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Christine has always been weak in mathematics. Based on her performance prior to the final exam in Calculus, there is a 51% chan

Charra [1.4K]

Answer:

a) 0.3270 = 32.70% probability that Christine fails the course

b) 0.3917 = 39.17% probability that she had found someone to help

Step-by-step explanation:

We have these following probabilities:

61% probability that Christine find a coach.

100 - 61 = 39% probability that Christine does not find a coach.

With a coach, a 21% probability of failing.

Without a coach, a 51% probability of failing.

a. What is the probability that Christine fails the course?

21% of 61%(finds a tutor) plus 51% of 39%(does not find a tutor). So

P = 0.21*0.61 + 0.51*0.39 = 0.327.

0.327 = 32.7% probability that Christine fails the course

b. Christine ends up failing the course. What is the probability that she had found someone to help?

This can be formulated by the Bayes formula:

$P(B|A) = \frac{P(B)*P(A|B)}{P(A)}$

In which P(B|A) is the probability of B happening given that A happened and P(A|B) is the probability of A happening given that B happened.

In this question.

P(B|A) is the probability of finding a tutor given that she failed.

P(B) is the probability of finding a tutor. So P(B) = 0.61.

P(A|B) is the probability of failing when finding a tutor. So P(A|B) = 0.21

P(A) is the probability of failing. So P(A) = 0.327.

So

$P(B|A) = \frac{P(B)*P(A|B)}{P(A)} = \frac{0.61*0.21}{0.327} = 0.3917$

0.3917 = 39.17% probability that she had found someone to help

4 0

3 years ago

Exercise 7.3.5 The following is a Markov (migration) matrix for three locations        1 5 1 5 2 5 2 5 2 5 1 5 2 5 2 5 2

fredd [130]

Answer:

Both get the same results that is,

$\left[\begin{array}{ccc}140\\160\\200\end{array}\right]$

Step-by-step explanation:

Given :

$\bf M=\left[\begin{array}{ccc}\frac{1}{5}&\frac{1}{5}&\frac{2}{5}\\\frac{2}{5}&\frac{2}{5}&\frac{1}{5}\\\frac{2}{5}&\frac{2}{5}&\frac{2}{5}\end{array}\right]$

and initial population,

$\bf P=\left[\begin{array}{ccc}130\\300\\70\end{array}\right]$

a) - After two times, we will find in each position.

$P_2=[P].[M]^2=[P].[M].[M]$

$M^2=\left[\begin{array}{ccc}\frac{1}{5}&\frac{1}{5}&\frac{2}{5}\\\frac{2}{5}&\frac{2}{5}&\frac{1}{5}\\\frac{2}{5}&\frac{2}{5}&\frac{2}{5}\end{array}\right]\times \left[\begin{array}{ccc}\frac{1}{5}&\frac{1}{5}&\frac{2}{5}\\\frac{2}{5}&\frac{2}{5}&\frac{1}{5}\\\frac{2}{5}&\frac{2}{5}&\frac{2}{5}\end{array}\right]$

$=\frac{1}{25} \left[\begin{array}{ccc}7&7&7\\8&8&8\\10&10&10\end{array}\right]$

$\therefore\;\;\;\;\;\;\;\;\;\;\;P_2=\left[\begin{array}{ccc}7&7&7\\8&8&8\\10&10&10\end{array}\right] \times\left[\begin{array}{ccc}130\\300\\70\end{array}\right] = \left[\begin{array}{ccc}140\\160\\200\end{array}\right]$