Bayes' Rule with a binomial distribution

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

$$\mathbb{P}(O\mid L)=\frac{\mathbb{P}(O\cap L)}{\mathbb{P}(L)}=\frac{{\textstyle (}\genfrac{}{}{0ex}{}{6}{1}{\textstyle )}\left(\frac{1}{4}\right){\left(\frac{3}{4}\right)}^{5}}{\frac{1}{5}}\approx 1.7798$$

and

$$\mathbb{P}(O\mid {L}^{C})=\frac{\mathbb{P}(O\cap {L}^{C})}{\mathbb{P}({L}^{C})}=\frac{{\textstyle (}\genfrac{}{}{0ex}{}{10}{1}{\textstyle )}\left(\frac{1}{4}\right){\left(\frac{3}{4}\right)}^{9}}{\frac{4}{5}}\approx .2346.$$

I would plug this into the Bayes' Rule equation, but I got a probability above 1 in one of the cases. Why?

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

$$\mathbb{P}(O\mid L)=\frac{\mathbb{P}(O\cap L)}{\mathbb{P}(L)}=\frac{{\textstyle (}\genfrac{}{}{0ex}{}{6}{1}{\textstyle )}\left(\frac{1}{4}\right){\left(\frac{3}{4}\right)}^{5}}{\frac{1}{5}}\approx 1.7798$$

and

$$\mathbb{P}(O\mid {L}^{C})=\frac{\mathbb{P}(O\cap {L}^{C})}{\mathbb{P}({L}^{C})}=\frac{{\textstyle (}\genfrac{}{}{0ex}{}{10}{1}{\textstyle )}\left(\frac{1}{4}\right){\left(\frac{3}{4}\right)}^{9}}{\frac{4}{5}}\approx .2346.$$

I would plug this into the Bayes' Rule equation, but I got a probability above 1 in one of the cases. Why?