# probability and combinations

I have just started probablity and I am trying to understand these question.

A study has found that 60% of all companies say that the cost of health care is too high. If we choose 8 companies at random, find the probability that the given number of businesses say the cost of health care is too high.

a) All 8 companies

This is my working.

$8\choose8$ x 0.6 = 0.6 = 60%

b) at most 2 companies

I am not sure how to do this.

I am a beginner to probablity so forgive me if this suppose to be obvious.

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How many companies are there in total ? – Damien L Feb 12 '13 at 16:07
Your answers aren't correct yet. What is the chance for the first company saying the costs are too high? What is the chance for the 2nd one and so on... and think about en.wikipedia.org/wiki/Chain_rule_%28probability%29 – ulead86 Feb 12 '13 at 16:08

Since you didn't specify this in your question, let's assume that the answer of each company is independent of the answers of other companies.

As you've probably shown in class (or in the text you're reading), if event A and event B are independent, then $P(A,B)=P(A)P(B)$. For instance, if you toss a fair coin twice, the probability that you'll get two heads is $P(HEAD,HEAD)=P(HEAD)P(HEAD)=\frac{1}{2}\cdot\frac{1}{2}=\frac{1}{4}$.

In your case, the "head" means that the company says the cost of health care is too high, and its probability is 0.6. Can you answer question (a) now?

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a) You have 8 companies $(X_1,X_2,\ldots,X_8)$ with $p=0.6$ chance, they say, the costs for health care are too high.

Every company has the same chance (0.6) to say this, so you get

$$0.6 \cdot 0.6 \cdot 0.6 \cdot 0.6 \cdot 0.6 \cdot 0.6 \cdot 0.6 \cdot 0.6=0.6^8\approx 0.0016$$

So the chance, that all 8 companies say the costs are too high ist about 0.16%.

b) At most 2 companies is the same as "$1-max. \; 1 \; company$". So you get

up to you

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.0016 would be 0.16% as the 1 is in the thousandths place. – JB King Feb 12 '13 at 16:58