Answer:
The value of E(X) is 1.56
Step-by-step explanation:
Number of Boxes=4
Number of Balls =3
If X is the number of box in which the balls are than the number of balls in the box are given as ((4-X)+1)^3
Number of ways such that all the balls are in the box number 4, no balls in the first three boxes is given as n(x=4)=1^3=1
Number of ways such that all the balls are in the box number 3, is given as n(x=3)=n(x=2)-n(x=4)=2^3-1^3=7
Number of ways such that all the balls are in the box number 2, is given as n(x=2)=n(x=1)-n(x=3)=3^3-2^3=19
Number of ways such that all the balls are in the box number 1, is given as n(x=1)=n(x=4)-n(x=3)=4^3-3^3=37
So the total number of ways are
n(total)=n(x=4)+n(x=3)+n(x=2)+n(x=1)
n(total)=1+7+19+37=64
So Probabilities are given as
p(x=1)=n(x=1)/n(total)=37/64
p(x=2)=n(x=2)/n(total)=19/64
p(x=3)=n(x=3)/n(total)=7/64
p(x=4)=n(x=4)/n(total)=1/64
So the E(x) is given as
[tex]E(X)=\sum^{n}_{i=1}p_ix_i\\E(X)=\sum^{4}_{i=1}p_ix_i\\E(X)=x_1p_1+x_2p_2+x_3p_3+x_4p_4\\E(X)=1\times\dfrac{37}{64}+2\times\dfrac{19}{64}+3\times\dfrac{7}{64}+4\times\dfrac{1}{64}\\E(X)=\dfrac{25}{16}=1.56[/tex]
So the value of E(X) is 1.56
