Poisson Distribution - Poor Understanding

Question

So I have this question about poisson distribution:

"The number of computers bought during one day from shop A is given a Poisson distribution mean of 3.5, while the same for another shop B is 5.0, calculate the probability that a total of fewer than 10 computers are sold from both shops in 4 out of 5 consecutive days"

I proceded to calculate the net probability which came to $0.653$, I then realised you'd need to use Binomial Distributiopn, so I put in the given and needed values giving me $0.315$, this however is where I get confused, I thought this was the answer but the markscheme says add on $(0.635^5)$ and I have no idea why.

Could someone explain this to me? Many thanks.

Answer 1

If you sell fewer than 10 computers in 5 out of 5 consecutive days then you must have also sold 4 out of 5.

It would be nice if questions like this made it explicit if they mean "in exactly 4 out of 5 days" or "in at least 4 out of 5 days" but there we are!

Here they mean "in at least 4 out of 5 days" so it's the probability of exactly 4 out of 5 days $+$ the probability of exactly 5 out of 5 days. That extra bit should be $0.653^5$ so you or they have made a typo...

Answer 2

It sounds like instead of exactly 4 out of 5 days, the system wants at least four out of five days. Hence we have \begin{align*} P(X \geq 4) &= P(X=4)+P(X=5) \\ &= \binom{5}{4}p^{4}(1-p)+\binom{5}{5}p^5(1-p)^{0}\\ &=0.3154388+0.1187075\\ &=0.4341463 \end{align*} where $p=0.6529737$ is the chance that they sold fewer than 10 items in one day and $X$ follows the binomial distribution you used.