The question is as follows:
Each morning, an intern is supposed to call $10$ people and ask them to take part in a survey. Every person called has a $1/4$ probability of agreeing to take the survey, independent of any other people’s decisions. Eighty percent of the days, the intern does what they are supposed to; however, the other twenty percent of the time, the intern is lazy and only calls $6$ people. One day, exactly $1$ person agree to take the survey. What is the probability the intern was lazy that day?
My solution is partially shown using Bayes' Rule with the binomial distribution. Let $L$ be the event he is lazy and let $O$ be the event of one person agreeing to the survey. Then,
$$\mathbb{P}(L\mid O)=\frac{\mathbb{P}(O\mid L)\mathbb{P}(L)}{\mathbb{P}(O\mid L)\mathbb{P}(L)+ \mathbb{P}(O\mid L^{C})\mathbb{P}(L^{C})}.$$
We know
$$\mathbb{P}(O\mid L)=\frac{\mathbb{P}(O \cap L )}{\mathbb{P}(L)}= \frac{\binom{6}{1}\left ( \frac{1}{4} \right )\left ( \frac{3}{4} \right )^5}{\frac{1}{5}} \approx 1.7798$$
and
$$\mathbb{P}(O\mid L^C)=\frac{\mathbb{P}(O \cap L^C )}{\mathbb{P}(L^C)} =
\frac{\binom{10}{1}\left ( \frac{1}4{} \right )\left ( \frac{3}{4} \right )^9}{\frac{4}{5}} \approx .2346.$$
I would plug this into the Bayes' Rule equation, but I got a probability above $1$ in one of the cases. Why?
Thanks in advance!
$P(O \cap L) = P(X = 1)P(L)$ where X ~ Bin(0.25, 6).
so $P(O | L)$ should be about $1.7798/5 \approx 0.35$ instead.
Just think about it logically. The probability of both events being true means that he has to be lazy first, and given he is lazy, what is the probability of him getting one successful survey taker.