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marishachu [46]

3 years ago

12

One student from a music class is randomly selected on the first day of class. There is an 80% chance that the student arrived o

n time, and a 20% chance that the student arrived on time and is a female student.
What is the probability that the chosen student is a male student, given that the student arrived at class on time?
Write as a percent and round to the nearest tenth.

1 answer:

sattari [20]3 years ago

8 0

Denote A the event "The student arrivent on time"
F the event "The student arrivent on time and is a female"
M the event "The student arrivent on time and is a male"
We have the formula:
$P(A)=\times P(F)+\times P(M)$
In numbers we get:
$0.8=\times 0.2+\times P(M)$
Solving for P(M) we get:
$P(M)=0.8-0.5\times 0.2=0.7$

Answer 70%

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a) 91.33% probability that at most 6 will come to a complete stop

b) 10.91% probability that exactly 6 will come to a complete stop.

c) 19.58% probability that at least 6 will come to a complete stop

d) 4 of the next 20 drivers do you expect to come to a complete stop

Step-by-step explanation:

For each driver, there are only two possible outcomes. Either they will come to a complete stop, or they will not. The probability of a driver coming to a complete stop is independent of other drivers. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

20% of all drivers come to a complete stop at an intersection having flashing red lights in all directions when no other cars are visible.

This means that $p = 0.2$

20 drivers

This means that $n = 20$

a. at most 6 will come to a complete stop?

$P(X \leq 6) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6)$

In which

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{20,0}.(0.2)^{0}.(0.8)^{20} = 0.0115$

$P(X = 1) = C_{20,1}.(0.2)^{1}.(0.8)^{19} = 0.0576$

$P(X = 2) = C_{20,2}.(0.2)^{2}.(0.8)^{18} = 0.1369$

$P(X = 3) = C_{20,3}.(0.2)^{3}.(0.8)^{17} = 0.2054$

$P(X = 4) = C_{20,4}.(0.2)^{4}.(0.8)^{16} = 0.2182$

$P(X = 5) = C_{20,5}.(0.2)^{5}.(0.8)^{15} = 0.1746$

$P(X = 6) = C_{20,6}.(0.2)^{6}.(0.8)^{14} = 0.1091$

$P(X \leq 6) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) = 0.0115 + 0.0576 + 0.1369 + 0.2054 + 0.2182 + 0.1746 + 0.1091 = 0.9133$

91.33% probability that at most 6 will come to a complete stop

b. Exactly 6 will come to a complete stop?

$P(X = 6) = C_{20,6}.(0.2)^{6}.(0.8)^{14} = 0.1091$

10.91% probability that exactly 6 will come to a complete stop.

c. At least 6 will come to a complete stop?

Either less than 6 will come to a complete stop, or at least 6 will. The sum of the probabilities of these events is decimal 1. So

$P(X < 6) + P(X \geq 6) = 1$

We want $P(X \geq 6)$ . So

$P(X \geq 6) = 1 - P(X < 6)$

In which

$P(X < 6) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) = 0.0115 + 0.0576 + 0.1369 + 0.2054 + 0.2182 + 0.1746 = 0.8042$

$P(X \geq 6) = 1 - P(X < 6) = 1 - 0.8042 = 0.1958$

19.58% probability that at least 6 will come to a complete stop

d. How many of the next 20 drivers do you expect to come to a complete stop?

The expected value of the binomial distribution is

$E(X) = np = 20*0.2 = 4$

4 of the next 20 drivers do you expect to come to a complete stop

4 0

4 years ago