Question 1082877
There are 14 girls
There are 15 boys
Making 15+14 = 29 students total


Let's define two events
K = event of picking Kate
F = event of picking a girl


The probability of event F is
P(F) = 14/29
since there are 14 girls out of 29 students total


The probability of event K is
P(K) = 1/29 
since there's only one way to pick Kate out of 29 students total


Because Kate is a girl, this means that the event "K and F" is really just K. 
Selecting Kate is the same as selecting a girl, but not the other way around. 
So that's why P(K and F) = 1/29 as well.


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*[Tex \LARGE P(K | F) = \frac{P(K \text{ and } F)}{P(F)}]


*[Tex \LARGE P(K | F) = \frac{1/29}{14/29}]


*[Tex \LARGE P(K | F) = \frac{1}{29} \div \ \frac{14}{29}]


*[Tex \LARGE P(K | F) = \frac{1}{29} \times \ \frac{29}{14}]


*[Tex \LARGE P(K | F) = \frac{1*29}{29*14}]


*[Tex \LARGE P(K | F) = \frac{1}{14}]


*[Tex \LARGE P(K | F) \approx 0.07142857]


*[Tex \LARGE P(K | F) \approx 0.071] Round to 3 decimal places


This implies that if we're guaranteed information that we picked a female, then the chances of picking Kate are roughly 0.071 (converts to 7.1%). Since we know the selected person is female, we can ignore the males completely. 


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*[Tex \LARGE P(F | K) = \frac{P(K \text{ and } F)}{P(K)}]


*[Tex \LARGE P(F | K) = \frac{1/29}{1/29}]


*[Tex \LARGE P(F | K) = 1]


This says "if you know you selected Kate, then the probability of picking a girl is 100%". This is due to the nature of how the sets K and F are related. K is a subset of F.


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In summary the answers are
*[Tex \LARGE P(K | F) \approx 0.071]
*[Tex \LARGE P(F | K) = 1]
The first result is approximate to 3 decimal places. The second result is exactly 1 indicating "100% chance of happening" or "guaranteed to happen".