Question

Given three balls and four boxes. Assume the boxes are numbered 1, 2 ,3 and 4.Place the ball one by one into the boxes randomly. Among the boxes that have at least aball, letXdenote the smallest number associated with one of these boxes. For example,X= 3 means box 1 and box 2 are empty while box 3 has at least a ball. Please calculateE(X).

Answer 1

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

$E(X)=\sum^{n}_{i=1}p_ix_i\\E(X)=\sum^{4}_{i=1}p_ix_i\\E(X)=x_1p_1+x_2p_2+x_3p_3+x_4p_4\\E(X)=1\times\dfrac{37}{64}+2\times\dfrac{19}{64}+3\times\dfrac{7}{64}+4\times\dfrac{1}{64}\\E(X)=\dfrac{25}{16}=1.56$

So the value of E(X) is 1.56