Question

Choose the one alternative that best completes the statement or answers the question. A shipment of twenty radios contains six defective radios. Two radios are randomly selected from the shipment. 14) Find the probability that both radios selected are defective

Answer 1

The probability that both radios are defective  = $(\frac{3}{38} )$

Step-by-step explanation:

Here, the total number of radios  = 20

The number of defective radios  = 6

The number of radios selected at random = 2

P(Selecting first defective radio)  = $\frac{\textrm{Total favorable outcomes}}{\textrm{Total outcomes}} = \frac{6}{20} = (\frac{3}{10} )$

Now, after selecting first defective radio, total radios left = 20 - 1 = 19

Also, total defective radios  = 6 - 1 = 5

P(Selecting second defective radio)  = $\frac{\textrm{Total favorable outcomes}}{\textrm{Total outcomes}} = (\frac{5}{19} )$

So, the combined probability of selecting 2 defective radios

$= (\frac{3}{10} ) \times (\frac{5}{19} ) = (\frac{3}{38})$

Hence, the probability that both radios are defective  = $(\frac{3}{38} )$