Two distinct integers are selected at random from the list of integers

Question

Two distinct integers are selected at random from the list of integers from 10 to 60, inclusive. What is the probability that the chosen numbers are divisible by 6 or 9?

A-  \frac{3}{85}

B-  \frac{9}{245}

C- \frac{11}{255}

D- \frac{26}{425}

E-  \frac{156}{2450}

IanStewart · Answer

Between 10 and 60 inclusive we have:

- five multiples of 9
- nine multiples of 6

We can't just add these numbers (five plus nine) to work out how many numbers are multiples of six or nine, because some numbers are in both groups: if a number is a multiple of 18 (the LCM of 9 and 6) it is divisible by both. If we just add 5+9, we'll be counting every multiple of 18 twice. We have three multiples of 18 between 10 and 60, so we must subtract three to avoid counting those multiples of 18 twice, and we thus have 5+9-3 = 11 distinct numbers between 10 and 60 inclusive which are divisible by six or by nine (or by both).

So, when we pick our first number, 11 numbers 'work' out of the 51 we are choosing from. Once we pick that number, we have 50 numbers left in total, and 10 of them 'work'. So the probability both numbers 'work' is:

(11/51)(10/50) = (11/51)(1/5) = 11/255

JBN96 · Answer

Numbers that are devisable by 6 or 9 are 12, 18, 24, 27, 30, 36, 42, 45, 48, 54, 60. So 11 numbers in total. Both numbers have to be distinct from each other so 11/51*10/50=110/2550=11/255 so C

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Hazel1134 · Answer

Numbers that satisfy the above conditions: 
1) div by 6 : 60/6 - 1 = 9 
2) div by 9 : 60 / 9 - 1 = 5 (count only the integer) 
3) satisfy both 6 and 9 (multiples of 18) : 60 / 18 = 3 (count only the integer).

You subtract 1 from 1) & 2), but not from 3) because, 6 and 9 is under the given range, but 18 is not. ;D

9 + 5 - 3 = 11 numbers satisfy the above conditions.

Now, there are 51 numbers from 10 ~ 60 and we are to pick numbers twice that satisfy the condition stated without replacement, hence it becomes as follows:

(11/51) * (10/50) = 11 / 255

Answer C

exc4libur · Answer

assuming that two distinct integers are selected without replacement and the probability is that any of the chosen numbers is divisible by 6 or 9, then:

total numbers: 60-10+1=51
divisible by 6: (54-12)/6+1=9
divisible by 9: (54-18)/9+1=5
divisible by 18  : (54-18)/18+1=3
favorable numbers: 9+5-3=11 (-3 is the overlap: ie. {18,36,54} are both multiples 6 and 9)

P(6,9)=11/51•10/50=11•2•5/51•2•5•5=11/51•5=11/255

Answer (C)

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Two distinct integers are selected at random from the list of integers

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