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A jar contains 12 red marbles numbered 1 to 12 and 4 blue marbles numbered 1 to 4. a marble is drawn at random from the jar. find the probability of the given event. (a) the marble is red your answer is : 3/4 preview (b) the marble is odd-numbered your answer is : 1/2 preview (c) the marble is red or odd-numbered your answer is : 7/8 preview (d) the marble is blue and even-numbered your answer is : 5/8

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kritter1
a) P(getting red)= total number of red marbles possible/all marbles 
P(red)=12/(12+4) 
P(red)=12/16 
P(red)=3/4 

b) Between 1-12, 6 possible odd values. Between 1-4, 2 possible odd values 
P(odd number) =(6+2)/16 
P(odd number)=8/16
P(odd number)=1/2 

c) Find probability of getting red OR getting odd number (when 'OR' is used, you add the probabilities) 
P(odd or red)=(8/16)+(12/16) 
P(odd or red)=20/16 or 1.25 possibility of occurring 

d) Probability of getting blue OR getting even number; 

P(Blue)=1-(12/16)
P(Blue)=(16/16)-(12/16) 
P(Blue)=4/16 
P(Blue)=1/4 

P(even)=1-(1/2)
P(even)=1/2 

Add them together; 

P(odd or blue)=(1/4)+(1/2) 
P(odd or blue)=(1/4)+(2/4) 
P(odd or blue)=3/4 

Hope I helped :) 

Answer:

A jar contains 12 red marbles numbered 1 to 12

And 4 blue marbles numbered 1 to 4.

Total number of marbles = 12+4 =16

(a) A marble is drawn at random from the jar. The probability that the marble is red will be = [tex]\frac{12}{16}=\frac{3}{4}[/tex]

(b) In all, the odd number marbles are: 1,3,5,7,9,11 (in red) and 1,3(in blue) so total odd number marbles are = 8

So, probability that the marble is odd-numbered is : [tex]\frac{8}{16}=\frac{1}{2}[/tex]

(c) The probability that the marble is red or odd-numbered your answer is :

[tex]\frac{12}{16}+\frac{8}{16}-\frac{6}{16}=\frac{14}{16}=\frac{7}{8}[/tex]

(d) The probability that the marble is blue and even-numbered =

[tex]\frac{4}{16}+\frac{8}{16}-\frac{2}{16}=\frac{10}{16}=\frac{5}{8}[/tex]

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