the two - way table shows the number of sport utility vehicles with certain features for sale at the car…

the two - way table shows the number of sport utility vehicles with certain features for sale at the car lot.\n| | 4 - wheel drive | no 4 - wheel drive | total |\n|--|--|--|--|\n| third - row seats | 18 | 12 | 30 |\n| no third - row seats | 7 | 28 | 35 |\n| total | 25 | 40 | 65 |\nwhat is the probability that a randomly selected car with no 4 - wheel drive has third - row seats?\n0.3\n0.4\n0.7\n0.8

the two - way table shows the number of sport utility vehicles with certain features for sale at the car lot.\n| | 4 - wheel drive | no 4 - wheel drive | total |\n|--|--|--|--|\n| third - row seats | 18 | 12 | 30 |\n| no third - row seats | 7 | 28 | 35 |\n| total | 25 | 40 | 65 |\nwhat is the probability that a randomly selected car with no 4 - wheel drive has third - row seats?\n0.3\n0.4\n0.7\n0.8

Answer

Explanation:

Step1: Identify relevant values

We want probability of third - row seats given no 4 - wheel drive. Number of cars with no 4 - wheel drive and third - row seats is 12, and number of cars with no 4 - wheel drive is 40.

Step2: Use conditional probability formula

The formula for conditional probability $P(A|B)=\frac{P(A\cap B)}{P(B)}$. In terms of counts, if $A$ is the event of having third - row seats and $B$ is the event of having no 4 - wheel drive, the probability is $\frac{\text{Number of cars with }A\text{ and }B}{\text{Number of cars with }B}$. So the probability $P=\frac{12}{40}$.

Step3: Simplify the fraction

$\frac{12}{40}=\frac{3}{10}=0.3$

Answer:

0.3