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ANTONII [103]
3 years ago
15

Ryan’s mom randomly chooses two days each week for Ryan to do his chores. What is the probability that she picks Saturday and Su

nday?
Mathematics
2 answers:
torisob [31]3 years ago
4 0
There are a total of 7C2 of options for Ryan's mom to choose.

If Ryan's mom picks Saturday and Sunday, this is the only 1 option in those 7C2 of options. Thus, the propability is (1 / 7C2) × 100% = 4.762%.
Alona [7]3 years ago
4 0

Answer:

Probability that she picks Saturday and Sunday = 1/21 = 0.0476

Step-by-step explanation:

Total number of days in a week = 7

Number of days selected = 2

Total number of combinations = 7C₂ =\frac{7*6}{1*2} =21

Total number of favorable ways = 1 that is  Sunday and  Saturday.

Probability that she picks Saturday and Sunday = 1/21 = 0.0476

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Complete Question

The complete question is shown on the first uploaded image

Answer:

a

The probability that he or she would choose the pair of socks in the center position is   p =\frac{1}{5}

The correct answer choice is

X has a binomial distribution with parameters n=100 and p=1/5  

b

The mean is  \mu = 20

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c

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d

The correct answer is

The experiment supports the center stage effect. If participants were truly picking the socks at random, it would be highly unlikely for 34 or more to choose the center pair.

Using the R the probability Pe = 0.0003

The probabilities P \approx Pe

Step-by-step explanation:

Since the person selects his or her desired pair of socks at random , then the probability that the person would choose the pair of socks in the center position from all the five identical pair is mathematically evaluated as

                  p =\frac{1}{5}

                    =0.2

The mean of this distribution is mathematical represented as

           \mu = np

substituting the value

         \mu = 100 * 0.2

             \mu = 20

The standard deviation is mathematically represented as

         \sigma = \sqrt{np (1-p)}

substituting the value

           = \sqrt{100 * 0,2 (1-0.2)}

           \sigma=4

Applying normal approximation the probability that 34 or more subjects would choose the item in the center if each subject were selecting his or her preferred pair of socks at random would be mathematically represented as

               P=P(X \ge 34 )

By standardizing the normal approximation we have that

              P(X \ge 34) \approx P(Z \ge z)

Now z is mathematically evaluated as

               z = \frac{x-\mu}{\sigma }

Substituting values

             z = \frac{34-20}{4}

               =3.5

So  using the z table the P(Z \ge 3.5) is  0.0002

The probability P and Pe that 34 or more subject would choose the center pair is very small  So

The correct answer is

The experiment supports the center stage effect. If participants were truly picking the socks at random, it would be highly unlikely for 34 or more to choose the center pair.

 

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