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Question 2 (1 point) Saved A survey found that 83% of Canadians have a smartphone, 39% of Canadians have a tablet and 28% of Canadians have both. What is the probability that a randomly selected Canadian has a smartphone or a tablet? 94% 55% 122% 67%

**94%**. Your answer is correct.

First, we define the following events:

S = event that Canadians have a smartphone

T = event that Canadians have a tablet

Given are the following probabilities:

$P(S)=83%P(T)=39%P(S∩T)=28%$ (In probabilistic terms, the term 'and' refers to an intersection of events or ∩.)

Now we have to find the probability that a randomly selected Canadian has a smartphone or a tablet. We write it in symbols as

$P(S∪T)=?$ (In probabilistic terms, the term 'or' refers to a union of events or ∪.)

Using the general probability addition rule for the union of two events which is given by the formula

$P(A∪B)=P(A)+P(B)−P(A∩B)$

we have

$P(S∪T)=P(S)+P(T)−P(S∩T)$

$P(S∪T)=83%+39%−28%$

$P(S∪T)=94%$

I hope this helps!