Question

Consider the following contingency table.

	B	Bc
A	26	26
Ac	20	28

a. Convert the contingency table into a joint probability table. (Round intermediate calculations to at least 4 decimal places and final answer to 4 decimal places.)

	B	Bc	Total
A
Ac
Total

b. What is the probability that A occurs? (Round intermediate calculations to at least 4 decimal places and final answer to 4 decimal places.)
  

c. What is the probability that A and B occur? (Round intermediate calculations to at least 4 decimal places and final answer to 4 decimal places.)
  

d. Given that B has occurred, what is the probability that A occurs? (Round intermediate calculations to at least 4 decimal places and final answer to 4 decimal places.)
  

e. Given that A^c has occurred, what is the probability that B occurs? (Round intermediate calculations to at least 4 decimal places and final answer to 4 decimal places.)
  

f. Are A and B mutually exclusive events?
  

• Yes because P(A | B) ≠ P(A).

• Yes because P(A ∩ B) ≠ 0.

• No because P(A | B) ≠ P(A).

• No because P(A ∩ B) ≠ 0.

g. Are A and B independent events?
  

• Yes because P(A | B) ≠ P(A).

• Yes because P(A ∩ B) ≠ 0.

• No because P(A | B) ≠ P(A).

• No because P(A ∩ B) ≠ 0.

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Answer 1

	B	Bc	total
A	26	26	52
Ac	20	28	48
total	46	54	100

( b ) P (  A occurs ) = 52 /100 = 0.5200

( c ) p ( A and B occur ) = 26 /100 = 0.2600

( d ) P (  A occurs /  B has occurred ) =p ( A and B occur ) / P( B ) = 26 / 46 = 0.5652

( e ) P ( B occurs /  A^c has occurred ) = P ( B and  A^c) / P ( A^c )

= 20 / 48

= 0.4167

( f )  No because P(A ∩ B) ≠ 0.

( g ) No because P(A | B) ≠ P(A).