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snow_tiger [21]
2 years ago
14

From a group of 13 women and ​12 men, a researcher wants to randomly select 8 women and 8 men for a study. In how many ways can

the study group be​ selected?
Mathematics
1 answer:
Fantom [35]2 years ago
4 0

The total number of ways the study can be selected is: 637065

Given,

Total number of women in a group= 13

Total number of men in a group = 12

Number of women chosen = 8

Number of men chosen = 8

∴ the total number of ways the study group can be selected = 13C₈ and 12C₈.

This in the form of combination factor = nCr

                                                     ∴ nCr = n!/(n₋r)! r!

13C₈ = 13!/(13 ₋ 8)! 8!

        = 13!/5!.8!

        = 1287

12C₈ = 12!/(12₋8)! 8!

        = 12!/5! 8!

        = 495

Now multiply both the combinations of men and women

= 1287 × 495

= 637065

Hence the total number of ways the study group is selected is 637065

Learn more about "Permutations and Combinations" here-

brainly.com/question/11732255

#SPJ10

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Answer:

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Answer:

P(X \geq 8)= P(X=8) +P(X=9) +P(X=10)

And we can find the individual probabilities like this:

P(X=8)=(10C8)(0.65)^0 (1-0.65)^{10-8}=0.176  

P(X=9)=(10C9)(0.65)^1 (1-0.65)^{10-9}=0.072  

P(X=10)=(10C10)(0.65)^2 (1-0.65)^{10-10}=0.013  

And adding we got:

P(X \geq 8)= P(X=8) +P(X=9) +P(X=10)= 0.176+0.072+0.013=0.262

Step-by-step explanation:

Previous concepts

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".

Let X the random variable of interest, on this case we now that:

X \sim Binom(n=10, p=0.65)

The probability mass function for the Binomial distribution is given as:

P(X)=(nCx)(p)^x (1-p)^{n-x}

Where (nCx) means combinatory and it's given by this formula:

nCx=\frac{n!}{(n-x)! x!}

Solution to the problem

For this case we want to find this probability:

P(X \geq 8)= P(X=8) +P(X=9) +P(X=10)

And we can find the individual probabilities like this:

P(X=8)=(10C8)(0.65)^0 (1-0.65)^{10-8}=0.176  

P(X=9)=(10C9)(0.65)^1 (1-0.65)^{10-9}=0.072  

P(X=10)=(10C10)(0.65)^2 (1-0.65)^{10-10}=0.013  

And adding we got:

P(X \geq 8)= P(X=8) +P(X=9) +P(X=10)= 0.176+0.072+0.013=0.262

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