Question

a jar contains 6 red marbles numbered 1 to 6 and 12 blue marbles numbered 1 to 12. a marble is drawn at random from the jar. find the probability of the given event, please show your answers as reduced fractions.
(a) the marble is red. p(red)=
(b) the marble is odd - numbered. p(odd)=
(c) the marble is red or odd - numbered. p(red or odd)=
(d) the marble is blue or even - numbered. p(blue or even)=

Explanation:

Step1: Calculate total number of marbles

The jar has 6 red marbles and 12 blue marbles. So the total number of marbles is $6 + 12=18$.

Step2: Calculate P(red)

The number of red marbles is 6. Using the probability formula $P(A)=\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}}$, we have $P(\text{red})=\frac{6}{18}=\frac{1}{3}$.

Step3: Calculate number of odd - numbered marbles

Among the 6 red marbles, 3 are odd - numbered (1, 3, 5). Among the 12 blue marbles, 6 are odd - numbered (1, 3, 5, 7, 9, 11). So the total number of odd - numbered marbles is $3 + 6=9$. Then $P(\text{odd})=\frac{9}{18}=\frac{1}{2}$.

Step4: Calculate P(red and odd)

The red and odd - numbered marbles are 3 (red marbles numbered 1, 3, 5). So $P(\text{red and odd})=\frac{3}{18}=\frac{1}{6}$.

Step5: Calculate P(red or odd)

Using the formula $P(A\cup B)=P(A)+P(B)-P(A\cap B)$, we substitute $A$ as the event of getting a red marble and $B$ as the event of getting an odd - numbered marble. So $P(\text{red or odd})=P(\text{red})+P(\text{odd})-P(\text{red and odd})=\frac{6}{18}+\frac{9}{18}-\frac{3}{18}=\frac{6 + 9-3}{18}=\frac{12}{18}=\frac{2}{3}$.

Step6: Calculate P(blue)

The number of blue marbles is 12. So $P(\text{blue})=\frac{12}{18}=\frac{2}{3}$.

Step7: Calculate P(even)

The number of even - numbered marbles is $18 - 9 = 9$. So $P(\text{even})=\frac{9}{18}=\frac{1}{2}$.

Step8: Calculate P(blue and even)

Among the 12 blue marbles, 6 are even - numbered (2, 4, 6, 8, 10, 12). So $P(\text{blue and even})=\frac{6}{18}=\frac{1}{3}$.

Step9: Calculate P(blue or even)

Using the formula $P(A\cup B)=P(A)+P(B)-P(A\cap B)$, we substitute $A$ as the event of getting a blue marble and $B$ as the event of getting an even - numbered marble. So $P(\text{blue or even})=P(\text{blue})+P(\text{even})-P(\text{blue and even})=\frac{12}{18}+\frac{9}{18}-\frac{6}{18}=\frac{12 + 9-6}{18}=\frac{15}{18}=\frac{5}{6}$.

Answer:

(a) $\frac{1}{3}$
(b) $\frac{1}{2}$
(c) $\frac{2}{3}$
(d) $\frac{5}{6}$