There are 5 people at a party each person must shake hands once with every other person at the party how many handshakes dose it
1 answer:
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Answer:
(a) A y P(A) = 0.4 (b) y
=0.5 (c)
∪
y P(
∪
) = 0.9 (d)
∩
y P(
∩
)=0.2
Step-by-step explanation:
A was defined as the event that a potential customer, randomly chosen, buys from outlet 1 in the original problem statement. We know that B denotes the event that a randomly chosen customer buys from outlet 2. So
P() = 0.3, P(
) = 0.4 and P(
) = 0.1
(a) P(A) = P() = P(
) + P(
) = 0.1 + 0.3 = 0.4
(b) P(B) = P() = P(
) + P(
) = 0.1 + 0.4 = 0.5
P( ) = 1-P(B) = 1-0.5 = 0.5
(c) The customer does not buy from outlet 1 is the complement of A, i.e., , and the customer does not buy from outlet 2 is the complement of B, i.e.,
, so, the customer does not buy from outlet 1 or does not buy from outlet 2 is
∪
and P(
∪
) = P(
) by De Morgan's laws
P() = 1-P(A∩B)=1-0.1=0.9
(d) The customer does not buy from outlet 1 is the complement of A, and the customer does not buy from outlet 2 is the complement of B, so we have that the statement in (d) is equivalent to ∩
and P(
∩
) = P(
) by De Morgan's laws, and
P() = 1-P(A∪B)=1-[P(A)+P(B)-P(A∩B)]=1-[0.4+0.5-0.1]=1-0.8=0.2