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melamori03 [73]
2 years ago
7

There are 26 prize tickets in a bowl, labeled A to Z. What is the probability that a prize ticket with a vowel will be chosen, n

ot replaced, and then another prize ticket with a vowel will be chosen? Does this represent an independent or dependent event? Explain.
Mathematics
1 answer:
AfilCa [17]2 years ago
7 0

The probability that a prize ticket with a vowel will be chosen, not replaced, and then another prize ticket with a vowel will be chosen is 0.031 or 3.1%. The events are dependent events.

Given that there are five vowels, which come from the letters A, E, I, O, and U, the likelihood of drawing a vowel is:

5/26

since there are a total of 26 alternatives available. After choosing that, we are left with 25 alternatives, 4 of which are vowels, thus the probability is:

4/25

The final likelihood is thus:

5/26 * (4/25) = 0.031

In other words, the likelihood is 3.1% when choosing a vowel and then another (without replacement).

The first event has an impact on the second event since there are fewer vowels and overall possibilities, hence the events are dependent.

Learn more about probability here :

brainly.com/question/16812448

#SPJ1

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Answer:

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100 π inches^3

Step-by-step explanation:

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4 0
3 years ago
heights of statistics students were obtained by a teacher as part of an experiment conducted for the class. The last digit of th
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Answer:

1) B. The height appear to be reported because there are disproportionately more 0s and 5s.

2) A. They are likely not very accurate because they appear to be reported.

Step-by-step explanation:

The distribution table is shown below:

Last Digit           Frequency

     0                          9

     1                           1

     2                          1

     3                          3

     4                          1

     5                         11

     6                          1

     7                          0

     8                          3

     9                          1

1. Based on the distribution table, we see a very disproportionate distribution. There is a high frequency of 0's and 5's. This lays credence to the heights being reported rather than measured. As such, option B is the correct answer

<u>B. The height appear to be reported because there are disproportionately more 0s and 5s</u>.

2. Since the heights were reported and not measured, they are most certainly not accurate. The conclusion is that the result is not accurate. As such, option A is the correct answer

<u>A. They are likely not very accurate because they appear to be reported</u>.

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4 years ago
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