The exact problem statement: A car trader usually sells 70% of cars during the last four months of the year and 50% of cars during other months. The car trades are independent and identically distributed. What is the probability that the trader will sell 6 out of 12 cars in August or 2 out of 10 in September (round the answer to the nearest hundredth, i.e. second digit after decimal point)?
My attempt: The distribution of cars sold in a fixed period of time follows the Poisson distribution since the trades are iid.
In the month of August, $E[\text{no. of cars sold in Aug}] = 50\% \times 12=6$ because we started with $12$ cars and are expected to sell half of them. Thus, $$p(k)=\frac{6^ke^{-6}}{k!} ~~~\text{so} ~~~p(6)=\frac{6^6e^{-6}}{6!}\approx 16\%.$$
Similarly, $E[\text{no. of cars sold in Sep}] = 70\% \times 10=7$. So, $$p(k)=\frac{7^ke^{-7}}{k!} ~~~\text{so} ~~~p(2)=\frac{7^2e^{-7}}{2!}\approx 2\%.$$
Thus the desired answer is $\approx 18\%$, but this isn't any of the given choices (the largest probability in the choices is 12%). We could use Binomial distribution because the numbers are small but still the answer would be 22%. Tell me where I'm going wrong.
EDIT: Choices are: 12%, 6%, 1%, 0%.
My understanding is that the number of car trades in a given month is modeled by a binomial random variable with parameter $n$ representing the available inventory at the beginning of the month, and $p$ representing the probability of selling a given car in that inventory in that month. So the sale of a car in the inventory is modeled as an independent and identically distributed Bernoulli trial with probability of sale $p$.
The probability that exactly $X = 6$ out of $n = 12$ cars are sold in August when the probability of an individual sale is $p = 0.5$, is $$s_1 = \Pr[X = 6 \mid n = 12, p = 0.5] = \binom{12}{6}(0.5)^{6}(1 - 0.5)^{12-6} \approx 0.225586.$$
Similarly, the probability that exactly $X = 2$ out of $n = 10$ cars are sold in September when the probability of an individual sale is $p = 0.7$, is $$s_2 = \Pr[X = 2 \mid n = 10, p = 0.7] = \binom{10}{2}(0.7)^2(1 - 0.7)^{10-2} \approx 0.0014467.$$
Therefore the probability that either of these two events occurs is $$1 - (1-s_1)(1-s_2) \approx 0.226706.$$
However, if instead the question asked for $2$ out of $10$ cars sold in August and $6$ out of $12$ cars sold in September, then we would have $$s_1 \approx 0.0792479, \\ s_2 \approx 0.0439453,$$ and the desired probability is $0.119711$, which when rounded is $12\%$. Check the wording of the question and consider whether it might have been posed incorrectly.