Round all answers to 4 decimal places.a. A bag contains 4 black marbles, 10 white marbles, and 9 red marbles. If a marble is drawn from the
bag, replaced, and another marble is drawn, what is the probability of drawing first a black marble and
then a red marble?
b. A bag contains 10 blue marbles, 9 red marbles, and 4 white marbles. If two different marbles
are drawn from the bag , what is the probability of drawing first a blue marble and then a white marble?

Answer:

a. P(A) = P(B)

c. P(A) = 1 - P(B)

a and c are true . The rest are false.

Step-by-step explanation:

Two events A and B are said to be equally likely when one event is as likely to occur as the other. In other words each event should occur in equal number in repeated trials. For example when a fair coin is tossed the head is likely to appear as the tail, and the proportion of times each side is expected to appear is 1/2.

So when the events A= {1,3,4} B = {2,4,5} are equally likely then suppose their probability is 1/2.

a. P(A) = P(B)   True

1/2= 1/2

b. P(A) = 2P(B)  False

1/2 is not equal to 1

c. P(A) = 1 - P(B)  True

1/2= 1-1/2= 1/2

d. P(A) + P(B) > 1   False

1/2 + 1/2 is not greater than 1

e. P(A) - P(B) < 0   False

1/2-1/2= 0  is not less than 0

f. P(A) - P(B) > 1   False

1/2-1/2= 0 is not greater than 1

Final answer:

The relationships between the probabilities are evaluated and explained.

Explanation:

a. P(A) = P(B) could possibly hold if P(A) = 1/3 and P(B) = 1/3.

b. P(A) = 2P(B) could not hold, as probabilities cannot exceed 1.

c. P(A) = 1 - P(B) could possibly hold if P(A) = 2/3 and P(B) = 1/3.

d. P(A) + P(B) > 1 could possibly hold if P(A) = 1/3 and P(B) = 1/2.

e. P(A) - P(B) < 0 could not hold, as the difference between probabilities cannot be negative.

f. P(A) - P(B) > 1 could not hold, as the difference between probabilities cannot exceed 1.

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Answer: 1:6

Step-by-step explanation:

Reduce each side of the ratio

Answer:

For every 1 squirrel there are 6 acorns

Step-by-step explanation:

u just divide 18 by 3

Answer:

Lola used 3/4 more ink cartridges than her friends

Step-by-step explanation:

Answer:

See explanation below

Step-by-step explanation:

Here a coin was tossed three times.

Let H = head &  T = tail

Find the following:

a) The sample space:

Since a coin is tossed thrice, all possible outcome would be:

S = { HHH, HHT, HTH, HTT, TTT, TTH, THH, THT}

b) i) A = Exactly 2 tails: Here exactly 2 tails were recorded.

A = {HTT, TTH, THT}

ii) B = at least two tails: Here 2 or more tails were recorded.

B = {HTT, TTT, TTH, THT}

iii) C = the last two tosses are heads:

C = { HHH, THH}

c) List the elements of the following events:

i) A. This means all outcomes in A

= {HTT, TTH, THT}

ii) A∪B. A union B, means all possible outcomes present in A or B or in both

= {HTT, TTH, THT, TTT}

iii) A∩B. This means all possible outcomes of A that are present in B.

= {HTT, TTH, THT}

iv) A∩C. All outcomes A that are present in B

= {∅}

The sample space of tossing a coin three times consists of eight possible outcomes: HHH, HHT, HTH, THH, TTH, THT, HTT, and TTT. Events A, B, and C can be determined by listing the appropriate outcomes. The intersection and union of events A and B can also be determined.

(a) The sample space, Ω, of tossing a coin three times can be determined by listing all the possible outcomes: HHH, HHT, HTH, THH, TTH, THT, HTT, and TTT.

(b) i. A = {HHT, HTH, THH}

ii. B = {TTT, TTH, THT, HTT, HHT, HTH, THH}

iii. C = {HTH, TTH}

(c) i. A = {HHT, HTH, THH}

ii. A∪B = {HHT, HTH, THH, TTT, TTH, THT, HTT, HHT}

iii. A∩B = {HHT, HTH, THH}

iv. A∩C = {HHT, HTH}

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Answer: B

Step-by-step explanation: