4.5 is 8.2 what is the sum

A sequence of Bernoulli trials consists of choosing components at random from a batch of components. A selected component is eit

Drupady [299]

Answer:

a) 0.0287 = 2.87% probability that three non-defective components in a batch of seven components.

b) 0.0336 = 3.36% probability that 8 non-defective components are drawn before the first defective component is chosen.

Step-by-step explanation:

A sequence of Bernoulli trials composes the binomial distribution.

Binomial probability distribution

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

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

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

And p is the probability of X happening.

The probability that a selected component is non-defective is 0.8

This means that $p = 0.8$

a) Three non-defective components in a batch of seven components.

This is P(X = 3) when n = 7. So

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 3) = C_{7,3}.(0.8)^{3}.(0.2)^{4} = 0.0287$

0.0287 = 2.87% probability that three non-defective components in a batch of seven components.

b) 8 non-defective components are drawn before the first defective component is chosen.

Now the order is important, so the we just multiply the probabilities.

8 non-defective, each with probability 0.8, and then a defective, with probability 0.2. So

$p = (0.8)^8*0.2 = 0.0336$

0.0336 = 3.36% probability that 8 non-defective components are drawn before the first defective component is chosen.

4 0

2 years ago

If the second number is subtracted from the sum of the first number and 2 times the third number, the result is 1. The thrid num

weeeeeb [17]

Answer:

Step-by-step explanation:

$x,\ y,\ z-\text{three numbers}\\\\\left\{\begin{array}{ccc}(x+2z)-y=1&(1)\\z+2x=3&(2)\\x+3y+z=18&(3)\end{array}\right\\\\(2)\\z+2x=3\qquad\text{subtract}\ 2x\ \text{from both sides}\\z=3-2x\qquad(*)\\\\\text{Substitute}\ (*)\ \text{to (1) and (3)}\\\\\left\{\begin{array}{ccc}x+2(3-2x)-y=1&\text{use the distributive property}\\x+3y+(3-2x)=18\end{array}\right$

$\left\{\begin{array}{ccc}x+(2)(3)+(2)(-2x)-y=1\\x+3y+3-2x=18&\text{subtract 3 from both sides}\end{array}\right\\\left\{\begin{array}{ccc}x+6-4x-y=1&\text{subtract 6 from both sides}\\(x-2x)+3y=15\end{array}\right\\\left\{\begin{array}{ccc}(x-4x)-y=-5\\-x+3y=15\end{array}\right\\\left\{\begin{array}{ccc}-3x-y=-5&\text{multiply both sides by 3}\\-x+3y=15\end{array}\right$

$\underline{+\left\{\begin{array}{ccc}-9x-3y=-15\\-x+3y=15\end{array}\right}\qquad\text{add all sides of the equations}\\.\qquad-10x=0\qquad\text{divide both sides by (-10)}\\.\qquad\boxed{x=0}\\\\\text{Put it to the second equation:}\\-0+3y=15\\3y=15\qquad\text{divide both sides by 3}\\\boxed{y=5}\\\\\text{Put the value of}\ x\ \text{to}\ (*):\\\\z=3-2(0)\\\boxed{z=3}$