Which two numbers should be placed in the blanks below so that the difference between consecutive numbers is the same?

manufacturing company produces digital cameras and claim that their products maybe 3% defective. A video company, when purchasin

alexdok [17]

Answer:

P(X>17) = 0.979

Step-by-step explanation:

Probability that a camera is defective, p = 3% = 3/100 = 0.03

20 cameras were randomly selected.i.e sample size, n = 20

Probability that a camera is working, q = 1 - p = 1 - 0.03 = 0.97

Probability that more than 17 cameras are working P ( X > 17)

This is a binomial distribution P(X = r) $nCr q^{r} p^{n-r}$

$nCr = \frac{n!}{(n-r)!r!}$

P(X>17) = P(X=18) + P(X=19) + P(X=20)

P(X=18) = $20C18 * 0.97^{18} * 0.03^{20-18}$

P(X=18) = $20C18 * 0.97^{18} * 0.03^{2}$

P(X=18) = 0.0988

P(X=19) = $20C19 * 0.97^{19} * 0.03^{20-19}$

P(X=19) = $20C19 * 0.97^{19} * 0.03^{1}$

P(X=19) = 0.3364

P(X=20) = $20C20 * 0.97^{20} * 0.03^{20-20}$

P(X=20) = $20C20 * 0.97^{20} * 0.03^{0}$

P(X=20) = 0.5438

P(X>17) = 0.0988 + 0.3364 + 0.5438

P(X>17) = 0.979

The probability that there are more than 17 working cameras should be 0.979 for the company to accept the whole batch

6 0

3 years ago

A. One player places 1 red, 5 green and 3 blue tiles in Bag A, and 6 red, 4 green, and 2 blue in Bag B. What is the probability

cupoosta [38]

Answer:

$\frac{8}{27}$ is the probability that a player draws out two tiles of the same color assuming they are drawing one tile from each bag.

Step-by-step explanation:

In each bag there are red, green, and blue tiles, meaning that no matter which color is pulled out first there is always some probability that the second tile will be the same color. So, we can set up three possible outcomes:

Red: The player pulls out a red tile first. This has a $\frac{1}{9}$ probability of happening. Then in order to succeed for the problem, the next tile also needs to be red which has a $\frac{6}{12}$ probability attached to it. $\frac{1}{9}$ × $\frac{6}{12}$ = $\frac{1}{18}$ probability of happening.

Green: There is a $\frac{5}{9}$ probability of the player pulling out a green tile first. In this case we want to calculate the probability of the second tile being green, which would be $\frac{4}{12}$ . $\frac{5}{9}$ × $\frac{4}{12}$ = $\frac{5}{27}$ .

Blue: There is a $\frac{3}{9}$ probability of the first tile being blue in which case we are hoping for the second tile to be blue as well. The probability of the second tile being blue is $\frac{2}{12}$ on its own, and them both being blue is $\frac{3}{9}$ × $\frac{2}{12}$ = $\frac{1}{18}$