probability word problem (prob. that no more than two die)

Folks, I am trying to help my daughter with this probability problem and really have no clue. Please help.

Suppose that 10% of the patients who have a certain disease die from it. If 5 patients have the disease, what is the probability that no more than 2 patients will die from it?

a) 0.0729

b) 0.5905

c) 0.9914

d) 0.8761

Thank you so much!

Calculate this:
\(\displaystyle \L
\sum\limits_{k = 0}^2 {\left( {\begin{array}{c}
5 \\
k \\
\end{array}} \right)\left( {0.1} \right)^k \left( {0.9} \right)^{5 - k} }\)

If I may expand on what pka posted. Just to make sure you understand the what it means. Hope that's OK, pka.

What you have is a classic example of a binomial distribution.

If p=the percentage of patients who die from the disease

q=the percentage of patients who do not die from the disease.

n=the number of patients who have the disease.

We use \(\displaystyle \L\\(p+q)^{5}\)

Now expand that and get: \(\displaystyle \L\\p^{5}+5p^{4}q+10p^{3}q^{2}+10p^{2}q^{3}+5pq^{4}+q^{5}\)

Now, because it asks for no more than 2, you use the 'p exponents' up to

\(\displaystyle p^{2}\). Start at the right of the expression and use

\(\displaystyle \L\\10p^{2}q^{3}+5pq^{4}+q^{5}\). Disregard the rest.

Now, enter in your values, p=.1, q=.9

\(\displaystyle \L\\10(.1)^{2}(.9)^{3}+5(.1)(.9)^{4}+(.9)^{5}=\)one of your choices.

This is just a brief explanation of the binomial distribution.

pka gave it to you in the mathematical form. I went 'around the Horn'.

Re: probability word problem (prob. that no more than two di

Hello, Goods!

I'll try to give you a straight answer . . .

Suppose that 10% of the patients who have a certain disease die from it.
If 5 patients have the disease, what is the probability that no more than 2 patients will die from it?

\(\displaystyle a)\;0.0729\;\;\;b)\;0.5905\;\;\;c)\;0.9914\;\;\;d)\;0.8761\)

Click to expand...

We have: \(\displaystyle \,P(die)\,=\,0.1\) and \(\displaystyle P(live) \,= \,0.9\)

"No more than 2 die" means: 0 die or 1 dies or 2 die.

We will compute these probabilities and add them.


\(\displaystyle P(0\text{ die})\;=\;\begin{pmatrix}5\\0\end{pmatrix} (0.1)^0(0.9)^5 \;=\;0.59049\)

\(\displaystyle P(1\text{ die})\;=\;\begin{pmatrix}5\\1\end{pmatrix}(0.1)^1(0.9)^4\;=\;0.32805\)

\(\displaystyle P(2\text{ die})\;=\;\begin{pmatrix}5\\2\end{pmatrix}(0.1)^2(0.9)^3\;=\;0.0729\)


Therefore: \(\displaystyle \,P(\text{0, 1, or 2 die})\;=\;0.059049\,+\,0.32805\,+\,0.0729\;=\;\)0.99144

Soroban, you and I have a completely different notion of what ‘straightforward’ means. This question was about helping a man’s daughter find a simple answer.
galactus gave very good answer to expand upon mine.