How Do Different Approaches to Conditional Probability Affect Problem Solving?

In summary, the first problem involves calculating the probability of picking one red and one green ball from a set of 9 balls without replacement. The correct calculation is P(R)P(G|R)+P(G)P(R|G) = (2/9)(3/8)+(3/9)(2/8). The second problem involves finding the probability that a couple watches a show given that the man and woman have individual probabilities of watching the show. Since the
  • #1
mohamed el teir
88
1

Homework Statement


suppose we have 9 balls : 2 red, 3 green, 4 yellow. and we draw 2 balls without replacement, the probability that one of them is red and the other is green is : P(R)P(G\R)+P(G)P(R\G) = (2/9)(3/8)+(3/9)(2/8)

i faced a problem in the textbook which says: the probability that a married man watches a show is 0.4, and the prob. that a married woman watched the same show is 0.5, and the prob. the man watches this show given that his wife does is 0.7. what is the probability that a couple watches the show ?

the answer to this problem is the intersection which is P(W)P(M\W) = 0.5*0.7 = 0.35

my question: why didn't we treat this problem like the first one, i mean giving the answer as P(W)P(M\W)+P(M)P(W\M) ?

Homework Equations


probability of A given B = P(A\B) = P(A∩B) / P(B)

The Attempt at a Solution


my thinking about the solution contradiction is merged with the problem statement to be relevant
 
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  • #2
We can't use the first method in the second problem because we do not know P(W|M): the probability that the woman watches the show given that the man does. Fortunately, we can do it as above, which is even easier.

That might lead one to ask - then why can't we do the first problem in the same way as the second one? The answer is that we can. The calculation using that approach is
Probability of one red and one green = P(G1 | R1) P(R1) where
G1 is the event that exactly one green is picked, and
R1 is the event that exactly one red is picked.

P(R1) is the prob that the first ball is red and the second isn't, plus the prob that the second ball is red and the first isn't. That's equal to:
$$P(R1) =\left(\frac{2}{9}\cdot\frac{7}{8}+\frac{7}{9}\cdot\frac{2}{8}\right)$$
We also have
$$P(G1 | R1)=\frac{3}{7}$$
Multiplying these out, you'll see it gives the same answer as the other method.

However, doing it this way is at least as complex as the way it was done above, so no efficiency is gained by doing it this way.

In probability there are often a number of different ways of approaching a problem. With practice, one develops intuitions about which ways are likely to be the quickest for a given sort of problem.
 
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  • #3
andrewkirk said:
We can't use the first method in the second problem because we do not know P(W|M): the probability that the woman watches the show given that the man does.
but can't we say that this = P(M∩W) / P(M) and we already have the intersection and P(M) ? by the way the second part of the question was asking about this ( P(W|M) )
 
  • #4
mohamed el teir said:
we already have the intersection
No, we are not given that. The intersection is the answer to the problem, not an input to it.
the second part of the question was asking about this ( P(W|M) )
Once we have done the first part, which is to calculate P(M∩W), we can then calculate P(W|M) in the way you suggest.
 
  • #5
andrewkirk said:
No, we are not given that. The intersection is the answer to the problem, not an input to it.

i may have a conflict in something, regarding this :
mohamed el teir said:
P(R)P(G\R)+P(G)P(R\G)

is P(R∩G) = P(R)P(G\R) + P(G)P(R\G) as a whole, or is it = P(R)P(G\R) = P(G)P(R\G) as if we are adding P(R∩G) to P(R∩G) giving 2P(R∩G) ?
 
  • #6
It's the second one
P(R∩G)= P(R)P(G|R) = P(G)P(R|G)
As you guess at the end of your post, the first one gives double the value of the correct answer.
 
  • #7
i think I've figured what what was wrong in my thinking about this.

for the colored balls example we could normally treat with their tree diagram as follows:
upload_2016-2-25_13-45-21.png


because red, green, yellow are the all possible outcomes of this experiment, so we can normally say: probability of both green and red is : P(R)P(G|R)+P(G)P(R|G)

while in the married man/woman problem, i was treating with the experiment like this:
Untitled.png


and this is wrong because man and woman are not all possible outcomes of this experiment, i mean it is not like a coin which will give you a tail or head, so P(M)P(W|M)+P(W)P(M|W) is wrong because it is based on a wrong tree, while this:
Untitled.png

is a right tree and can be used to get the correct values
 
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Related to How Do Different Approaches to Conditional Probability Affect Problem Solving?

What is conditional probability?

Conditional probability is the likelihood of an event occurring given that another event has already occurred. It is expressed as P(A|B), where A is the event of interest and B is the given condition.

How is conditional probability calculated?

Conditional probability is calculated by dividing the probability of the intersection of the two events (P(A∩B)) by the probability of the given condition (P(B)). This can also be written as P(A|B) = P(A∩B) / P(B).

What is the difference between conditional probability and joint probability?

Conditional probability is the likelihood of an event occurring given a specific condition, while joint probability is the likelihood of two events occurring simultaneously. In conditional probability, the given condition is already assumed to have occurred, whereas in joint probability, both events are considered independent of each other.

How is conditional probability used in real life?

Conditional probability is used in a variety of fields such as medical research, economics, and weather forecasting. In real life, it can be used to predict the likelihood of a disease given certain risk factors, or the probability of a stock market crash based on current economic conditions.

What are some common misconceptions about conditional probability?

One common misconception is that conditional probability and joint probability are the same thing. Another is the belief that the given condition has a direct influence on the event of interest, when in reality it may just be correlated. It is also often assumed that conditional probability always results in a smaller probability, when in fact it can be greater or equal to the original probability.

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