Question

The two-way table shows the number of sport utility vehicles with certain features for sale at the car lot. a 4-column table has 3 rows. the first column has entries third-row seats, no third-row seats, total. the second column is labeled 4-wheel drive with entries 18, 7, 25. the third column is labeled no 4-wheel drive with entries 12, 28, 40. the fourth column is labeled total with entries 30, 35, 65. what is the probability that a randomly selected car with no 4-wheel drive has third-row seats? 0.3 0.4 0.7 0.8

Answer 1

The probability that a randomly selected car with no 4-wheel drive has third-row seats is given by: Option B: 0.4

How to calculate the probability of an event?

Suppose that there are finite elementary events in the sample space of the considered experiment, and all are equally likely.

Then, suppose we want to find the probability of an event E.

Then, its probability is given as

$P(E) = \dfrac{\text{Number of favorable cases}}{\text{Number of total cases}} = \dfrac{n(E)}{n(S)}$

where favorable cases are those elementary events who belong to E, and total cases are the size of the sample space.

What is chain rule in probability?

For two events A and B, by chain rule, we have:

$P(A \cap B) = P(B)P(A|B) = P(A)P(B|A)$

where P(A|B) is probability of occurrence of A given that B already occurred.

For this case, the table given is:

Entries                          4-wheel drive        No 4-wheel drive        Total

Third Row Seats                  18                                12                        30

No Third Row Seats             7                                28                        35

Total                                     25                               40                        65

Let we take

A = event that a randomly selected car has no-4 wheel drive

B = event that a randomly selected car has third row seats

The total ways a car can be selected = 65 = n(S)

Total ways A can happen = n(A) = 40 (from the table).

Similarly, n(B) = 30

P(A) = n(A)/n(S) = 40/65

P(B) = n(B)/n(S) = 30/65

P(A∩B) = n(A∩B)/n(S) = 12/65

As by chain rule, we have:

P(A∩B) = P(A)P(B|A) = P(B)P(A|B)

We need P( A randomly selected car has three seats given that the selected car is with no 4-wheel drive)

which is symbolically P( B | A)

Thus, we use: P(A∩B) = P(A)P(B|A)

or

$12/65 = (30/65)(P(B|A))\\\dfrac{12/65}{30/65} = P(B|A)\\\\P(B|A) = \dfrac{12}{30} = \dfrac{2}{5} = 0.4$

Thus, the probability that a randomly selected car with no 4-wheel drive has third-row seats is given by: Option B: 0.4

Learn more about probability here:

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