Probability Question Using Percentages

[Tygra]

Dear all

I have the following problem:

In a group of students
55% boys
65% prefer to watch film A
10% are girls that prefer film B

One of the students are picked at random. Workout the probability that the student is a boy who prefers to watch film A.

Here is my working:



This is not the correct answer. Could someone show me where I am going wrong?

Many thanks

 

[Dr.Peterson]

In a group of students
55% boys
65% prefer to watch film A
10% are girls that prefer film B

One of the students are picked at random. Workout the probability that the student is a boy who prefers to watch film A.

You misinterpreted "10% are girls that prefer film B" as "10% of the girls prefer film B". The 10% is a fraction of all the students, not just of the girls.

That should be enough to get you to the right answer.

I'll tell you that I myself made a table with boys and girls as rows, and movies A and B as columns, rather than a tree. That fits better with the nature of the problem as it really is (no conditional probabilities). But your approach should work.

 

[Tygra]

Thanks Dr Peterson, I am getting the correct answer now.

 

[Agent Smith]

I wonder if we can solve this problem without drawing a table.

Too, the table we draw, is it a 2-way table? I mean is that its name?

@Dr.Peterson , you said "(no conditional probabilities)". I wonder what you mean.

If M = Random person is a boy and A = Random person likes film A,

$P(B \text{ and } A) = P(B) \times P(A|B)$

 

[Dr.Peterson]

I wonder if we can solve this problem without drawing a table.

Of course. There are many ways to do the work.

Too, the table we draw, is it a 2-way table? I mean is that its name?

Yes, I've seen that term used for it. I've also seen it called a contingency table. I don't think of it in terms of any name, but just do it.

@Dr.Peterson , you said "(no conditional probabilities)". I wonder what you mean.

I meant what I said: The given data are not conditional probabilities (such as P(prefers B | girl)). If they were, you could still fill in the table, but not directly. On the other hand, a tree diagram is mostly about conditional probabilities.

If [ B ] = Random person is a boy and A = Random person likes film A,

$P(B \text{ and } A) = P(B) \times P(A|B)$

True; but there is no "probability that they like film A given that they are a boy" explicitly stated in this problem.

The error I pointed out was reading "10% [of the students] are girls that prefer film B", which means P( girl and prefers B), as if it were conditional, "10% of the girls prefer film B", which would be P(prefer B | girl) = 0.10.

 

[Agent Smith]

True; but there is no "probability that they like film A given that they are a boy" explicitly stated in this problem.

Gracias for the response.

Is the table below ok?

	Girls	Boys	Total
Likes film A	35	30	65
Likes film B	10	25	35
Total	45	55	100

P(Boy & Likes A) = P(Boy) * P(Likes A | Boy) = $\frac{55}{100} \times \frac{30}{55} = \frac{30}{100} = \frac{3}{10}$. Correct?

 

[Dr.Peterson]

Gracias for the response.

Is the table below ok?

	Girls	Boys	Total
Likes film A	35	30	65
Likes film B	10	25	35
Total	45	55	100

P(Boy & Likes A) = P(Boy) * P(Likes A | Boy) = $\frac{55}{100} \times \frac{30}{55} = \frac{30}{100} = \frac{3}{10}$. Correct?

Or, with less work, straight from the table (see red numbers): P(Boy, and likes A) = 30/100.

 

[Agent Smith]

Is the tree diagram in the OP a simple Markov chain?