Question

Involving selecting balls from a box of numbered balls. Assume that the box contains balls numbered from 1 through 28, and that 3 are selected. A random variable X is defined as 3 times the number of odd balls selected, plus 4 times the number of even.How many different values are possible for the random variable X?

Answer 1

There are 4 different values possible for the random variable X which is defined as 3 times the number of odd balls selected plus 4 times the number of even balls when selecting 3 balls from a box containing balls numbered from 1 through 28.

Step-by-step explanation:

To determine how many different values are possible for the random variable X in the scenario of selecting balls from a box, we have to consider the total outcomes of selecting odd and even numbered balls. Since the box contains balls numbered from 1 through 28, we can infer there are 14 even and 14 odd balls. When three balls are selected, we can have the following scenarios: all odd balls (OOO), all even balls (EEE), two odd and one even (OOE), two even and one odd (EEO), and any combination of odd and even balls selected. Each selection scenario contributes to a different value of X calculated as 3 times the number of odd balls selected plus 4 times the number of even balls selected.

Let's illustrate all possibilities:

OOO: X=3*3+4*0=9 (All three are odd)

OOE or EOO or OEO: X=3*2+4*1=10 (Two are odd, one is even)

EE0 or EEO or OEE: X=3*1+4*2=11 (One is odd, two are even)

EEE: X=3*0+4*3=12 (All three are even)

Hence, there are 4 different values possible for X based on the selection of balls.

Answer 2

Answer:4

Step-by-step explanation:

Let a denotes the no of odd ball selected and b denotes the no of even ball selected

then
`X=3 a+4 b`

for
`a=0,b=3, X=12`

for
`a=1,b=2, X=11`

for
`a=2,b=1, X=10`

for
`a=3,b=0, X=9`

thus X can take 4 values i.e. 9,10,11 and 12

`P(X=9)\ i.e. 3\ odd\ balls\ and\ 0\ even\ balls=(^(14)C_(3))/(^(28)C_3)=(1)/(9)`

`P(X=10)\ i.e. 2\ odd\ balls\ and\ 1\ even\ balls=(^(14)C_(2)* ^(14)C_1)/(^(28)C_3)=(7)/(18)`

`P(X=11)\ i.e. 1\ odd\ balls\ and\ 2\ even\ balls=(^(14)C_(1)* ^(14)C_2)/(^(28)C_3)=(7)/(18)`

`P(X=12)\ i.e. 0\ odd\ balls\ and\ 3\ even\ balls=(^(14)C_(3))/(^(28)C_3)=(1)/(9)`