A boy is asked to pick 3 marbles at random from a box containing 6 white, 5 red and 4 green marbles.?

Question

In how many ways can the set chosen contain at least 1 red marble?

Answer is 335. But i need the workings. Thanks

I tried using different cases where there is no red marble. But doesnt work. :(.

Answer 1

There are 6 white, 5 red and 4 green marbles.
If we randomly pick 3 marbles, at least one red marble.
So,

For 3 marbles, 1 red
no of way
= 5 C 1 x 10 C 2
= (5 / 1) x (10 x 9) / (2 x 1)
= 5 x 45
= 225

For 3 marbles, 2 red
no of way
= 5 C 2 x 10 C 1
= (5 x 4) / (2 x 1) x 10 / 1
= 10 x 10
= 100

For 3 marbles, 3 red
no of way
= 5 C 3
= (5 x 4 x 3) / (3 x 2 x 1)
= 10

Total number of way
= 225 + 100 + 10
= 335

Answer 2

its good you know the answer cuz that makes it so I can show you to that answer. I assume you know what factorial is, written as for example 2! which is 2 x 1. Factorial of 3 would be 3 x 2 x 1. A factorial of any number (n) would be n x (n-1),(n-2) all the way down to 1. I have done this same type of thing when for kicks I wanted to find out using the daily lottery number which is 4 digits to pick from 10 possibly number (0 thru 9) and they can repeat. Now without having to figure anything out if you go from 0000 to 9999 you have 10,000 which I am sure you would agree. Also to get that same 10,000 you take the placement spots (4) x possibly choices (10) gives same 10,000. Now to find how many there would be if all 4 digits were different say 1,2,3,4 or like 0,6,3,7 you would take the 10! divided by 4! but only going 4 numbers deep on the 10!. That gives you (10 x 9 x 8 x 7)/(1x2x3x4) you can multiply top and bottom through and then solve giving you 5040/24 = 210 and that is why they will pay you 208 if you choose 4 digits correctly, if you played $1 each for the 210 and it pays 208 you would lose $2 overall, and there is just slightly better than 50% they come up all 4 different. Going back to our problem. 6 + 5 + 4 = 15 so u know you want to pick 3 out of 15 possibles so u do 15*14*13/1*2*3
multiply and divide and you get 455. Knowing this is total using all 15 you can figure what the total would be without the 5 red and the difference of the two gives you answer you want. Take 10 * 9 * 8/1 x 2 x 3 gives you 720/6 leaving you with 120. So you would have 455 using all 3 colors and 120 without the red gives you the 335. To prove this using smaller amount pick 4 numbers from a possible of 5. Total to figure all different would be (2 x 3 x 4 x 5)/(1 x 2 x 3 x 4) gives you 120/24 leaves 5 possible scenarios of picking 4 numbers out of 5 choices and all being different, 1234,1235,1245,1345, and 2345. I hope it helps. As I said it was something I looked up awhile back for fun and have kept the formulations and computations memorized. Email me with any questions, [e-mail]

Answer 3

It SHOULD work....

Total ways for 3 marbles: 15!/(12!3!) = 15*14*13/6 =

= 5*7*13 = 455

Ways to pick 3 marbles that are all non-red:

10!/(7!3!) = 10*9*8/6 = 10*3*4 = 120

455 - 120 = 335

Answer 4

Here, u have to find the number of ways the set can be provided at least 1 red marble.
So u first calculate the total number of set containing 3 marbles regardless their color.
15C3 = 455
Now, calculate when all 3 outcomes should be any outcome of red marble.
10C3 = 120
Subtract above two numbers will get th set containing at least 1 red marble.(355).

The concept behind is.....
Let us say
C0 - No outcome of red colour
C1 - exactly 1 marble of red color
C2 - exacly 2 marble of red color
S - Total outcome regardless of the color

S = C0 + C1 + C2
and, C1+C2 - means outcome containing at least 1 red color marble.
then,
C1+C2 = S - C0;
this is what i have applied above.