# Probability question who can explain it><

#### longcrystal

Joined Mar 22, 2010
20
Of 25 calculators, 4 have faulty displays. A random sample of 2 calculators is selected
without replacement.
Let C1 be the event in the first calculator selected the display works
Let N2 be the event  in the second calculator selected the display fails
What is Pr(first display works and second display fails)?
= Pr(C1∩N2)
= Pr(C1).Pr(N2\C1) = (21/25).(4/24) = 0.14. (using the multiplication rule for the probability
of two events)

*actually i wanna ask about Pr(N2\C1)=(4/24) this value 4/24 i cannot understand it someone else can explain to me how the value come from.and show me the easy step can get 4/24 i really want to understand it about this ~~thank all><

grateful~someone can help me thank a lot

#### retched

Joined Dec 5, 2009
5,208
This kind of stuff smashes my brain into mush.

It seems to me that the 4/24 should actually be 4/25. If that is the 4 bad displays out of the 25 calculators.

But then again, I have NO IDEA.

It seems everytime I think I am on to something I know nothing about, someone will post a reply putting me in my place.

But, hey... Whats the probability of that happening?

#### Ghar

Joined Mar 8, 2010
655
You do the events in order.
The situation is you first pick a good calculator followed by a bad calculator. It's all about reading those sentences a million times.

First event, you have 25 calculators with 4 bad ones in the bunch, giving you 25-4 = 21 good ones.
The chance of picking a good one is then 21/25.

Second event, you now have only 24 calculators because you're holding a good one in your hand already. There are still 4 bad ones.
That means the chance of picking a bad one is 4/24.

Since it's a 'random' sample the events are independent, which lets you multiply them together.
21/25 * 4/24 = 0.14

#### retched

Joined Dec 5, 2009
5,208
Ooohhh. That's how it works.

I told ya someone would jack up my idea and give you the correct answer!

#### longcrystal

Joined Mar 22, 2010
20
sorry mr,ghar i cannot understand one thing ~why at the second event , i already holding a good one calculator.~~~can u explain to me other way????sorry><..issit this sentence------> ( A random sample of 2 calculators is selected
without replacement.).so that at the second event must holding a good one calculator already???

#### jpanhalt

Joined Jan 18, 2008
11,088
sorry mr,ghar i cannot understand one thing ~why at the second event , i already holding a good one calculator.~~~can u explain to me other way????sorry><..issit this sentence------> ( A random sample of 2 calculators is selected
without replacement.).so that at the second event must holding a good one calculator already???
Yes, I believe you have the right concept. Ghar's comment is right on:

Let me see, if I can explain it another way. Upon reading the question multiple times, you see that the question is not simply, what is the probability the second calculator is bad. The question is, what is the probability the first is good and the second is bad. In that case, you know the first one must be good, and the probability of that is 21/25, but that reduces the pool of calculators to 20 good plus 4 bad. Thus, the probability of selecting a bad calculator from that pool is 4/24.

This problem is different from one in which the pool is very large with 16% faulty calculators (i.e., 4/25) or in which there is replacement.

John