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Sagot :
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
[tex]P(E) = \dfrac{\text{Number of favorable cases}}{\text{Number of total cases}} = \dfrac{n(E)}{n(S)}[/tex]
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:
[tex]P(A \cap B) = P(B)P(A|B) = P(A)P(B|A)[/tex]
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
[tex]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[/tex]
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:
brainly.com/question/1210781
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