not able to understand a probability concept

Question

Suppose there are patients suffering from a particular lung disease. Either lung is diseased with a probability of 0.1. How to find the probability of exactly n lungs being diseased ? n = { 0,1,2 } If I think of binomial distribution then what will be the probability of success, p ? Will it be 0.1 ? If yes,Why ?

Answer 1

"Either lung" means. Either the left lung, the right, lung or both.

So probability of either lung diseased is equal to $1 - P(\text{neither lung diseased})$.

And the probability of neither lung is $P(\text{left lung healthy})P(\text{right lung healthy}) = P(\text{a specific lung healthy})^2$.

And probability of $P(\text{specific lung diseased}) = 1 -P(\text{a specific lung healthy})$

So we have $P(\text{either lung diseased}) = 1-P(\text{a specific lung healthy})^2=.1$ so $ P(\text{a specific lung healthy})= \sqrt{.9}\approx 0.949$

And $P(\text{a specific lung diseased}) = 1 -\sqrt{.9} \approx 0.051$

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The probability of $0$ lungs diseased is $1- P(\text{either lung diseased}) = 1 - .1 = .9$.

Or we could say it is $P(\text{specific lung healthy})^2 = \sqrt{.9}^2 = .9$

The probability of $1$ lung diseased i $P(\text{left lung diseased and right healty}) + P(\text{right lung diseased and left healthy}) = 2P(\text{specific lung diseased})P(\text{specific lung healthy}) = 2\sqrt{.9}(1 - \sqrt{.9})\approx 0.097$

The probability of $2$ lungs diseased is $P(\text{specific lung diseased})^2 = (1 - \sqrt{.9})^2 = 1 + .9 -2\sqrt{.9} \approx 0.003$

We can generalize as exactly $n$ lungs diseased as

$(\text{# of ways to chose which lungs are diseased})P(\text{specific n lungs are diseased})P(\text{specific 2-n lungs are healthy})=$

${2 \choose n}P(\text {specific lung diseased})^nP(\text{specific lung healthy})^{2-n}=$

${2 \choose n} (1-\sqrt{.9})^n \sqrt{.9}^{2-n}$.

I have no idea what you mean by "probability of success". Success of what?