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ankoles
2 months ago
14

The probability of drawing two aces from a standard deck is 0.0059. We know this probability, but we don't know if the first car

d was replaced. If the two draws are defined as event A and event B, are the events dependent or independent?
A. They are dependent because, based on the probability, the first ace was replaced before drawing the second ace.
B. They are dependent because, based on the probability, the first ace was not replaced before drawing the second ace.
C. They are independent because, based on the probability, the first ace was replaced before drawing the second ace.
D. They are independent because, based on the probability, the first ace was not replaced before drawing the second ace.
Mathematics
1 answer:
Inessa [12.5K]2 months ago
4 0

Answer:Option C is correct.

C. The events are independent because the first ace was replaced before drawing the second ace.

Step-by-step explanation:

It is given that the probability of drawing two aces from a standard deck is 0.0059

If the first card is drawn and replaced, it alters the probability. Performing draws with replacement makes each event independent of one another

The chance of getting an ace on the first draw is 4/52, and with the replacement, the probability remains 4/52 for the second draw

Therefore, if we consider the first and second draws as event A and event B respectively, these events are independent

C. The events are independent because the first ace was replaced before drawing the second ace.

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According to a survey by Bankrate, of adults in the United States save nothing for retirement (CNBC website). Suppose that adult
tester [12383]

Complete Question

The complete question appears in the first uploaded image

Answer:

a) Yes, selecting 15 corresponds to a binomial experiment

b)

c)

d) P(r = 15) = 3.2768 *10^{-11}

Step-by-step explanation:

Regarding question a:

For an experiment to qualify as binomial

the trials have to be independent

each trial must yield one of two possible outcomes

Given that the selection of 15 individuals is random, we ascertain that the trials are independent and the outcomes are “either the individual saves for retirement or does not save for retirement.”

Therefore, we conclude that the selection of 15 people at random is indeed a binomial experiment.

In question b:

The probability that all selected adults do not save for retirement is mathematically modeled as

P(r = n) = ^nC_r * p^r * q^{n-r}

Here C signifies combination

r = 15 implies all selected adults

n refers to the population size equating to 15

From the problem, p = 0.20

and q can be calculated as

=>

=> q = 1 - p

Thus

P(r = 15) = ^{15}C_{15} * p^{15} * q^{15-15}

P(r = 15) = 3.2768 *10^{-11} Regarding question c:

The probability that exactly five of the selected adults do not save for retirement is mathematically modeled as

P(r = 5) = ^{15} C_5 * (0.20)^5 * (0.80)^{15}

P(r = 5) = 0.1032

In relation to question d:

The probability that at least one of the selected adults opts not to save for retirement can be mathematically expressed as

P(r \ge 1 ) = 1 - P (r = 0 )

P(r \ge 1 ) = 1 - [ ^{15} C _ 0 * (0.20)^{0} * (0.80 )^{15}]

P(r \ge 1 ) = 1 - 0.0352

P(r \ge 1 ) = 0.9648

4 0
3 months ago
Elias’s teacher has given him a rectangle that is 15 feet by 20 feet. She has asked him to create a new rectangle using a scale
Inessa [12570]
No, as the fraction equates to a whole number. Thus, it would actually be an increase.
8 0
3 months ago
Read 2 more answers
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