.A box contains four blue and three red marbles. Michelle picks four marbles at random from this box. If Z is the random variable representing the number of blue marbles picked from the box, construct the probability distribution of the random variable Z

Solution:

four blue and three red marbles

Total = 4+3 = 7

Z = the number of blue marbles picked

Z = {0,1,2,3,4}

$p=\frac 17, q=\frac 67, n=4$ (p is the success getting a blue marble).

$P(Z=0)=^4C_0(\frac 17)^0(\frac 67)^4=\frac {1296}{2401} \\ P(Z=1)=^4C_1(\frac 17)^1(\frac 67)^3=\frac {864}{2401} \\ P(Z=2)=^4C_2(\frac 17)^2(\frac 67)^2=\frac {216}{2401} \\P(Z=3)=^4C_3(\frac 17)^3(\frac 67)^1=\frac {24}{2401} \\ P(Z=4)=^4C_4(\frac 17)^4(\frac 67)^0=\frac {1}{2401}$

Thus, the probability distribution of Z is:

Z P(Z)

0 1296/2401

1 864/2401

2 216/2401

3 24/2401

4 1/2401