## mathmath333 one year ago Probablity question

1. 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}}

2. 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}}

3. ganeshie8

say the $$3$$ numbers selected are $$\{a,b,c\}$$

4. ganeshie8

also suppose that $$a\lt b\lt c$$

5. mathmath333

ok

6. ganeshie8

then you want $$a*b\le 10$$

7. ganeshie8

notice that if $$a*b\le 10$$, then $$a \le \sqrt{~10~}$$

8. ganeshie8

so $$a$$ can only be either $$2$$ or $$3$$

9. ganeshie8

when $$a=2$$, check what all $$b$$ values will work

10. mathmath333

how does the condition $$a=\sqrt{10}$$ came.

11. ganeshie8

good question, that is called seive of eratosthenes https://en.wikipedia.org/wiki/Sieve_of_Eratosthenes

12. mathmath333

ok

13. ganeshie8

we have $$ab\le 10$$ and you know that $$a\lt b$$ whats the maximum value that $$a$$ can take ?

14. mathmath333

2

15. ganeshie8

why not 3 ?

16. mathmath333

cuz $$3\times 4=12$$

17. ganeshie8

Perfect! so the only value that $$a$$ can take is $$2$$

18. mathmath333

why not $$1$$ ?

19. ganeshie8

you tell me why

20. mathmath333

$$1*2\le 10$$

21. ganeshie8

remember you want to pick all 3 "different" numbers

22. ganeshie8

what happens if any one number is $$1$$ ?

23. mathmath333

it satisfies the condition example $$1,2,3$$

24. ganeshie8

it doesn't, 1*2 is not 3

25. ganeshie8

the smaller two numbers must multiply to the third number

26. mathmath333

oh i see, sry

27. mathmath333

yea , only $$2$$ for $$a$$

28. ganeshie8

yes let $$a=2$$ and find all $$b$$ such that $$a*b\le 10$$

29. mathmath333

$$b=3,4,5$$

30. ganeshie8

Yes, so the numbers in favor are : (2, 3, 6) (2, 4, 8) (2, 5, 10)

31. ganeshie8

thats 3 in favor

32. ganeshie8

save that

33. ganeshie8

next, find how many total ways are there to choose 3 different numbers from the given 10 numbers

34. mathmath333

Is their any condition to choose 3 different numbers from the given 10 numbers.

35. ganeshie8

thats the sample space so no conditions, just find the total number of ways of choosing 3 numbers from 10

36. mathmath333

$$\dfrac{10!}{7!}$$

37. ganeshie8

nope

38. ganeshie8

its just $$\large \dbinom{10}{3}$$

39. ganeshie8

notice that here order doesn't matter... so it is a combination

40. mathmath333

why not $$^{10}P_{3}$$ ??

41. mathmath333

ok

42. ganeshie8

favor : $$\large 3$$ total : $$\large \dbinom{10}{3}$$

43. ganeshie8

take the ratio for the probability

44. mathmath333

so answer=$$\dfrac{1}{40}$$

45. ganeshie8

Yes, did u get why this is a combination and not a permutation problem ?

46. mathmath333

cuz we choosing 3 numbers one by one and not forming 3 numbers like $$abc_{10}$$

47. ganeshie8

yes, we never bothered about order while doing this problem

48. ganeshie8

we could also do it using permutations but it is painful here

49. ganeshie8

if we use permutations, the count of favor also changes

50. ganeshie8

(2, 3, 6) (2, 4, 8) (2, 5, 10)

51. 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)

52. ganeshie8

after all that mess, you will get the same answer

53. 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