Harder Binomial Distribution problem

[Guest]

A factory has seven machines - four of model A, which are in use, on average, 80 percent of the time, and three of model B which are in use, on average, 60 percent of the time. If the supervisor walks into the factory at a randomly selected time, what is the probability that two machines of model A and one of model B will be in use?

This one has me beat!

 

[soroban]

Hello, americo74!

You have so many questions on Binomial Probability.
What part of it don't you understand?

A factory has seven machines:
four of model A, which are in use 80% of the time,
and three of model B which are in use 60% of the time.

If the supervisor walks into the factory at a randomly selected time,
what is the probability that two machines of model A and one of model B will be in use?

Two of the four model A's: \(\displaystyle \,\begin{pmatrix}4\\2\end{pmatrix}\,(0.8)^2(0.2)^2 \:=\:0.1536\)

One of the three model B's: \(\displaystyle \,\begin{pmatrix}3\\1\end{pmatrix}\,(0.6)^1(0.4)^2 \:= \:0.288\)

Therefore: \(\displaystyle \,P(\text{2A and 1B})\:=\0.1536)(0.288) \:=\:0.0442368\)