Ten coins are tossed; in how many ways can you have at least three tails?
768 ways.
I'm really not sure though.
how'd you do that?
i got 968 as well but i'm not exactly confident on how i got that
I basically found out how many possible combinations there are, then I found out how many way I can not get at least 3 tails (so how many ways I can get 2 tails), and figured out the fraction of the time I'd get only two tails, and divided the first number by the denominator of the fraction, and took that number 256 and subtracted it from the original number,
It's really weird, and I'm pretty sure it's wrong, but that's all I can think of other than writing down every possibility.
that's okay. thank you so much for trying to help. really appreciate it. It's a multiple choice question. here are the options: a. 1024 b. 971 c. 968
Hmmm... no clue. sorry :/
easier use the complement event, so we must calculate all events : (2 trails , 8 rest) + (1 trail , 9 rest) + (no trail) 2 trails, 8 rest = 10!/(2!*8!) = (10*9*8!)/(2*1*8!) = 45 1 trail, 9 rest = 10!/(9!*1!) = 10 no trail = 10!/10! = 1 total event = 45+10+1=56 so, complement of events above is 2^10-56=1024-56=968 ways
sorry didn't mean to type that. RadEn thank you so much! if you have time for one more question?? this one's also een driving me crazy: A bin contains individual salt and pepper packets. If 17 of the packets are chosen, how many ways are there to select at least 1 packet of salt? a. 131233 b. 131062 c. 131271
do same idea it means : 2^17 - 1 = 131072 - 1 = 131071 are u sure the option C like that ?
i'm positive and that's exactly what I did. it's an online homework. i can't figure out with it's like that.
thank you so much for your help, though
welcome
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