A car rental agency currently has 43 cars available, 27 of which have a GPS navigation system. Two cars are selected at random from these 43 cars. Find the probability that both of these cars have GPS navigation systems.
Round to four decimal places.
P(27chose 2) (43 choose 2)
27/43 * 26/42 = 702/1806 = .3887
Yup
To find the probability that both selected cars have GPS navigation systems, we need to divide the number of favorable outcomes (both cars having GPS systems) by the total number of possible outcomes.
First, let's find the number of possible outcomes. We are selecting 2 cars from a total of 43 cars, so the total number of possible outcomes is given by the combination formula:
C(43, 2) = 43! / (2!(43-2)!) = 43! / (2!41!) = (43 * 42) / (2 * 1) = 903.
Next, let's find the number of favorable outcomes. Since there are 27 cars with GPS systems, the number of favorable outcomes is given by selecting 2 cars from these 27:
C(27, 2) = 27! / (2!(27-2)!) = 27! / (2!25!) = (27 * 26) / (2 * 1) = 351.
Finally, we can calculate the probability by dividing the number of favorable outcomes by the total number of possible outcomes:
P(both cars have GPS systems) = 351 / 903 ≈ 0.3883 (rounded to four decimal places).
Therefore, the probability that both selected cars have GPS navigation systems is approximately 0.3883.