My question is "if the probability of a company’s pen manufacturing defects were 1/10, and if 12 such pens were manufactured, what would be the probability of the following:
1.)exactly two would be defective??
2.)at least two will be defective??
3.))none will be defective??
I am not asking you to do my homework for you but i dont want to show you all the solutions i tried and take up space. just to prove i tried working on it i tackled it the following way (i am sure i am wrong).:
ans 1 => mean = 2*(1/10),variance = 4*.1, and s.d = .2 - .4 = -.2…now how do i get p(x=2)??/?
This is a really cool website and it also addresses permutatons and combinations which is my worst topic ever. Kalid I never seem to understand this topic. no matter what. The best so far has been your small intro to this topic but even after reading your explanations whenever I try new questions on this i get stuck. Can you please give a detailed post on this topic of combinatorics. I will be very very very grateful.
I dont know why but whenever I start doing these questions its like a wall comes up in my mind…do you think i need to think more on these questions? My basics?What could be the problem?
@Frank: Picking 4 from 20 would be a combination because you don’t care the order. In this case, you’d plug in k=4 and n=20 into the combination formula above n!/[(n-k)!k!]. k is the number of items you want, and n is the number of total items.
@Benjamin: This is more of a stats problem, but I’ll give some high-level points. The 3) is easiest: you need to find the chances that all pens worked well. The chance of 1 pen working is 9/10, so the chance of every one working is (9/10)^12.
For the other questions, it helps to invert. For the chance that at least 2 are defective, you can think about the chance exactly 0 or 1 pens are defective, and take the opposite probability. These can get a little tricky to compute – I’ll probably have to do a post on it.
@Mohammad: Thanks for the suggestion, it seems people would like a more detailed look at these. I’m not an expert but have found a few techniques that work for me. A lot of familiarity comes with practice – start with easier problems and work your way up. I’ll be sure to do a post on this topic in the future :).
----Lining up marbles ----
Let’s say you have 3 bags of marbles. Bag 1 has m different marbles, bag 2 n different marbles, bag3 l different marbles. You may
How many different ways to line up the marbles in a row of 3 (you may only use 1 marble from each bag?
Hi Kalid…It’s been a while. I just wanted to say thanks again. With a final on Friday 21st I’m a little nervous. I do plan to read through the site a few more times.
Kalid, with regards to Aztral’s post (Don’t go by me Aztral)can we say:
There is a total of 3 positions.
Choosing form bag 1, 3 choices to place m marbles i.e. (3m)
Choosing form bag 2, 2 choices to place n marbles i.e. (2n)
Choosing form bag 3, 1 choice to place l marbles i.e. (l)
Leaving us with (3m)(2n)(l)?
This sounds like something similar to what I might see…don’t know
@Dennis/Aztral: Yep, you guys are on the right track. There’s a few different ways to think about problems like these, I really need to do a follow-up
I first forget about the order the marbles. If you have 3 bags (M, N, L), then the total choices are
M * N * L
Using real numbers: If I have 10 Maroon marbles, 5 Navy Blue, and 3 Lime, there are 10 * 5 * 3 = 150 choices.
But we didn’t talk about the order. For any 3 marbles, ABC, we can re-arrange them 3 * 2 * 1 = 6 times:
ABC
ACB
BAC
BCA
CBA
CAB
So, we have to multiply our 150 arrangements (where M was picked first) by 6, to get 900.
Similarly, you’d have 6 * M * N * L. You got the same result through a different path, which is great. The key is to recognize the impact of the permutation (ordering).
I’ve been reading up on set theory ever since, and came up with this (I also realized I didn’t state that a) the marbles are unique, b) the selected marble from bag1 always goes in the first position, marble from bag2 in the second…ie. no need to consider arrangement since they’re already arranged)
Let’s call the bags “sacks” now ;), so that sack1 is S1, sack2 is S2,…
Then basically we’re just creating a new set S=S1xS2xS3. |S| = |S1|x|S2|x|S3|
A drawer contains eight red, eight yellow, eight green and eight black socks. What is the probability of getting at least one pair of matching socks when five socks are randomly pulled from the drawer?
Great website!
Following up on your response to #41 above…
I am looking for a generalized formula for combinations when one has to select r items out of n items, where there can be z items that are similar in the original n items, with frequencies k1, k2, … kz.
I saw a formula on-line that says the answer is
n!/(k1! k2! … kz!) but this doesn’t take into account r.
Please help!
Also, does this class of “similar items” apply to permutations as well? If so, is there a generalized formula for permutations too, when there are z similar items in n original items, and one is taking them r at a time?
Here is one that a number of us have been pondering for some time. Suppose I was just dealt two hearts from a standard deck of cards. What are the odds that exactly 3 of the next 5 cards dealt will also be hearts? There are 11 hearts remaining in a deck of 50 cards and I want exactly 3 of them in the next 5 cards, and the ‘set’ seems to be boolean, Heart or Not. It seems like quite a different problem from standard combinations. I’ll keep working on it and let you know if I solve it.
Number of possible hands matching my criteria is 11_C_3 * 39_C_2 = 165 * 741 = 122,265 possible hands with three more of the same suit. Divide that by the number of possible hands 50_C_5 = 2,118,760 and we see that I have 5.77% chance of getting exactly 3 more hearts. Additionally there are 12,870 remaining hands with 4 hearts and 462 remaining hands with all 5 hearts, so starting with two hearts in my hand I seem to have (122,265 + 12,870 + 462) / 2,118,760 = 6.4% chance of making a flush with suited hole cards.
“I’ve always confused ‘permutation’ and ‘combination’ — which one’s which?”
I was working at a quick service restaurant (we had combo meals) when I first learned combinations/permutations in school. I found it helped me to think that when a customer ordered a combo meal, just like with combinations it didn’t matter what order they received each item – just that they were all present.
