Is there any difference in the test order in a binomial distribution?

Question

I have this question

It is known that during the pandemic, if people in a given city are tested at random, 85% are likely to be free from contamination, that is, to be negative.

20 people from that city are waiting to receive the test result. What is the probability of:

(i) A maximum of 1 person tests positive.

(ii) Only the fourth person tests positive.

The (i) question i answered using the Binomial distribution concept: $$ P(X\leq 1)=P(X=0)+P(X=1) $$ and then using the values $$ {{20}\choose{0}}\cdot (0,15)^0\cdot(1-0,15)^{20-0}+{{20}\choose{1}}(0,15)^1\cdot(1-0,15)^{20-1} $$ I reached (using some approximation) $$ 0,039+0,137=0,175 $$

But then, it asks about the fourth person tested.

In this case, I use $P(X=1)$ 'cause there is no difference if is the fourth ou the tenth person or i need to use some conditionality concept that I'm not getting here?

Answer 1

The probability mass function for a binomial random variable, such as $\mathsf P(X\,{=}\,1)$, consists of two factors.

$$\mathsf P(X\,{=}\,1) = \binom{20}1 \cdot p^1(1-p)^{19}$$

1. the probability for obtaining a particular sequence of success and failures. That is the factor of $p^1(1-p)^{19}$
2. the binomial coefficient which counts distinct ways to rearrange such a sequence. That is the factor of $\binom{20}1$.

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You seek the probability for obtaining a particular sequence of 19 failures and 1 success, with that lone success being the fourth trial. So...