Respuesta :
Answer:
The probability that the test is taken an even number of times is 0.30.
Step-by-step explanation:
The probability that a student passes the driving test at any attempt is,
p = 4/7.
The event of a student passing in any attempt is independent of each other.
The probability that the test is taken an even number of times is:
P (even number of tests) = P (Passing in the 2nd attempt)
+ P (Passing in the 4th attempt)
+ P (Passing in the 6th attempt) ...
If a student passed in the 2nd attempt it implies that he failed in the first.
Then, P (Passing in the 2nd attempt) = [tex](\frac{3}{7}) \times (\frac{4}{7})[/tex]
Similarly, P (Passing in the 4th attempt) = [tex](\frac{3}{7})^{3} \times (\frac{4}{7})[/tex], since he failed in the first 3 attempts.
And so on.
Compute the probability of an even number of tests as follows:
P (even number of tests) = [tex](\frac{3}{7}) \times (\frac{4}{7})+(\frac{3}{7})^{3} \times (\frac{4}{7})+(\frac{3}{7})^{5} \times (\frac{4}{7})+...[/tex]
The result follows a Geometric progression for infinite values.
The sum of infinite GP is:
[tex]S=\frac{a}{1-r^{2}}[/tex]
The probability is:
P (even number of tests) = [tex](\frac{3}{7}) \times (\frac{4}{7})+(\frac{3}{7})^{3} \times (\frac{4}{7})+(\frac{3}{7})^{5} \times (\frac{4}{7})+...[/tex]
[tex]=\frac{(\frac{3}{7})(\frac{4}{7} ) }{1-(\frac{3}{7})^{2}}\\=\frac{12}{49}\times\frac{49}{40}\\ =\frac{12}{40}\\ =0.30[/tex]
Thus, the probability that the test is taken an even number of times is 0.30.
