Question

1. Suppose that the table below represents the results of a survey of 250 residents in an apartment  that were asked whether they prefer chips, cookies, or fruit and whether they were pregnant or not. Let A denote the event that a person selected at random prefers chips and B denote the event that a person selected at random is pregnant.
   1. Calculate P(A)
   2. Calculate P(B)
   3. Calculate P(A and B)
   4. Calculate P(A or B)
   5. Calculate P (B - A)
   6. Calculate P (A|B)
   7. Calculate P(B|A^c)

	Chips	Cookies	Fruit	Total
Pregnant	54	23	42	119
Not pregnant	22	11	98	131
Total	76	34	140	250

Answer 1

Let A denote the event that a person selected at random prefers chips and B denote the event that a person selected at random is pregnant.

Hence, n(A) = Number of person prefer chips = 76. n(B) = number of person is pregnant = 119,

n(S) = total number of person. = 250

a) P(A) = n(A) / n(S) = 76/250 = 0.304

b) P(B) = n(B) / n(S) = 119 / 250 = 0.476

c) P( A and B) = n(A and B) / n(S) = 54/250 = 0.216

d) P(A or B) = P(A) + P(B) - P(A and B) = 0.304+0.476-0.216 = 0.564

e) P(B - A) = P(B) - P(A and B) = 0.476 - 0.216 = 0.260

f) P(A |B) = 54/119 = 0.454

g) P(B | A^c ) = P( B and A^c ) / P(A^c ) = P(B-A) / (1-P(A)) = 0.260 / (1-0.304) = 0.855

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