Question

Suppose you just received a shipment of six televisions. four of the televisions are defective. if two televisions are randomly​ selected, compute the probability that both televisions work. what is the probability at least one of the two televisions does not​ work?

Answer 1

The probability that both TVs work is 1/15. 
The probability that at least one does not work is 14/15.

Explanation
4/6 of the TVs are defective, and 2/6 are working.  The probability that two TVs drawn from this sample would be working is given by

2/6(1/5) = 2/30 = 1/15.  Getting one that works the first time changes the probability of the second one working.

The probability that at least one of them is defective is the same as 1 minus the probability that neither of them are defective.  This probability was 1/15; then we have

1-1/15 = 14/15.

Answer 2

The computed probabilities that both televisions selected are not defective and that atleast one is defective are :1/15 and 14/15 respectively.

• Total Number televisions = 6
• Number of defective televisions = 4

Non - defective televisions = Total - defective = 6 - 4 = 2

P(non - defective) = P(1st non-defective) × P(2nd non-defective)

P(1st non-defective) = 2 / 6 = 1/3

P(2nd non-defective) = 1 / 5

P(2 non-defective) = 1/3 × 1/5 = 1/15

2.) Probability that atleast one is defective :

P(Atleast 1) = 1 - P(2 non-defective)

P(Atleast 1) = 1 - 1 /15 = 14/15

Therefore, the probability that atleast one of the television is defective is 14/15.

Learn more : https://brainly.com/question/18405415