Question

Probablity question

Answer 1

\(\large \color{black}{\begin{align}
& \normalsize \text{Three different numbers are selected from the set}\ X=\{1,2,3,4...10\}\hspace{.33em}\\~\\
& \normalsize \text{What is the probablity that the product of the two numbers is equal to }\hspace{.33em}\\~\\
& \normalsize \text{the third }\hspace{.33em}\\~\\
\end{align}}\)

Answer 2

\(\large \color{black}{\begin{align}
& a.)\ \dfrac{3}{10}\hspace{.33em}\\~\\
& b.)\ \dfrac{1}{40}\hspace{.33em}\\~\\
& c.)\ \dfrac{1}{20}\hspace{.33em}\\~\\
& d.)\ \dfrac{4}{5}\hspace{.33em}\\~\\
\end{align}}\)

Answer 3

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

Answer 4

also suppose that \(a\lt b\lt c\)

Answer 5

ok

Answer 6

then you want \(a*b\le 10\)

Answer 7

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

Answer 8

so \(a\) can only be either \(2\) or \(3\)

Answer 9

when \(a=2\), check what all \(b\) values will work

Answer 10

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

Answer 11

good question, that is called seive of eratosthenes

https://en.wikipedia.org/wiki/Sieve_of_Eratosthenes

Answer 12

ok

Answer 13

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

Answer 14

2

Answer 15

why not 3 ?

Answer 16

cuz \(3\times 4=12\)

Answer 17

Perfect! so the only value that \(a\) can take is \(2\)

Answer 18

why not \(1\) ?

Answer 19

you tell me why

Answer 20

\(1*2\le 10\)

Answer 21

remember you want to pick all 3 "different" numbers

Answer 22

what happens if any one number is \(1\) ?

Answer 23

it satisfies the condition example \(1,2,3\)

Answer 24

it doesn't, 1*2 is not 3

Answer 25

the smaller two numbers must multiply to the third number

Answer 26

oh i see, sry

Answer 27

yea , only \(2\) for \(a\)

Answer 28

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

Answer 29

\(b=3,4,5\)

Answer 30

Yes, so the numbers in favor are :

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

Answer 31

thats 3 in favor

Answer 32

save that

Answer 33

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

Answer 34

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

Answer 35

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

Answer 36

\(\dfrac{10!}{7!}\)

Answer 37

nope

Answer 38

its just \(\large \dbinom{10}{3}\)

Answer 39

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

Answer 40

why not \(^{10}P_{3}\) ??

Answer 41

ok

Answer 42

favor : \(\large 3\)

total : \(\large \dbinom{10}{3}\)

Answer 43

take the ratio for the probability

Answer 44

so answer=\(\dfrac{1}{40}\)

Answer 45

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

Answer 46

cuz we choosing 3 numbers one by one and not forming 3 numbers like \(abc_{10}\)

Answer 47

yes, we never bothered about order while doing this problem

Answer 48

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

Answer 49

if we use permutations, the count of favor also changes

Answer 50

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

Answer 51

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)

Answer 52

after all that mess, you will get the same answer

Answer 53

In these probability problems, it doesn't matter whether you use permutations or combinations..
the final answer wont change if you do it correctly