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A number is chosen at random from the positive, even integers from 2 to 50. Find the probability that the number chosen is divisible by 5 or 8.


1/25


2/5


11/25

Verified Answer :

konrad509

There are 25 even integers between 2 and 50 (2,4,6,...,50).
Among them, there are 5 numbers divisible by 5 (10,20,30,40,50), 6 divisible by 8 (8,16,24,32,40,48), and 1 divisible by 5 and 8 simultaneously (40).

The last means these two events are not mutually exclusive. So, if

A - the event of choosing a number divisible by 5,
B - the event of choosing a number divisible by 8,

the probability of choosing a number divisible by 5 or 8 is equal to
\(P(A\cup B)=P(A)+P(B)-P(A\cap B)\)

So,
\(P(A\cup B)=\dfrac{5}{25}+\dfrac{6}{25}-\dfrac{1}{25}\\ P(A\cup B)=\dfrac{10}{25}\\ \boxed{P(A\cup B)=\dfrac{2}{5}}\)

^^ What this guy above said ^^

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