Question

Level: HARD + Super medal

On average, a noob gets 1 kill per 800 games.
He plays 200 games in a day.
What is the probability that the noob gets 2 or more kills per day?

Answer 1

ideas: This is a poisson distribution

Answer 2

I just solved it.
1 - poissCdf(0.25,0,1)
Where 0.25 is 200/800

Answer 3

anyone like medal?

Answer 4

ok using the poisson distribution
we know that
n = 200
\[\lambda = 200* \frac{1}{800} = \frac{1}{4}\]
So the expected num of kills per day is 1/4
\[P(X = k) = \frac{e^{-\lambda} \lambda^k}{k!}\]
To find probability of more than 2, we need to find prob of 0 kills and 1 kill
\[P(X \ge 2) = 1 - [P(X=0) + P(X=1)]\]

Answer 5

\(\lambda =200/800=1/4=0.25\)

and yeah then we subtract probability of getting 0 and 1 from 1
so
\(P(0)=\large\frac{(0.25)^0 \times e^{-0.25}}{0!}\)

\(P(1)=\large\frac{(0.25)^1 \times e^{-0.25}}{1!}\)

P(needed)=1-0.7788007 -0.1947 =0.0264

Answer 6

Geez :b I underestimated you guys.

Answer 7

just learned this stuff today
thanks!

Answer 8

Nice! we must be in the same class then.

Answer 9

We started probability stuff 3 lectures ago and next up is conics