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OleMash [197]
4 years ago
9

A bowl contains 4 yellow marbles and 3 red marbles. Michael randomly draws 4 marbles from the bowl. He replaces each marble afte

r it is drawn. What is the probability that he draws 3 yellow marbles and 1 red marble.
A. 12/35
B. 3/35
C. 192/2401
D. 768/2401
Mathematics
1 answer:
ra1l [238]4 years ago
5 0
So each event is indipentent of each other
chance=(desired outcomes)/(total possible outcomes)
3+4=7
7=total outcomes
3red
4yellow
so each time there is a 3/7 chance of drawing a red marble and a 4/7 chance of
red=3/7
yellow=4/7

so 3 yellow marbles and 1 red=4/7 times 4/7 times 4/7 times 3/7=(4^3 times 3)/(7^4)=192/2401 which is C

the answer is C 192/2401
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1 a  p -value  =   0.030054

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Step-by-step explanation:

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    p -value  =  P(Z > 1.88) = 0.030054

Considering question b

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Considering question 2a

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     The sample size is  n=16

     The  test statistic is  t =  3.733

Generally the degree of freedom is mathematically represented as

        df =  n - 1

=>     df =  16 - 1

=>     df =  15

Generally from the t distribution table  the probability of   t =  3.733 at a degree of freedom of  df =  15 for a right tailed test is  

       p-value  =  t_{3.733 ,  15} = 0.00099966

Considering question 2b

    The alternative hypothesis is H1:μ<μ0

     The degree of freedom is df=23

     The  test statistic is ,t= −2.500

Generally from the t distribution table  the probability of   t= −2.500 at a degree of freedom of  df=23 for a left  tailed test is  

       p-value  =  t_{-2.500 ,  23} = 0.00999706

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     The sample size is  n= 7

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       t_{-2.2500 , 6} = 0.03272060

Generally the p-value  for t= −2.2500 for a two tailed test is

     p-value  =  2 *  0.03272060 = 0.0654412

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