2. ans C
using set theory
1-P(a union B)
1 - [p(A) + P(B) - P(A int B)]
80/80 - [24 + 40 - 20] / 80
(80 - 44) / 80
36 / 80
I'm struggling with probability problems ...can anyone recommend a good book/site for probability? It'd be deeply appreciated
The probability that events E and F will both occur is
a. The probability that event E will occur.............................b.0.58
30. In a group of 80 students, 24 are enrolled in
geometry, 40 in biology, and 20 in both. If a
student were randomly selected from the 80
students, what is the probability that the student
selected would not be enrolled in either course?
Expecting the world to treat you fairly because you are good is like expecting the bull not to charge because you are a vegetarian.
2. the answer i am getting is ( C ). 0.45 .
prob( choosing a student doing neither course ) = 36 / 80 .
i m uploading an image file , a venn diagram .
1. at most one can say is that p( E) + p( F) = 1.42 .
bcoz , p( E or F ) = p( E) + p( F) - p( E and F ) = 1 .
=> prob( E ) >= 0.42 . also prob( F ) >= 0.42 .
so , in my view , a and b cannot be compared ( lack of info ) .
note even if we don't assume that E & F are exhaustive like above , what p( E and F ) = 0.42 at best gives us is that prob that E occurs is at least 0.42 , and nothing more info .
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