Binomial Probability with Complements

Hello everyone,

I am having trouble answering the multiple choice question at the bottom of this message.

As I am not sure how to express the complement of the desired probability, I calculated the values of the multiple choices and found that none match that of the given solution. Therefore, I'm tempted to answer d), but because the following equation seems true to me, I'd like to ask if someone could see whether d) is really correct:

P(at least four bulbs last more than 1 year) = 1 - P(exactly four do NOT last more than 1 year).

Thank you very much!

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At least 4 means 4, 5, 6, 7, or 8 last more than a year.

\(\displaystyle \sum_{k=4}^{8}\binom{8}{k}(.3679)^{k}(.6321)^{8-k}=.3327\)

Using complements. The opposite of at least 4 is 0 to 3.

\(\displaystyle 1-\sum_{k=0}^{3}\binom{8}{k}(.3679)^{k}(.6321)^{8-k}=.3327\)

Thanks for your reply, galactus.

May I ask why the complement is:
\(\displaystyle 1-\sum_{k=0}^{3}\binom{8}{k}(.3679)^{k}(.6321)^{8-k}?\)

Since P(at least four last more than a year) = 1 - P(all four do NOT last more than a year),

why would P(all four do NOT last more than a year) = \(\displaystyle \sum_{k=0}^{3}\binom{8}{k}(.3679)^{k}(.6321)^{8-k}?\)

Shouldn't P(all four do NOT last more than a year) be an intersection of events, and not an union of events as the summation implies?

There are 8 bulbs. 'At least 4' means 4,5,6,7, or 8.

What is left out of the 8?. 0,1,2,3

If it said at least 3. The complement would be 0,1,2

See?.

Hello galactus,

Thanks for your reply.

I understand that the complement of "at least 4" = 0, 1, 2, 3. But doesn't this mean the event that ALL 4 do NOT last more than 1 year? If this is the case, shouldn't this event be an intersection \(\displaystyle \Longleftrightarrow\) P(0 bulbs do not last more than 1 year) AND P(1 bulb do not last more than 1 year) AND ... AND P(4 bulbs do not last more than 1 year) \(\displaystyle \Longleftrightarrow\) choice a) \(\displaystyle = 1 - \binom{8}{4}(.3679)^{4}(.6321)^{4} ?\)

Thanks again.