Question

4.2 theoretical probability. mutually exclusive/not mutually exclusive. a jar contains 32 red marbles numbered 1 to 32 and 52 blue marbles numbered 1 to 52. a marble is drawn at random from the jar. find the probability of the given event. type as a fraction. (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) = question help: message instructor

Explanation:

Step1: Calculate total number of marbles

The total number of marbles is the sum of red and blue marbles. So, $32 + 52=84$.

Step2: Calculate P(red)

The probability of getting a red marble is the number of red marbles divided by the total number of marbles. So, $P(\text{red})=\frac{32}{84}=\frac{8}{21}$.

Step3: Calculate number of odd - numbered marbles

Among 32 red marbles, 16 are odd - numbered, and among 52 blue marbles, 26 are odd - numbered. So the total number of odd - numbered marbles is $16 + 26 = 42$. Then $P(\text{odd})=\frac{42}{84}=\frac{1}{2}$.

Step4: Calculate P(red or odd)

Use the formula $P(A\cup B)=P(A)+P(B)-P(A\cap B)$. $P(\text{red})=\frac{32}{84}$, $P(\text{odd})=\frac{42}{84}$, and $P(\text{red}\cap\text{odd})=\frac{16}{84}$. Then $P(\text{red or odd})=\frac{32 + 42-16}{84}=\frac{58}{84}=\frac{29}{42}$.

Step5: Calculate P(blue or even)

First, find $P(\text{blue})=\frac{52}{84}$, $P(\text{even}) = 1 - P(\text{odd})=1-\frac{1}{2}=\frac{1}{2}$, and $P(\text{blue}\cap\text{even})=\frac{26}{84}$. Then $P(\text{blue or even})=\frac{52+42 - 26}{84}=\frac{68}{84}=\frac{17}{21}$.

Answer:

(a) $\frac{8}{21}$
(b) $\frac{1}{2}$
(c) $\frac{29}{42}$
(d) $\frac{17}{21}$