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stepladder [879]
2 years ago
10

A balloon artist claims the probability a randomly selected balloon pops while being molded is 0.10 and that balloons pop indepe

ndent of other balloons. Part A. If we select a random sample of 3 balloons, what’s the probability that all of them pop while being molded?
Mathematics
2 answers:
jok3333 [9.3K]2 years ago
7 0

Answer:

0.001

Step-by-step explanation:

This can be modeled as a binomial distribution as:

  • There is a fixed number (n) of trials
  • Each trial is either a success or failure
  • All trials are independent
  • The probability of success (p) is the same in each trial
  • The variable is the total number of successes in the n trials

Let "success" be the balloon popping.

⇒ trials (n) = 3

⇒ probability of success (p) = 0.10

To find the probability that all 3 of the balloons pop, use the binomial distribution with x = 3:

\begin{aligned} \displaystyle P(X=x) & =\binom{n}{x} \cdot p^x \cdot (1-p)^{n-x}\\\\\implies P(X=3) & =\binom{3}{3} \cdot 0.1^3 \cdot (1-0.1)^{3-3}\\ & = 1 \cdot 0.001 \cdot 1\\ & =0.001 \end{aligned}

Nat2105 [25]2 years ago
4 0
<h3>Answer:  0.001</h3>

Work Shown:

(0.10)^3 = (0.10)*(0.10)*(0.10) = 0.001

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logₐ(3) = x

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Percentage increase in volume will be

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Determine whether a probability distribution is given. If a probability distribution is given, find its mean and standard deviat
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Answer:

E(X) = \sum_{i=1}^n X_i P(X_i) = 0*0.031 +1*0.156+ 2*0.313+3*0.313+ 4*0.156+ 5*0.031 = 2.5

We can find the second moment given by:

E(X^2) = \sum_{i=1}^n X^2_i P(X_i) = 0^2*0.031 +1^2*0.156+ 2^2*0.313+3^2*0.313+ 4^2*0.156+ 5^2*0.031 =7.496

And we can calculate the variance with this formula:

Var(X) =E(X^2) -[E(X)]^2 = 7.496 -(2.5)^2 = 1.246

And the deviation is:

Sd(X) = \sqrt{1.246}= 1.116

Step-by-step explanation:

For this case we have the following probability distribution given:

X          0            1        2         3        4         5

P(X)   0.031   0.156  0.313  0.313  0.156  0.031

The expected value of a random variable X is the n-th moment about zero of a probability density function f(x) if X is continuous, or the weighted average for a discrete probability distribution, if X is discrete.

The variance of a random variable X represent the spread of the possible values of the variable. The variance of X is written as Var(X).  

We can verify that:

\sum_{i=1}^n P(X_i) = 1

And P(X_i) \geq 0, \forall x_i

So then we have a probability distribution

We can calculate the expected value with the following formula:

E(X) = \sum_{i=1}^n X_i P(X_i) = 0*0.031 +1*0.156+ 2*0.313+3*0.313+ 4*0.156+ 5*0.031 = 2.5

We can find the second moment given by:

E(X^2) = \sum_{i=1}^n X^2_i P(X_i) = 0^2*0.031 +1^2*0.156+ 2^2*0.313+3^2*0.313+ 4^2*0.156+ 5^2*0.031 =7.496

And we can calculate the variance with this formula:

Var(X) =E(X^2) -[E(X)]^2 = 7.496 -(2.5)^2 = 1.246

And the deviation is:

Sd(X) = \sqrt{1.246}= 1.116

6 0
3 years ago
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