Just another probability question... (1 Viewer)

[lookoutastroboy]

Hey guys,

Just one question which I had trouble with lately:

Question: Three numbers are chosen at random from the numbers 1 to 100. Calculate the probability of:

b) the first number being divisible by 8 and the second being even
c) the first number being even and the second being divisible by 8
d) the three numbers are divisible by 3, 6 and 12 in any order
e) two of the numbers being greater than 50

Answers by the way are
b) 49/825
c) 49/825
d) 62/2695
e) Assuming the question means only two of the numbers are greater than 50 the probability is 75/198. If the question means at least two, then the probability is 1/2.




Thanks in advance, l.a.

 

[AlexJB]

I'll do the first two, and I'll do the other two tomorrow if nobody has done them.

Probability of a number being divisible by 8 is the same as saying the probability of a number being a multiple of 8. Between 1-100, that would be 8, 16, 24, ..., 96. Twelve possibilities. Therefore we start by saying:

Pr(Div by 8) * Pr(Even)
= 12/100 * Pr(even)

Now, if no numbers were removed, the probability of a number being even would be 1/2. However, we have to account for the even number being taken out in the first selection (if the selected number was divisible by 8). If a number divisible by 8 were selected, then there would be one less even number. Instead of 50/99, there would be 49/99

Therefore, equal to: 12/100 * 49/99 = your answer?

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Second is kind of the same.

= Pr(even) * Pr(divisible by 8)
Think the answer above would be enough, but to wrap your head around the order you could do:

= 1/2 * Pr(div.by 8)
= 1/2 * (12/99 * 49/50) [theres one less even number, so instead of 50, theres 49/50]

= your answer?