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drek231 [11]
4 years ago
13

You want to make five-letter codes that use the letters A, F, E, R, and M without repeating any letter. What is the probability

that a randomly chosen code starts with M and ends with E?
Mathematics
1 answer:
Feliz [49]4 years ago
4 0

Answer:

The probability that a randomly chosen code starts with M and ends with E is 0.05 ....

Step-by-step explanation:

According to the given statement we have to make five letter code from  A, F, E, R, and M without repeating any letter. We have to find that what is probability that a randomly chosen code starts with M and ends with E.

Thus the probability of picking the first letter M = 1/5

After that we require the sequence (not E, not E, not E) which is equal to:

= 3/4 * 2/3* *1/2

= 1/4

Now multiply 1/5 and 1/4

1/5 * 1/4

= 1/20

= 0.05

Therefore the probability that a randomly chosen code starts with M and ends with E is 0.05 ....

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Using the hypergeometric distribution, it is found that there is a 0.7568 = 75.68% probability that neither can wiggle his or her ears.

The people are chosen from the sample without replacement, which is why the <u>hypergeometric distribution</u> is used to solve this question.

Hypergeometric distribution:

P(X = x) = h(x,N,n,k) = \frac{C_{k,x}C_{N-k,n-x}}{C_{N,n}}

C_{n,x} = \frac{n!}{x!(n-x)!}

The parameters are:

  • x is the number of successes.
  • N is the size of the population.
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  • k is the total number of desired outcomes.

In this problem:

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  • Two are selected, thus n = 2.

The probability that neither can wiggle his or her ears is P(X = 0), thus:

P(X = x) = h(x,N,n,k) = \frac{C_{k,x}C_{N-k,n-x}}{C_{N,n}}

P(X = x) = h(0,1000,2,130) = \frac{C_{130,0}C_{870,2}}{C_{1000,2}} = 0.7568

0.7568 = 75.68% probability that neither can wiggle his or her ears.

A similar problem is given at brainly.com/question/24826394

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444.3 = (4 x 100) + (4 x 10) + (4 x 1) + (3 x 1 /10 )

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