Respuesta :
Answer:
(a)P(Red)= [TeX]\frac{1}{3}[/TeX]
(b)P(odd)=[TeX]\frac{1}{2}[/TeX]
(c)P(red or odd)=[TeX]\frac{2}{3}[/TeX]
(d) P(blue or even)=[TeX]\frac{5}{6}[/TeX]
Step-by-step explanation:
The contents of the jar are:
6 red marbles numbered 1 to 6.
12 blue marbles numbered 1 to 12.
Total=6+12=18
(a)Probability that the marble is Red
P(Red)= [TeX]\frac{6}{18}=\frac{1}{3}[/TeX]
(b)The marble is odd-numbered.
Number of Odd Numbered Red Marbles(1,3,5)=3
Number of Odd Numbered Blue Marbles(1,3,5,7,9,11)=6
Total Odd Numbered Marbles=3+6=9.
P(odd)=[TeX]\frac{9}{18}=\frac{1}{2}[/TeX]
(c)The marble is red or odd-numbered.
Number of Red Marbles=6
Number of Odd Numbered Marbles=9
Number of Red Odd Numbered Marbles=3
The two events are not mutually exclusive, therefore:
P(red or odd)=P(Red)+P(Odd)-P(Red and Odd)
=[TeX]\frac{6}{18}+\frac{9}{18}-\frac{3}{18}=\frac{12}{18}=\frac{2}{3}[/TeX]
(d) The marble is blue or even-numbered.
Number of Blue Marbles=12
Number of Even Numbered Marbles=6+3=9
Number of Blue Even Numbered Marbles=6
The two events are not mutually exclusive, therefore:
P(blue or even)=P(blue)+P(even)-P(blue and even)
=[TeX]\frac{12}{18}+\frac{9}{18}-\frac{6}{18}=\frac{15}{18}=\frac{5}{6}[/TeX]
