Question

Suppose Kristin buys 10 tickets out of 150 which will award 5 prizes on the basis of a random drawing. What are the chances that Kristin will win:

a.) no prize?
b.) exactly 3 prizes?
c.) all the 5 prizes?
d.) at least one prize?

Answer 1

Hints:

This is an appropriate application of the binomial probability distribution.

let
n=number of prizes (5)
p=probability of winning (10/150)
r=number of prizes won
P(r)=probability of winning r prizes

then
\(P(r)=nCr~ p^r (1-p)^{n-r}\)

and
\(nCr = \dfrac{n!}{r!(n-r)!}\)

You can then proceed to calculate the P(r) for the following cases:
a) r=0
b) r=3
c) r=5
d) 1-P(0)

Answer 2

Ohhhh. I don't understand binomial probability distribution. Can you help me a little further? :)

Answer 3

@killua_vongoladecimo
Do you mean you are Not supposed to use binomial, or you need help with binomial?

Answer 4

I need help with binomial. Hmm So let me give it a try for A.)

\[P \left( r \right) = 5C0 (\frac{ 1 }{ 15 })^{0} (\frac{ 14 }{ 15 })^{5}\]

Is this right?

Answer 5

So here are the values I got:
a.) 0.708245596
b.) 0.002581069959
c.) 0.000001316872428
d.) 0.291754403

Answer 6

All correct and well done. :)