what is the probability that at least one does not work?

Question

Suppose that you just received a shipment of six televisions and two are defective. If two televisions are randomly selected, compute the probability that both televisions work.

here is my work and steps to solve it I believe you should set up a fraction of the working televisions over the total televisions like this (4/6)*(3/5) E=(4/6) and F=(3/5) P(E and F) 6=total televisions 2=don't work 4=work

Welcome to MSE. Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or closed. To prevent that, please [edit] the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers. — José Carlos Santos, Mar 04 '20 at 07:21
Here's a tutorial and reference for typesetting math on this site. — joriki, Mar 04 '20 at 08:15

score 1 · Answer 1 · answered Mar 04 '20 at 13:17

1

First pick one tv. Then pick another.Probability(both good) =Pr(first good)Pr(second tv good| given that first was good) =$\frac{4}{6} \times \frac {3}{5}=\frac{2}{5}$

answered Mar 04 '20 at 13:17

P. Lawrence

5,674

thanks I will do that – 23Jerry8Kob Mar 05 '20 at 03:48

what is the probability that at least one does not work?

1 Answers1