Question

determine whether the event described is mutually exclusive or not. then determine the probability of the event. type these as fractions.
a) a die is rolled. what is the probability of rolling 6 or 3?
choose: select an answer
probability: select an answer
mutually exclusive
not mutually exclusive
b) a die is rolled. what is the probability of rolling at least 3?
choose: select an answer
probability:
c) a card is drawn from a standard deck of cards. what is the probability of drawing 10 or jack?
choose: select an answer
probability:
d) a card is drawn from a standard deck of cards. what is the probability of drawing ace or spade?
choose: select an answer
probability:

Explanation:

Step1: Define mutually - exclusive events

Two events are mutually exclusive if they cannot occur at the same time.

Step2: Analyze part a

When rolling a die, getting a 6 and getting a 3 cannot happen simultaneously. So, it's mutually exclusive. The probability of rolling a 6 is $\frac{1}{6}$, and the probability of rolling a 3 is $\frac{1}{6}$. Using the formula for mutually - exclusive events $P(A\cup B)=P(A)+P(B)$, we have $P(6\ or\ 3)=\frac{1}{6}+\frac{1}{6}=\frac{2}{6}=\frac{1}{3}$.

Step3: Analyze part b

Rolling a die, the event of rolling at least 3 means getting 3, 4, 5, or 6. These outcomes are not mutually exclusive to each other in the sense of the question's context. The total number of possible outcomes is 6. The number of favorable outcomes (3, 4, 5, 6) is 4. So the probability $P=\frac{4}{6}=\frac{2}{3}$.

Step4: Analyze part c

When drawing a card from a standard deck of 52 cards, a card cannot be a 10 and a Jack at the same time. So, it's mutually exclusive. There are 4 tens and 4 Jacks. The probability of drawing a 10 is $\frac{4}{52}$, and the probability of drawing a Jack is $\frac{4}{52}$. Using the formula for mutually - exclusive events $P(A\cup B)=P(A)+P(B)$, we get $P(10\ or\ Jack)=\frac{4 + 4}{52}=\frac{8}{52}=\frac{2}{13}$.

Step5: Analyze part d

In a standard deck of 52 cards, an Ace of Spades is both an Ace and a Spade, so the events of drawing an Ace and drawing a Spade are not mutually exclusive. There are 4 Aces, 13 Spades, and 1 Ace of Spades. Using the formula $P(A\cup B)=P(A)+P(B)-P(A\cap B)$, we have $P(A)=\frac{4}{52}$, $P(B)=\frac{13}{52}$, $P(A\cap B)=\frac{1}{52}$. Then $P(Ace\ or\ Spade)=\frac{4 + 13-1}{52}=\frac{16}{52}=\frac{4}{13}$.

Answer:

a) Mutually Exclusive, $\frac{1}{3}$
b) Not Mutually Exclusive, $\frac{2}{3}$
c) Mutually Exclusive, $\frac{2}{13}$
d) Not Mutually Exclusive, $\frac{4}{13}$