Question

Three phones in a shipment of eighteen are known to be broken. A randomly selected phone is removed from the shipment and tested. It is
found to be fine and not broken. If a second phone is randomly selected from those remaining, what is the probability of it NOT being broken?
1/6
1/9
2/17
3/17
14/17

Answer 1

Answer:

$\frac{14}{17}$

Step-by-step explanation:

Number of phones broken = 3

Number of phones not broken = 15

Total number of phone in the shipment = 18

Since a not broken phone has been randomly selected, the number of phones not broken reduces to 14. And the total number of phones would be 17.

Pr(of newly selected phone NOT broken) = $\frac{number of not broken phones}{total number of phones}$

                                           = $\frac{14}{17}$

Therefore, the probability of the selected phone not being broken is $\frac{14}{17}$.