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yanalaym [24]
3 years ago
9

If you pick a gumball at random, put it back, and then pick another gumball at random, what is the probability of picking a blue

gumball then picking a yellow gum ball? A) 1/25 B) 1/50 C)3/100 D) 3/10 There are two blues, one yellow, one white, one orange, two red, one pink, one purple, one green
Mathematics
2 answers:
Monica [59]3 years ago
5 0

Answer:  The correct option is (B) \dfrac{1}{50}.

Step-by-step explanation:  Given that there are two blue, one yellow, one white, one orange, two red, one pink, one purple and one green gumball.

We are to find the probability of picking a blue gumball then a yellow gum ball after replacing the first gumball.

The total number of gumballs is equal to the number of elements in the sample space for the event of picking a ball from the collection.

So,

n(S)=2+1+1+1+2+1+1+1\\\\\Rightarrow n(S)=10.

Let, E be the event of picking a blue gumball.

Then, n(E) = 2.

and let F be the event of picking a yellow gumball after putting the first ball back.

Then, n(F)=1.

Therefore, the probability of picking a blue gumball then a yellow gum ball after replacing the first gumball will be

P\\\\=P(E)\times P(F)\\\\\\=\dfrac{n(E)}{n(S)}\times \dfrac{n(F)}{n(S)}\\\\\\=\dfrac{2}{10}\times\dfrac{1}{10}\\\\\\=\dfrac{1}{50}.

Thus, the required probability is \dfrac{1}{50}.

Option (B) is CORRECT.

AleksAgata [21]3 years ago
4 0
The answer is B) 1/50

Probability of picking Blue-2/10 which is 1/5

Probability of picking Yellow-1/10

1/5x1/10=1/50
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Level of significance α is the probability of committing a type 1 error. The area under the distribution is known as the rejection region and it is the area towards the right of the distribution.

The table I'm using is towards the left of the distribution.

But z>2.45 + z<2.45 = 1

z> 2.45 = 1 - z<2.45

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Answer:

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The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

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Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

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When we are approximating a binomial distribution to a normal one, we have that \mu = E(X), \sigma = \sqrt{V(X)}.

In this problem, we have that:

p = 0.71

(a) For a sample of 16 Americans, what is the probability that at least 13 believe global warming is occurring?

Here n = 16, we want P(X \geq 13). So

P(X \geq 13) = P(X = 13) + P(X = 14) + P(X = 15) + P(X = 16)

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P(X \geq 13) = P(X = 13) + P(X = 14) + P(X = 15) + P(X = 16) = 0.1591 + 0.0835 + 0.0273 + 0.0042 = 0.2741

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(b) For a sample of 160 Americans, what is the probability that at least 110 believe global warming is occurring?

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Z = -0.71

Z = -0.71 has a pvalue of 0.2389

1 - 0.2389 = 0.7611

0.7611 = 76.11% probability that at least 110 believe global warming is occurring

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