A roulette wheel has 38 numbers of which 18 are read 18 are

A roulette wheel has 38 numbers of which 18 are read, 18 are black, and 2 are green. When the roulette wheel is spun, the ball is equally likely to fall on any of the 38 numbers, and the outcome from one spin of the wheel is independent of the outcome of another spin of the wheel.

Suppose the wheel is spun 10 times.

a. Find the probability of obtaining five reads, three blacks, and two greens.

b. Find the probability of obtaining five reds, three blacks, and two greens if the first four spins of the wheel result in all reds.

Solution

A roulette wheel has 38 numbers of which 18 are read, 18 are black, and 2 are green and the ball is equally likely to fall on any of the 38 numbers.

let X1 be the random variable denoting the number of reds obtained in n spins of the wheel

    X2 be the random variable denoting the number of blacks obtained in n spins of the wheel

    X3 be the random variable denoting the number of greens obtained in n spins of the wheel

since the outcome from one spin of the wheel is independent of the outcome of another spin of the wheel.

(X1,X2,X3) jointly follows a multinomial distributions with parameters (n,p1,p2,p3)

where n=total number of spins

p1=probability of obtaining a red=18/38

p2=probability of obtaining a black=18/38

p3=probability of obtaining a green=2/38

now the pmf is

f(x1,x2,x3)=n!/(x1!*x2!*x3!)*p1^x1p2^x2p3^x3          such that x1+x2+x3=n

a) here n=10 x1=5 x2=3 x3=2

then the probability of obtaining five reads, three blacks, and two greens is f(5,3,2)=10!/(5!*3!*2!)*p1⁵p2³p3²=0.017639 [answer]

b) the probability of obtaining five reads, three blacks, and two greens if the first four spins result in all reds

this means that from the rest 10-4=6 spins 1 red,three black and two greens are obtained.

since the outcome from one spin of the wheel is independent of the outcome of another spin of the wheel the required probability is

p1⁴*6!/(1!*3!*2!)*p1¹p2³p3²=0.00042 [answer]