I did binomcdf(22,.25,6) in my calculator. n=22, p=.25, x=6
Is this the correct process?
Suppose a fair four-sided die is tossed 22 times. What is the probability of getting at most 6 fours?
2 answers
I don't have that function on mine, so I have to actually think it through and do the math
prob(a four) = 1/4 , prob(not four) = 3/4
at most 6 fours means:
0 four, 1 four, 2 fours, 3 fours, 4 fours, 5 fours, 6 fours.
prob(your event)
= C(22,0)(3/4)^22 + C(22,1)(1/4) (3/4)^21 + ... + C(22,6) (1/4)^6 (3/4)^18
I will let you do the button-pushing
I have a feeling your binomcdf(22,.25,6)
will give you the probability of exactly 6 out of 22, but then again, I am not at all familiar with this function.
prob(a four) = 1/4 , prob(not four) = 3/4
at most 6 fours means:
0 four, 1 four, 2 fours, 3 fours, 4 fours, 5 fours, 6 fours.
prob(your event)
= C(22,0)(3/4)^22 + C(22,1)(1/4) (3/4)^21 + ... + C(22,6) (1/4)^6 (3/4)^18
I will let you do the button-pushing
I have a feeling your binomcdf(22,.25,6)
will give you the probability of exactly 6 out of 22, but then again, I am not at all familiar with this function.