Thomas Baynes has applied to both Harvard and Yale. Baynes has determined that the probability of getting into Harvard is 25 percent and the probability of getting into Yale (his father’s alma mater) is 42 percent. Baynes has also determined that the probability of being accepted at both schools is 2.8 percent. What is the probability of Baynes being accepted at either Harvard or Yale, but not both? From qbank. Answer is 64.2… I get 61.4 (not an option).
Additon Rule: P(A or B) = P(A) + P(B) - P(A and B) 25 + 42 - 2.8 = 64.2
Doesn’t that give probability of acceptance into at least 1 school?
P(Harvard) = 0.25 P(Yale) = 0.42 P(Harvard and Yale) = 0.028 For P(A or B) = P(A) + P(B) - P(A and B) = 0.25 + 0.42 - 0.028 = 0.642
It gives the probability being accepted into Harvard or Yale but not both (which is different than being accepted to both schools) Note in the question the key word is OR: Harvard OR Yale. Another hint is that they want the probability of one or the other schools but not both (the crossover in the Venn diagram). This gives the hint that you’re dealing with the addition rule
My question, Addition rule gives Prob (A or B or BOTH) They are asking for Prob(A or B but NOT both) So why aren’t we subtracting 2.8 from 64.2?
You see there are 2 categories of Addition rule, namely: (i) Addition Rule (for mutually exclusive events) (ii)Addtion Rule (for mutually not-exclusive events) For this question, it is referring to addition rule no (ii). ‘Mutually not-exclusive events’ means two or more events can occur together simultaneously. Therefore the formula: P(A or B) = P(A) + P(B) - P(AB) Now your confusion (i presumed) arise here, the misconception of a mutually exclusive and not-exclusive event. It will be easier if you were to draw a Venn diagram. For mutually non-exclusive events (Remember 2 events can occur at the same time) there will be an intersection of 2.8% between event A and event B. 'What is the probability of Baynes being accepted at either Harvard or Yale, but not both? ’ Looking at the question it fits the equation. Deducting 2.8% once is sufficient to reflect the intersection probabilty.
Sorry I’m still not getting it 
28% A 42% B 28 + 42 counts the intersection 2.8% twice, thus we subtract once for the double counting. But we’re still counting counting the overlap once. In other words, the first -2.8 removes for double counting, but in the vehn diagram we want the overlap to be empty so why don’t we subtract 2.8 a secon time?
copy the link in ur browser http://img253.imageshack.us/my.php?image=53843236jt0.jpg If that’s wrong someone please correct me.
draw a venn diagram- that will help you see it. The answer is 64.2= 25 + 42 - 2.8 you have two circles - one is 25, the other is 42 and they share the area 2.8. The probability of either one circle but not both is the total area less the shared area.
Thanks a lot guys. I see it now.