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Mice21 [21]
3 years ago
10

A carnival game allows a group of players to each draw and keep a marble from a bag. The bag contains 5 gold marbles, 25 silver

marbles, and 70 red marbles.
A player wins a large prize for drawing a gold marble and a small prize for drawing a silver marble. There is no prize for drawing a red marble.
At the start of the game, the probability of winning a large prize is 0.05 and the probability of winning a small prize is 0.25.
Suppose that the first player draws a silver marble and wins a small prize. What is the probability that the second player will also win a small prize?
If a group of four plays the game one at a time and everyone wins a small prize, which player had the greatest probability of winning a large prize?
How could the game be made fair for each player? That is, how could you change the game so that each player has an equal chance of winning a prize?
Mathematics
2 answers:
WINSTONCH [101]3 years ago
8 0
The probability is now .24

The first player had the highest probability of getting a silver marble 

The game could be made fair by adding back the marble that was drawn after each draw

I Hope this helped :)
I'm truly sorry if it didn't :(

I Hope you have the most amazing-est day :) 
BabaBlast [244]3 years ago
4 0

Answer:

you can give that guy brainliest now

Step-by-step explanation:

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Ostrovityanka [42]

Answer:

a) The study is an observational study

b) The proportion of children with significant social activity in children with acute lymphoblastic leukemia is approximately 0.802

The proportion of children with significant social activity in children without acute lymphoblastic leukemia is approximately 0.8565

c) The odds ratio is approximately 0.6780

d) The 95% CI is approximately 0.523 < OR < 0.833

e) i) Yes

ii) Children with more social activity have a higher occurrence of acute lymphoblastic leukemia

Step-by-step explanation:

a) An experimental study is a study in which the researcher adds inputs to the study and then monitors the outcomes

An observational study is one in which the researcher observes risk factors and does not intervene in the process

Therefore, the study is an observational study

b) The proportion of children with significant social activity in children with acute lymphoblastic leukemia is \hat p_1 = 1020/1272 = 85/106 = 0.8\overline {0188679245283} ≈ 0.802

The proportion of children with significant social activity in children without acute lymphoblastic leukemia is \hat p_2 = 5343/6238 ≈ 0.8565

c) We have the following two way table;

\begin{array}{ccc}Lymphoblastic \ Leukemia &With \ Social \ Activity & Without \ Social \ Activity\\With&1020 \ (a)&252 \ (b)\\Without&5343 \ (c)&895 \ (d) \end{array}

The \ odds \ ratio = \dfrac{1,020 \times 895}{5,343 \times 252} \approx 0.6780

The odds ratio ≈ 0.6780

d) The 95% confidence interval for the odds ratio is given as follows;

95 \% \ CI = OR \ \pm \ 1.96 \times \sqrt{\dfrac{1}{a} +\dfrac{1}{b}+\dfrac{1}{c}+\dfrac{1}{d}}

Where;

a = 1020, b = 252, c = 5343, d = 895

Therefore;

95 \% \ CI \approx  0.6780 \ \pm \ 1.96 \times \sqrt{\dfrac{1}{1,020} +\dfrac{1}{252}+\dfrac{1}{5,343}+\dfrac{1}{895}} \approx 0.678 \pm0.155

The 95% CI ≈ 0.523 < OR < 0.833

e) i) Given that the observed Odds Ratio is within the Confidence Interval, therefore, there is an indication that the amount of social activity is associated with acute lymphoblastic leukemia

ii) Based on the proportion of the study findings, children with more social activity have a higher occurrence of acute lymphoblastic leukemia

6 0
2 years ago
I begin the billing cycle with a $4,000 balance on my credit card. On the 13th day I pay off
IRISSAK [1]
$2,900 you have to subtract from the $4,000 and it is $1,100 you have left
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2 years ago
If you roll a die three times how many different sequences are possible?
Rama09 [41]
216 sequences are possible
8 0
3 years ago
Helen's house is located on a rectangular lot that is1 1-8 miles by 9/10 mile.Estimate the distance around the lot
klemol [59]

Answer: The distance around the lot is given by

4\frac{1}{20}\ miles

Step-by-step explanation:

Since we have given that

Length of rectangular lot is given by

1\frac{1}{8}\\\\=\frac{9}{8}\ miles

Width of rectangular lot is given by

\frac{9}{10}\ mile

We need to find the distance around the lot.

As we know the formula for "Perimeter of rectangle":

Perimeter=2(Length+Width)\\\\Perimeter=2(\frac{9}{8}+\frac{9}{10})\\\\Perimeter=2(\frac{45+36}{40})\\\\Perimeter=2\times \frac{81}{40}\\\\Perimeter=\frac{81}{20}\\\\Perimeter=4\frac{1}{20}\ miles

Hence, the distance around the lot is given by

4\frac{1}{20}\ miles


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3 years ago
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ANTONII [103]
25 to 75 in lowest terms:

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