In a statistics class there are 11 juniors and 6 seniors; 4 of the seniors are females, 6 of the juniors are males. If a student is selected at random, what is the probability that the student is either a senior or a male?

Answer 1

#65.74%#

let #A# be the event of being a senior and #B# the event of male We know #p(A) = 6/17# #p(notB|A) = 4/6# #p(B|notA) = 6/11#

What we can figure out is

#p(notA) = 11/17# #p(B|A) = 2/6# #p(notB|notA) = 5/11#

We are then asked to sovle #p(A or B) = p(A)+p(B) - p(A)*p(B)#. we just need to figure out #p(B)# and solve. Based on the law of total probability #p(B) = sum_i^n p(B,A_i)# in our case

#p(B) = p(B|A)p(A) +p(B|notA)p(notA) # #p(B) =2/6 * 6/17 +6/11*11/17 = 8/17 #

now we solve #p(A or B) = 6/17+8/17- 6/17*8/17 = .6574 #

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.