Probability confusion

[Jess16]

I am trying to help ne of my friends with her stats class and I don't understand why I am not getting an answer listed on the practice test problem...

1) You are hospitalized with a certain disease for the first time. There is a 20% chance that you will die during your hospitalization. Let X be the number of people who lived. If we took a sample of 5 such paitents, what is is the probability that exactly 3 will die?

Now I am wondering why you cannot just multiply the probabilities: (.2)(.2)(.2)=0.008 That is not an answer choice.

The next question stems from that:
2) Same set up, with the 20% of death and 5 paitents being selected. Let X be the number of people who lived. What is the probability of less than 3 dying?

Now I am wondering if you take the probabiliity of 2 dying (.2)(.2)=0.04 and add that to the probabilty of 1 person dying (.2) and then add that to the probability of no one dying (.8)(.8)(.8)(.8)(.8)=0.328 Or do I need the answer for the first question to get the second?

Thank You for any help you can give

 

[galactus]

1) You are hospitalized with a certain disease for the first time. There is a 20% chance that you will die during your hospitalization. Let X be the number of people who lived. If we took a sample of 5 such paitents, what is is the probability that exactly 3 will die?

Now I am wondering why you cannot just multiply the probabilities: (.2)(.2)(.2)=0.008 That is not an answer choice.

This is a binomial. \(\displaystyle \binom{5}{3}(.20)^{3}(.80)^{2}\)

The next question stems from that:
2) Same set up, with the 20% of death and 5 paitents being selected. Let X be the number of people who lived. What is the probability of less than 3 dying?

[/quote]

Another binomial. \(\displaystyle \sum_{k=0}^{2}\binom{5}{k}(.20)^{k}(.80)^{5-k}\)

 

[soroban]

Hello, Jess16!

1) You are hospitalized with a certain disease for the first time.
There is a 20% chance that you will die during your hospitalization.
If we took a sample of 5 such paitents, what is is the probability that exactly 3 will die?

I am wondering why you cannot just multiply the probabilities: (0.2)(0.2)(0.2) = 0.008

\(\displaystyle \text{We have: }\;P(\text{die}) \,=\,0.2,\;\;P(\text{recover}) \,=\,0.8\)

You found: the probability that 3 patients die.


We have a sample of 5 patients.
. . of which 3 will die and 2 will recover.

. . \(\displaystyle \text{This would be: }\;(0.2)(0.2)(0.2)(0.8)(0.8)\)


But this imparts an order to the events.
It is the probability that the first 3 die and the last 2 survive.

\(\displaystyle \text{And there are: }\:{5\choose3} = 10\text{ possible orderings.}\) .**


\(\displaystyle \text{Therefore: }\;P(\text{3 die, 2 survive}) \;=\;10(0.2)^3(0.8)^2 \;=\;\boxed{0.0512}\)


~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

**

If you're not familiar with binomial coefficients,
. . you can list the possible orderings.

. . . . \(\displaystyle \begin{array}{c} DDDSS \\ DDSDS \\ DDSSD \\ DSDDS \\ DSDSD \\ DSSDD \\ SDDDS \\ SDDSD \\ SDSDD \\ SSDDD \end{array}\)