Question

question 37
4.2 theoretical probability. mutually exclusive/not mutually exclusive.
a jar contains 22 red marbles numbered 1 to 22 and 42 blue marbles numbered 1 to 42. 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)=
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Explanation:

Step1: Calculate total number of marbles

The total number of marbles is the sum of red and blue marbles. So, $22 + 42=64$.

Step2: Calculate P(red)

The probability of drawing a red marble is the number of red marbles divided by the total number of marbles. So, $P(\text{red})=\frac{22}{64}=\frac{11}{32}$.

Step3: Calculate number of odd - numbered marbles

Among the 22 red marbles, there are 11 odd - numbered ones (1, 3, …, 21). Among the 42 blue marbles, there are 21 odd - numbered ones (1, 3, …, 41). So the total number of odd - numbered marbles is $11 + 21=32$. Then $P(\text{odd})=\frac{32}{64}=\frac{1}{2}$.

Step4: Calculate P(red and odd)

The number of red and odd - numbered marbles is 11. Using the formula $P(A\cup B)=P(A)+P(B)-P(A\cap B)$, for $A$ = "red" and $B$ = "odd", $P(\text{red or odd})=P(\text{red})+P(\text{odd})-P(\text{red and odd})=\frac{22}{64}+\frac{32}{64}-\frac{11}{64}=\frac{22 + 32-11}{64}=\frac{43}{64}$.

Step5: Calculate P(blue and even)

The number of blue marbles is 42 and the number of even - numbered blue marbles is 21. The number of even - numbered red marbles is 11. The total number of marbles is 64. First, $P(\text{blue})=\frac{42}{64}$, $P(\text{even})=\frac{32}{64}$, $P(\text{blue and even})=\frac{21}{64}$. Then $P(\text{blue or even})=P(\text{blue})+P(\text{even})-P(\text{blue and even})=\frac{42}{64}+\frac{32}{64}-\frac{21}{64}=\frac{42 + 32-21}{64}=\frac{53}{64}$.

Answer:

(a) $\frac{11}{32}$
(b) $\frac{1}{2}$
(c) $\frac{43}{64}$
(d) $\frac{53}{64}$