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Having trouble with binomial distributions can't figure out part "a".

28. Frogs. A wildlife biologist examines frogs for a genetic trait he suspects may be linked to sensitivity to industrial toxins in the environment. Previous research had established that this trait is usually found in 1 of every 8 frogs. He collects and examines a dozen frogs. If the frequency of the trait has not changed, what’s the probability he finds the trait in
a) none of the 12 frogs?
b) at least 2 frogs?
c) 3 or 4 frogs?
d) no more than 4 frogs?

Is this correct so far for part "a"?
p=0
x=0
n=8
q=1
n-x=8

p = 1/8
q = 7/8
n = 12
x = 0
n-x = 12

Thanks for part "b" am I calculating just p(2) or 1-p(2)?

1 - ( P(0) + P(1) ) actually.

for part "c" I added p(1) + p(2) + p(3) + p(4) = 0.393 + 0.311 + 0.282 + 0273 = 1.259 I think something went wrong as it shouldn't be higher than 1.

For c it's just P(3) + P(4). Are you sure you're putting the numbers in right? I got 0.345 + 0.271 + 0.129 + 0.042 = 0.787

Well for p(1) I pressed [8] + [nCr] + [1] + [X] + [0.125] + [yx] + [1] + [X] + [0.875] + [yx] + [7] = [0.393] on my TI-30XA.

n is the number of trials, in this case the number of frogs caught.

 Dheneb wrote:
n is the number of trials, in this case the number of frogs caught.

OK, but am I putting in the right buttons?

Yeah. Right buttons but wrong numbers.