Please help! Will give fan and medal. When treated with antibiotic, 94% of all the dolphins are cured of an ear infection. if 7 dolphins are treated, find the probability that exactly 4 are cured. Probability is ____. Thank you in advance :)
i cuk at prob...x(
Oh, that's okay! Thanks anyway.
i can help
Okay! Yeh, that'd be great!
id say 0.134286
hold it did it wrong its actually 0.235
without the percent with dividing 94/4 its 23.5
might want to show your work @jjcoolja12
so i am guilty of being right
@inkyvoyd dude its just 94%/4 or 94/4
@Ray1998 , this is a binomial probability problem.
0.134286
without the percent its 13.428571
i've been proven guilty of being smart again
You are looking for the the probability that EXACTLY 4 dolphins are cured, right? let's look at it this way, C being cured and N being not cured... to get EXACTLY 4 dolphins cured, there are a lot of different ways. Dolphin 1 2 3 4 5 6 7 CCCCNNN CCCNCNN CCCNNCN CCNCCNN ... precisely how many ways are there? Well, this is a combinatorics problem... Let's think. We're trying to select 4 lucky dolphins from 7... does order matter?
What about the 7?
ok why do i have the feeling im wrong
Um, no?
so im not wrong eh probably in my other answered questions
So... do you know what you need to do to select 4 different dolphins from 7 dolphins? You said order doesn't matter.
No, I don't, sorry.
Do you recall permutations and combinations?
I don't think so, not from the top of my head.
You'll need them. take a quick look: http://www.mathwords.com/c/combination_formula.htm should be familiar; let me know if you have any problems
Unfortunately I must leave; I did not get the lead in to explaining the formula to do this problem as I wished, but here is a link to what you need.
you'll need the first link to apply the nCr formula so make sure you understand. n will be 7, because you have 7 trials, and r will be 4, because you want 4 successes. p is .94 as the probaiblity of a dolphin getting cured is 94%, , so q will be 1-.94=0.06. Plug and chug.
7c4*(0.94)^4*(0.06)^3
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