Statistics help ?

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baby_blue_188

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Jun 8, 2012
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A student selected at random probability .76 passing French101.
The probability is .94 for the student that'll pass French 102.
What's the probability that the student selected at random will pass both French 101 & 102?
 
Use the multiplication Rule.

To see the entire distributio of possibilities, multiply these without simplifying anything (0.76+0.24)(0.94 + 0.06).

There is one important problem with this simplistic setup. Does anyone who fails 101 even get to take 102? Hmmm.... An exellent thought question.
 

I have the same thoughts as tkhunny.
Mustn't one pass French 101 in order to be eligible to take French 102?

I suspect that this requirement is tacitly embedded in the problem.

The facts are actually: .\(\displaystyle \begin{array}{ccc}P(\text{pass 101}) \:=\:0.76 \\ P(\text{pass 102}\,|\,\text{passed 101}) \:=\: 0.94 \end{array}\)


Then: .\(\displaystyle P(\text{pass 101 & pass 102})\) is indeed \(\displaystyle (0.76)(0.94)\)
 
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