Drawing tree diagrams. after taking a course on statistics, 88% of students can successfully construct tree diagrams. of those who can construct tree diagrams, 90% passed, while only 60% of those students who could not construct tree diagrams passed. a) construct a tree diagram of this scenario. ? you can do it electronically using e. g. powerpoint, or draw freehand and scan the image. b) calculate the probability that a student is able to construct a tree diagram if it is known that he or she passed.

A. Check the image

B. The probability that a student is able to construct a tree diagram if it is known that he or she passed is 79.2%

Explanation:

a) A tree diagram is a tool that simplifies the distribution of probabilities. So we start making branches (lines) according to the probability of each succes to happen, so in this case we first divide the tree diagram between those who can construct tree diagrams that represent the 88% so it means it has a probability of 0.88 and those who can't construct tree diagrams, so if 88% can  just 12% can't construct tree diagrams that is a probability of 0.12.  This is the first part of the tree diagram.

After that tere is another condition that is passing so for those who construct it is the 90% (0.9 probability) for passing, so in this group it is just 10% (0.1 probability) that doesn't pass. Now for those who don't construct tree diagrams, 60% (0.6 probability) pass so 40% losses (0.4)

See in the diagram that if you add the numbers that compound each of the vertices the result is a 100% or 1.0 probability, this is what you have to take into account when you are constructing a tree diagram.

b) For calculating a probability after constructing the tree diagram you just need to multiply each of the factor in which you are interested. In this case:

Construct tree diagram X Pass = 0.88 x 0.9 = 0.792

So the probability for a student that can construct a tree diagram to pass is 0.792, now for having in percentage we just need to multiply by 100% then:

0.792 x 100 = 79.2%

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answer/// product differentiation;