Probability problem

[Guest]

Hey im having some problems with probability and here are the two homework questions.

1)What is the probability that you will pass a test and get at least 4 questions correct by randomly guessing on each question of a 6 question quiz multiple choice quiz with 5 choices for each question.

2) Every year at least 5 million people die of tobacco related causes, assume that the distribution is normal with a population mean of 5 million and a standard deviation of 2 million...find the probability that more than 4 million people will die of tobacco related causes and find the 75th percentile of the distribution of tobacco related deaths.

any help is appreciated!

 

[galactus]

1)What is the probability that you will pass a test and get at least 4 questions correct by randomly guessing on each question of a 6 question quiz multiple choice quiz with 5 choices for each question.

Use the binomial probability. \(\displaystyle \sum_{k=4}^{6}\binom{6}{k}\left(\frac{1}{5}\right)^{k}\left(\frac{4}{5}\right)^{6-k}\)

2) Every year at least 5 million people die of tobacco related causes, assume that the distribution is normal with a population mean of 5 million and a standard deviation of 2 million...find the probability that more than 4 million people will die of tobacco related causes and find the 75th percentile of the distribution of tobacco related deaths.

Use \(\displaystyle z=\frac{x-{\mu}}{\sigma}\) and look up the probability in a z table.

 

[soroban]

Hello, coheedfan1990!

I'll go through #1 and explain galactus' answer in baby steps . . .

1)What is the probability that you will pass a test and get at least 4 questions correct
by randomly guessing on each question of a 6-question multiple-choice quiz with 5 choices for each question?

\(\displaystyle \text{On any question: }\;\begin{Bmatrix}P(\text{right}) &=& \frac{1}{5} \\ \\[-3mm] P(\text{wrong}) &=& \frac{4}{5}\end{Bmatrix}\)


"Pass the test" means: "at least 4 right".
"At least 4 right" means: "4 right or 5 right or 6 right".


So we must consider the three cases.

[1] 4 right (and 2 wrong)
. . \(\displaystyle P(\text{4 right}) \:=\:{6\choose4}\left(\frac{1}{5}\right)^4\left(\frac{4}{5}\right)^2 \:=\:\frac{240}{15,\!625}\)

[2] 5 right (and 1 wrong)
. . \(\displaystyle P(\text{5 right}) \:=\:{6\choose5}\left(\frac{1}{5}\right)^5\left(\frac{4}{5}\right)^1 \:=\:\frac{24}{15,\!625}\)

[3] 6 right (and 0 wrong)
. . \(\displaystyle P(\text{6 right}) \;=\;{6\choose6}\left(\frac{1}{5}\right)^6\left(\frac{4}{5}\right)^0 \:=\:\frac{1}{15,\1625}\)


\(\displaystyle \text{Therefore: }\;P(\text{pass}) \;=\;\frac{240}{15,\!625} + \frac{24}{15,\1625} + \frac{1}{15,\!625} \;=\;\frac{265}{15,\!625} \;=\;\frac{53}{3125} \;\approx\;1.7\%\)

 

[Denis]

2) Every year at least 5 million people die of tobacco related causes, .....

Well, one of these years it'll be 5,000,001 ... cause I can't quit :roll: