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

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

$P(X=9)\ i.e. 3\ odd\ balls\ and\ 0\ even\ balls=\frac{^{14}C_{3}}{^{28}C_3}=\frac{1}{9}$

$P(X=10)\ i.e. 2\ odd\ balls\ and\ 1\ even\ balls=\frac{^{14}C_{2}\times ^{14}C_1}{^{28}C_3}=\frac{7}{18}$

$P(X=11)\ i.e. 1\ odd\ balls\ and\ 2\ even\ balls=\frac{^{14}C_{1}\times ^{14}C_2}{^{28}C_3}=\frac{7}{18}$

$P(X=12)\ i.e. 0\ odd\ balls\ and\ 3\ even\ balls=\frac{^{14}C_{3}}{^{28}C_3}=\frac{1}{9}$