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Zielflug [23.3K]
3 years ago
15

A box has 5 beads of the same size, but all are different colors. Tina draws a bead randomly from the box, notes its color, and

then puts the bead back in the box. She repeats this 3 times. What is the probability that Tina would pick a red bead on the first draw, then a green bead, and finally a red bead again?
1 over 625
1 over 180
1 over 150
1 over 125
Mathematics
2 answers:
Afina-wow [57]3 years ago
7 0

Answer:

Give the other gut brainly.

Step-by-step explanation:

lesya692 [45]3 years ago
5 0
Probability of first red = 1/5
and Probabiliiy  of the green =1/5 the red again is 1/5. 

Required probability is (1/5)^3  = 1/125

its D
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Answer:

The sample statistic for the proportion of voters surveyed who said they'd vote for Brown is 0.308.

The expected numbers of voters who would vote for Brown is 3604.

Step-by-step explanation:

The sample statistic is a numerical quantity used to represent a characteristic of a sample. For example, sample mean is a sample statistic representing the average of the sample. Or, sample proportion is a sample statistic representing the proportion of a particular variable in the sample. Or sample variance is a sample statistic representing the variance of the sample.

The sample statistic can be used to estimate the population parameter.

They are known as the point estimate of the parameter.

The sample proportion of a variable <em>X</em> is given by:

\hat p=\frac{X}{n}

It is provided that of the <em>n</em> = 250 registered voters selected, the number of voters who said they would vote for Brown is <em>X </em>= 77.

Compute the sample statistic for the proportion of voters surveyed who said they'd vote for Brown as follows:

\hat p=\frac{X}{n}

  =\frac{77}{250}\\

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Thus, the sample statistic for the proportion of voters surveyed who said they'd vote for Brown is 0.308.

Compute the expected number of people of the 11700 registered voters who would vote for Brown as follows:

E(X)=N\times \hat p

         =11700\times 0.308\\=3603.6\\\approx 3604

Thus, the expected numbers of voters who would vote for Brown is 3604.

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