Question

Imagine that you are about to play 10 rounds at a roulette table. Also assume that the roulette table has 36 numbers, thus the probability of success in each round is 1/36 = 0.02778. What is the probability that if you play 10 rounds you will win at least TWICE?

Answer 1

Solution:

Here, we have to use binomial distribution for finding a required probability.

We are given

Sample size = n = 10

Probability = p = 1/36 = 0.02778

We have to find P(X?2)

P(X?2) = 1 – P(X<2) = 1 – P(X?1)

P(X?1) = P(X=0) + P(X=1)

Now, we have to find P(X=0)

P(X=x) = nCx*p^x*(1 – p)^(n – x)

Where, nCx = n!/x!(n – x)!

Where, n! = 1*2*3*.....*(n – 1)*n

Using above formulas,

P(X=0) = 10C0*0.02778^0*(1 – 0.02778)^(10 – 0)

P(X=0) = 1*1* 0.97222^10

P(X=0) = 0.754476

Now, we have to find P(X=1)

P(X=1) = 10C1*0.02778^1*(1 – 0.02778)^(10 – 1)

P(X=1) = 10*0.02778*0.97222^9

P(X=1) = 0.215582

P(X?1) = P(X=0) + P(X=1)

P(X?1) = 0.754476 + 0.215582

P(X?1) = 0.970058

P(X?2) = 1 – P(X?1)

P(X?2) = 1 – 0.970058

P(X?2) = 0.029942

Required probability = 0.029942