I was asked a question; A student was late for college 0.25 of the time, what is the probability he is late 4 days in one college week. My answer was this:
L = Late, N = Not Late
L = 0.25, N = 0.75
if its a 7 day week
P(L+N)7 = L7 + (7C1)L6 N + (7C2)L5N2+(7C3)L4N3+(7C4)L3N4+…
P(4L, 3N) = (7C3)(0.25)4(0.75)3 = 0.057
If it’s a 5 day week
P(L+N)5 = L5+ (5C1)L4 N + (5C2)L3N2+(5C3)L2N3+(5C4)LN4+…
P(4L1N) = (5C1)(0.25)4(0.75) = 0.0146
Have I got this ok?
Yes, note that the binomial function is as follows:
$P(X=x) = \binom{n}{x}p^{x}(1-p)^{n-x}$
For example, if $Pr$(late)$=0.25$ and he is late $n=4$ times that week (assuming a week is 7 days), then
$P(X=4) = \binom{7}{4}(.25)^4(1-.25)^{3} \approx 0.057678$
Therefore, there is a 5.77% chance that the student is late 4 out of the 7 days that week.
Does this help?
Comment to show relevant graphs (made using R):
7 day week.
x = 0:7; pdf.7 = dbinom(x, 7, .25)
plot(x, pdf.7, ylab="PDF", type="h", lwd=2, col="blue",
main="PDF of BINOM(7, .25)")
lines(5,pdf.7[6], type="h", lwd=3, col="maroon")
abline(h=0, col="green2")
$P(X = 4)$ shown in maroon.
5 day week.
x = 0:5; pdf.5 = dbinom(x, 5, .25)
plot(x, pdf.5, ylab="PDF", type="h", lwd=2, col="blue",
main="PDF of BINOM(5, .25)")
lines(5,pdf.5[6], type="h", lwd=3, col="maroon")
abline(h=0, col="green2")
$P(X = 4)$ shown in maroon, but is (almost) too small to see.
dbinomis binomial PDF: $X$ = days late per wk. Five day week. P(X = 4).dbinom(4, 5, .25)returns 0.01464844. Seven day week:dbinom(4, 7, .25)returns 0.05767822. // If it's at least 4, then you have a different problem. – BruceET Apr 29 '20 at 04:03