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inna [77]
3 years ago
12

A jar contains 8 marbles numbered 1 through 8. an experiment consists of randomly selecting a marble from the jar, observing the

number drawn, and then randomly selecting a card from a standard deck and observing the suit of the card (hearts, diamonds, clubs, or spades). how many outcomes are in the sample space for this experiment? how many outcomes are in the event "an even number is drawn?" how many outcomes are in the event "a number more than 2 is drawn and a red card is drawn?" how many outcomes are in the event "a number less than 3 is drawn or a club is not drawn?"
Mathematics
1 answer:
zepelin [54]3 years ago
3 0
There are 8 possible outcomes for a marble being drawn and numbered. 
{1,2,3,4,5,6,7,8}
There are 4 possible outcomes for a card being selected from a standard deck.
{ <span>hearts, diamonds, clubs, spades}
So the number of outcomes in the sample space would be 8 x 4 = 32.

In the event "an even number is drawn", there are only 4 possible outcomes for a marble being drawn, {2,4,6,8}, whereas there are still 4 possible outcomes for a suit. So the number of outcomes in the event is 4 x 4 = 16.

</span><span>In the event "a number more than 2 is drawn and a red card is drawn", there are 6 possible outcomes for the marble being drawn, {3,4,5,6,7,8}, whereas there are only two possible suits for a card being selected as red, {heart, diamond}. So the number of outcomes in this event is 6 x 2 = 12.

In the event </span><span>"a number less than 3 is drawn or a club is not drawn", the number drawn could be 1 or 2 whereas a spade/heart/diamond could be selected. So the number of outcomes is 2 x 3 = 6.</span><span>

</span>
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2 years ago
We roll two fair 6-sided dice. Each of the 36 possible outcomes is assumed to be equally likely. (a) Given that the roll results
boyakko [2]

Answer:

a) \frac{1}{7}

b) \frac{2}{15}

Step-by-step explanation:

Given : We roll two fair 6-sided dice. Each of the 36 possible outcomes is assumed to be equally likely.

The outcomes are :

(1, 1) (1, 2)  (1, 3)  (1, 4)  (1, 5)  (1, 6)

(2, 1) (2, 2) (2, 3) (2, 4) (2, 5) (2, 6)

(3, 1) (3, 2) (3, 3) (3, 4) (3, 5) (3, 6)

(4, 1) (4, 2) (4, 3) (4, 4) (4, 5) (4, 6)

(5, 1) (5, 2) (5, 3) (5, 4) (5, 5) (5, 6)

(6, 1) (6, 2) (6, 3) (6, 4) (6, 5) (6, 6)

(a) Given that the roll results in a sum of 3 or more, find the conditional probability that doubles (the first and the second rolls result in the same number) are rolled.

Let P(A) be the probability of event that the roll results in a sum of 3 or more.

Except (1,1) rest sum is 3 or greater than 3.

So, P(A)=\frac{35}{36}

Now, P(A and B) is that the roll results in a sum of 3 or more and that doubles (the first and the second rolls result in the same number) are rolled.

i.e. (2,2), (3,3), (4,4), (5,5), (6,6) - 5

P(A\cap B)=\frac{5}{36}

The conditional probability is given by,

P(B|A)=\frac{P(A\cap B)}{P(A)}

P(B|A)=\frac{\frac{5}{36}}{\frac{35}{36}}

P(B|A)=\frac{5}{35}

P(B|A)=\frac{1}{7}

(b) Given that the two dice land on different numbers, find the conditional probability that the sum is 5.

Let P(A) be the probability of event that two dice land on different numbers.

Except (1,1),(2,2), (3,3), (4,4), (5,5), (6,6) rest two dice land on different numbers.

So, P(A)=\frac{30}{36}

Now, P(A and B) is that two dice land on different numbers and the sum is 5.

i.e. (1,4), (2,3), (3,2), (4,1) - 4

P(A\cap B)=\frac{4}{36}

The conditional probability is given by,

P(B|A)=\frac{P(A\cap B)}{P(A)}

P(B|A)=\frac{\frac{4}{36}}{\frac{30}{36}}

P(B|A)=\frac{4}{30}

P(B|A)=\frac{2}{15}

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