counting problem

[nwicole]

Q1
Two dice are rolled , one blue and one red. how many outcomes have either the blue die 3 or an even or both?

Q2
How many integers from 1 to 10,000 , inclusive , are multiples of 5 or 7 or both?

 

[stapel]

Q1. Two dice are rolled , one blue and one red. how many outcomes have either the blue die 3 or an even or both?

The red die is irrelevant, obviously, so you can ignore it. Assuming that the blue die is six-sided and that the numbers on the sides are 1 through 6, how many of the outcomes are either a 3 or a 2, 4, or 6? (Since 3 isn't even, there can't be a "both".)

Q2: How many integers from 1 to 10,000 , inclusive , are multiples of 5 or 7 or both?

How many are multiples of 5? How many are multiples of 7? Of course, these counts include multiples of 5*7 = 35, so some of each count are duplicates. How would you remove these? (In other words, the "or both" is already included, but is over-included. So don't worry about the "or both"; worry about removing the over-counting.)

If you get stuck, please reply showing your work so far. Thank you!

 

[pka]

Q1.Two dice are rolled , one blue and one red. how many outcomes have either the blue die 3 or an even or both.

Above is that a typo & you meant on both?

Q2 How many integers from 1 to 10,000 , inclusive , are multiples of 5 or 7 or both?

Here is a useful fact if \(\displaystyle N\) is a positive integer and \(\displaystyle 1\le k\le N\) then the number of multiples of 1 in 1 to N is \(\displaystyle \left\lfloor {\dfrac{N}{k}} \right\rfloor \). (That is the floor function.)

So the answer to this question is:
\(\displaystyle \left\lfloor {\dfrac{10^4}{5}} \right\rfloor +\left\lfloor {\dfrac{10^4}{7}} \right\rfloor -\left\lfloor {\dfrac{10^4}{35}} \right\rfloor \)

 

[nwicole]

The red die is irrelevant, obviously, so you can ignore it. Assuming that the blue die is six-sided and that the numbers on the sides are 1 through 6, how many of the outcomes are either a 3 or a 2, 4, or 6? (Since 3 isn't even, there can't be a "both".)


How many are multiples of 5? How many are multiples of 7? Of course, these counts include multiples of 5*7 = 35, so some of each count are duplicates. How would you remove these? (In other words, the "or both" is already included, but is over-included. So don't worry about the "or both"; worry about removing the over-counting.)

If you get stuck, please reply showing your work so far. Thank you!

I got stick on q1 is i dont get why the answer is 21.
step 1: If i pick blue , there are 6 ways
step 2: I can get 3 chances for even which is 2,4,6, so there are 3 ways
step 3: why shall I subtract 3?



q2, if i used 10,000/5 i got 2000; if i used 10,000/7=1428 , 10,000/75 i had 285?
so shall i just add 2000+1428-285?

thank you so much

 

[nwicole]

Above is that a typo & you meant on both?


Here is a useful fact if \(\displaystyle N\) is a positive integer and \(\displaystyle 1\le k\le N\) then the number of multiples of 1 in 1 to N is \(\displaystyle \left\lfloor {\dfrac{N}{k}} \right\rfloor \). (That is the floor function.)

So the answer to this question is:
\(\displaystyle \left\lfloor {\dfrac{10^4}{5}} \right\rfloor +\left\lfloor {\dfrac{10^4}{7}} \right\rfloor -\left\lfloor {\dfrac{10^4}{35}} \right\rfloor \)

Q 1 so the question should be on both? this is the part i dont get get why for both which is not happen lol

but i dont why the answer is 6+(6*3)-3 = 21?

thank you so much

 

[stapel]

Q 1 so the question should be on both?

The helper was asking, not informing. Please re-check the exercise. Is the word "on" or is the word "or"? The answer to the question will depend on what the actual question was. If "or", then follow my reply. If "on", then follow the other reply. :wink: