Question

Suppose that you randomly draw 18 cards from a standard deck of 52 cards with replacement. What is the probability of drawing at least 7 diamonds?

Answer 1

what an ugly calculation you have to do

Answer 2

at least 7 means 7 or 8 or 9 or 10 or...

you could compute 0 or 1 or 2 or 3 or 4 or 5 or 6 and subtract that total from 1

Answer 3

the probabilty you get a diamond is \(\frac{1}{4}\) and the probabilty you don't is \(\frac{3}{4}\) use the binomial to compute each one
the probability you get \(k\) diamonds is
\[P(X=k)=\binom{18}{k}(\frac{1}{4})^k\times (\frac{3}{4})^{18-k}\] and you have to repeat this for \(k=0,1,2,3,4,5,6\)

Answer 4

So K would equal (14 choose 0,1,2,etc)?