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Guest
Guest
I can't figure out how they canceled the factorials in this problem. Here it is:
How many bulbs should be examined so that the probability of finding at least 1 bad bulb exceeds 1/2?
I know I can do this problem an easier way, but I want to learn it this way:
****Just in case that is hard to read, this is what I mean:
What I get next is too hard to type up in here, but basically I'm wondering how I would expand (48 - n)! I can't find the formula for expanding it. I know that 3! would be 1*2*3, but how would I do (48 - n)! ? Would it be:
. . .(1 - n)(2 - n)(3 - n)...(48 - n)
Thanks
Take care,
Beckie
How many bulbs should be examined so that the probability of finding at least 1 bad bulb exceeds 1/2?
I know I can do this problem an easier way, but I want to learn it this way:
Code:
(2)(48) + (2)(48)
(1)(n ) (2)(n )
-------------------
(50)
(n )
Code:
C(2, 1) × C(48, n) + C(2, 2) × C(48, n)
---------------------------------------
C(50, n)
. . .(1 - n)(2 - n)(3 - n)...(48 - n)
Thanks
Take care,
Beckie