined that when one of its televisions is sold, there is $0.08$ chance that it will need service before the warranty period expires.
Suppose a retailer sells four televisions on a particular Saturday. What is the probability that none of the four will need service prior to the warranty expiring?
What I did is that I calculated the probability of at least one TV needs repair, (which is the opposite of none of four need repair): $0.08+0.08+0.08+0.08 = 0.32$
Then I calculate the probability of none of four need repair: $1 - 0.32 = 0.68$
On the other hand, my friend calculated using the percentage the TV does not need repair, which is $ 0.92 $. And got the probability of none of four need repair: $0.92x0.92x0.92x0.92 = 0.716$
I feel like both ways is right, but why are the answers different? Did I or my friend do it wrong?
$\endgroup$ 11 Answer
$\begingroup$Your friend is correct.
The probability that none of the four TVs needs repair is: each TV's individual probability of not being repaired, multiplied together.
$0.92 \cdot 0.92 \cdot 0.92 \cdot 0.92 = 0.92^4 = .71639296$
You cannot add the probabilities.
A helpful resource may be the responses to this Quora question "How to know when to add and when to multiply probabilities", particularly the response by Nisha Arora. She writes:
$\endgroup$If your problems has words like "or", "either", "atleast" or their synonyms, you need to 'ADD' favorable cases & hence the probabilities. [learn more about addition theorem & mutually exclusive cases]
If your problem has words like "and", "both", "all" or their synonyms, you need to 'MULTIPLY' favorable cases & hence the probabilities. [learn more about multiplication theorem & dependent events]
