1. ## intersection and union...

Hello,

I need help in these small problems. I know they are really stupid but I am getting wayy too confused with them...

1) Suppose that an assemble operation in a manufacturing plant involves four steps, which can be performed in any sequences. If the manufacturer wishes to compare assembly time for each of the sequences, how many different sequences will be involved in the experiment?
-----------I think it should be just 4! = 24...

2) A balanced coin is tossed four times. The coin tosses are independent, so the two outcomes per toss (H,T) are equally likely for each coin toss.
i) Calculate the probability that exactly two of the four tosses results in tails...
--------I think its just 6/16...not sure though!

ii) Calculate the probability that more than one toss results in heads..

3) A manufacturer has five seemingly identical computer monitors available for shipping. Unknown to him, two are defective. An order is place for one monitor. What is the probability that the defective monitor is shipped?
------------------I think its just... 5c1 / 5c2 = 1/2??

thanks for the help in advance!

2. Originally Posted by Vedicmaths
Hello,

I need help in these small problems. I know they are really stupid but I am getting wayy too confused with them...

1) Suppose that an assemble operation in a manufacturing plant involves four steps, which can be performed in any sequences. If the manufacturer wishes to compare assembly time for each of the sequences, how many different sequences will be involved in the experiment?
-----------I think it should be just 4! = 24...
That would be correct!

2) A balanced coin is tossed four times. The coin tosses are independent, so the two outcomes per toss (H,T) are equally likely for each coin toss.
i) Calculate the probability that exactly two of the four tosses results in tails...
--------I think its just 6/16...not sure though!
...which simplifies to $\tfrac{3}{8}$!

ii) Calculate the probability that more than one toss results in heads..
Since we're dealing with a binomial distribution: $X\sim \mathcal{B}\left(4,\tfrac{1}{2}\right)$, $P\!\left(X\geq1\right)=1-P\!\left(X=0\right)=1-{4\choose 0}\left(\tfrac{1}{2}\right)^0\left(\tfrac{1}{2}\ri ght)^4=1-\tfrac{1}{16}=\tfrac{15}{16}$.

3) A manufacturer has five seemingly identical computer monitors available for shipping. Unknown to him, two are defective. An order is place for one monitor. What is the probability that the defective monitor is shipped?
------------------I think its just... 5c1 / 5c2 = 1/2??

thanks for the help in advance!
It would be $\frac{{2\choose 1}{3\choose 0}}{{5\choose 1}}=\tfrac{2}{5}$

Does this make sense?

3. thanks a lot for the help...

I think I was thinking in a wrong direction about the manufacturer problem...but I guess I get it now!

cool!
I really appreciate it!