SOLUTION: If A and B are independent events, P(A) = 0.35, and P(B) = 0.35, find the probabilities below. (Enter your answers to four decimal places.) (a) P(A ᑎ B) (b) P(A

Since they are independent events, by definition of independent
events, P(A ᑎ B) = P(A)×P(B) = 0.35×0.35 = 0.1225 

P(A ᑌ B) = P(A) + P(B) - P(A ᑎ B) = 0.35 + 0.35 - 0.1225 = 0.5775

Since they are independent, knowing that B has occurred does not
affect the probability of A, so it's the same as P(A) = 0.35.
But you can do it by the formula for conditional probability, and
get the same thing:

P(A | B) = P(A ᑎ B)/P(B) = 0.1225/0.35 = 0.35 

(d) P(Ac ᑌ Bc) 

P(Ac) = 1-P(A) = 1-0.35 = 0.65
P(Bc) = 1-P(B) = 1-0.35 = 0.65

If two events are independent, so are their complements*, so
P(Ac ᑎ Bc) = P(Ac)×P(Bc) = (0.65)(0.65) = 0.4225

P(Ac ᑌ Bc) = P(Ac) + P(Bc) - P(Ac ᑎ Bc) = 0.65 + 0.65 - 0.4225 = 0.8775.

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*If you need to prove that,

P(Ac ᑎ Bc)              <-- use DeMorgan's law:
= P[(A ᑌ B)c] 
= 1 - P(A ᑌ B) 
= 1 - [P(A) + P(B) - P(A ᑎ B)] 
= 1 - [P(A) + P(B) - P(A)×P(B)]
= 1 - P(A) - P(B) + P(A)×P(B)    <--factor -P(B) out of last two terms 
= 1 - P(A) - P(B)×[1 - P(A)] 
= P(Ac) - P(B)×P(Ac)             <--factor P(Ac) out of both terms 
= P(Ac)×[1-P(B)] 
= P(Ac)×P(Bc) 

Edwin