# Conditional proabability from a tree diagram.

• Jun 10th 2011, 12:22 PM
tousif111
Conditional proabability from a tree diagram.
I am stuck on question number (3) below:

¾ of the population in Siniestra is left-handed and ¼ are right-handed. In an examination, it was shown that 80% of the left-handed passed and 90% of the right-handed passed.

1)Complete the tree diagram
2) Find the probability of
i) the people are left handed and they passed
ii) all passed.

3) A person was chosen at random from the population. He had passed the examination. Using the answers you provided in Question number 2. Find the probability that
i) the person chosen from the population is left handed.
• Jun 10th 2011, 12:44 PM
SpringFan25
L=Left Handed
Pa=Pass
$\displaystyle P(L|Pa) = \frac{P(L ~and~ Pa)}{P(Pa)}$

can you work out the terms on the right hand side

Hint: Perhaps you already worked them out in a previous question. (or perhaps not, you description of (2) appears to be missing words)
• Jun 10th 2011, 12:49 PM
tousif111
I did,

p(Left hand and passed)= p(Pa) X P(L and Pa)

Could you help me debug my mistake?
• Jun 10th 2011, 01:00 PM
SpringFan25
removing your abbreviations, you have written:

p(Left hand and passed)= p(Pa) X P(Left hand and passed)

Can you see your own mistake?

Moving on to the actual question, you are asked for P(L|Pa), not P(L and Pa).
• Jun 10th 2011, 01:09 PM
bryangoodrich
The first question is asking for P(L and Pa). This can be seen by the proportion of the population that is left-handed (3/4) weighted by the amount of them that would pass (80%). This is a direct product, no? Then P(Pa) is the the weighted sum of both proportions passing. The reason this is a direct product is because the population, I assume, is partitioned by handed-ness. If someone can be both left-handed and right-handed, we would have some issues. But this is not specified. Therefore, P(Pa) = P(is Left handed AND Passed OR is Right handed AND Passed). Does that make sense? If you calculate those, then understanding the definition of conditional probability, as SpringFan25 expressed earlier, you can solve the final question. Here P(Pa) is the total probabilty of the partitioned event space. Notice the key words I've emphasized. Their meaning is important to this problem.