mathmath333
  • mathmath333
Probablity question
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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mathmath333
  • mathmath333
\(\large \color{black}{\begin{align} & \normalsize \text{Three different numbers are selected from the set}\ X=\{1,2,3,4...10\}\hspace{.33em}\\~\\ & \normalsize \text{What is the probablity that the product of the two numbers is equal to }\hspace{.33em}\\~\\ & \normalsize \text{the third }\hspace{.33em}\\~\\ \end{align}}\)
mathmath333
  • mathmath333
\(\large \color{black}{\begin{align} & a.)\ \dfrac{3}{10}\hspace{.33em}\\~\\ & b.)\ \dfrac{1}{40}\hspace{.33em}\\~\\ & c.)\ \dfrac{1}{20}\hspace{.33em}\\~\\ & d.)\ \dfrac{4}{5}\hspace{.33em}\\~\\ \end{align}}\)
ganeshie8
  • ganeshie8
say the \(3\) numbers selected are \(\{a,b,c\}\)

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ganeshie8
  • ganeshie8
also suppose that \(a\lt b\lt c\)
mathmath333
  • mathmath333
ok
ganeshie8
  • ganeshie8
then you want \(a*b\le 10\)
ganeshie8
  • ganeshie8
notice that if \(a*b\le 10\), then \(a \le \sqrt{~10~}\)
ganeshie8
  • ganeshie8
so \(a\) can only be either \(2\) or \(3\)
ganeshie8
  • ganeshie8
when \(a=2\), check what all \(b\) values will work
mathmath333
  • mathmath333
how does the condition \(a=\sqrt{10}\) came.
ganeshie8
  • ganeshie8
good question, that is called seive of eratosthenes https://en.wikipedia.org/wiki/Sieve_of_Eratosthenes
mathmath333
  • mathmath333
ok
ganeshie8
  • ganeshie8
we have \(ab\le 10\) and you know that \(a\lt b\) whats the maximum value that \(a\) can take ?
mathmath333
  • mathmath333
2
ganeshie8
  • ganeshie8
why not 3 ?
mathmath333
  • mathmath333
cuz \(3\times 4=12\)
ganeshie8
  • ganeshie8
Perfect! so the only value that \(a\) can take is \(2\)
mathmath333
  • mathmath333
why not \(1\) ?
ganeshie8
  • ganeshie8
you tell me why
mathmath333
  • mathmath333
\(1*2\le 10\)
ganeshie8
  • ganeshie8
remember you want to pick all 3 "different" numbers
ganeshie8
  • ganeshie8
what happens if any one number is \(1\) ?
mathmath333
  • mathmath333
it satisfies the condition example \(1,2,3\)
ganeshie8
  • ganeshie8
it doesn't, 1*2 is not 3
ganeshie8
  • ganeshie8
the smaller two numbers must multiply to the third number
mathmath333
  • mathmath333
oh i see, sry
mathmath333
  • mathmath333
yea , only \(2\) for \(a\)
ganeshie8
  • ganeshie8
yes let \(a=2\) and find all \(b\) such that \(a*b\le 10\)
mathmath333
  • mathmath333
\(b=3,4,5\)
ganeshie8
  • ganeshie8
Yes, so the numbers in favor are : (2, 3, 6) (2, 4, 8) (2, 5, 10)
ganeshie8
  • ganeshie8
thats 3 in favor
ganeshie8
  • ganeshie8
save that
ganeshie8
  • ganeshie8
next, find how many total ways are there to choose 3 different numbers from the given 10 numbers
mathmath333
  • mathmath333
Is their any condition to choose 3 different numbers from the given 10 numbers.
ganeshie8
  • ganeshie8
thats the sample space so no conditions, just find the total number of ways of choosing 3 numbers from 10
mathmath333
  • mathmath333
\(\dfrac{10!}{7!}\)
ganeshie8
  • ganeshie8
nope
ganeshie8
  • ganeshie8
its just \(\large \dbinom{10}{3}\)
ganeshie8
  • ganeshie8
notice that here order doesn't matter... so it is a combination
mathmath333
  • mathmath333
why not \(^{10}P_{3}\) ??
mathmath333
  • mathmath333
ok
ganeshie8
  • ganeshie8
favor : \(\large 3\) total : \(\large \dbinom{10}{3}\)
ganeshie8
  • ganeshie8
take the ratio for the probability
mathmath333
  • mathmath333
so answer=\(\dfrac{1}{40}\)
ganeshie8
  • ganeshie8
Yes, did u get why this is a combination and not a permutation problem ?
mathmath333
  • mathmath333
cuz we choosing 3 numbers one by one and not forming 3 numbers like \(abc_{10}\)
ganeshie8
  • ganeshie8
yes, we never bothered about order while doing this problem
ganeshie8
  • ganeshie8
we could also do it using permutations but it is painful here
ganeshie8
  • ganeshie8
if we use permutations, the count of favor also changes
ganeshie8
  • ganeshie8
(2, 3, 6) (2, 4, 8) (2, 5, 10)
ganeshie8
  • ganeshie8
since (2, 3, 6) works, all the 6 permutations of it also work : (2, 6, 3) (3, 2, 6) (3, 6, 2) (6, 2, 3) (6, 3, 2)
ganeshie8
  • ganeshie8
after all that mess, you will get the same answer
ganeshie8
  • ganeshie8
In these probability problems, it doesn't matter whether you use permutations or combinations.. the final answer wont change if you do it correctly

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