Recent content by t_n_p

Of course. An oversight on my behalf. So that means D can be extended by 28, while B, G and E can only be extended by 8.

Homework Statement I have a critical path diagram (see below): The duration of each task is listed, as is the earliest start time (the number in the left side of the box) and the latest start time (the number in the right side of the box). Note that there is a dummy between G and H, and G...

Thanks for your help, much appreciated. All fairly straightforward in the end, just took me ages to make the connection!

10C4 and 13C4 respectively? So pr of 0 strawberries = (3C0 * 10C4) / (13C4) Etc I now see this as one of those type of questions that are like "A team of 6, comprised of 4 boys and 2 girls is to be selected. If there are 7 boys and 3 girls, how many combinations can be chosen? Now what if the...

OK. For zero strawberry cases we have the combinations {a,a,a,a} {a,a,a,c} {a,a,c,c}{a,c,c,c} {c,c,c,c}. The number of ways of obtaining each of these combinations is 10*9*8*7. Thus the total number of ways to obtain any combination with zero strawberries is 5*10*9*8*7. For 1 strawberry, the...

I get where you're coming from. Extremely convoluted for a high school question! So easy said and done with (as) and (sa), how will one shortcut to find number of ways of getting for example (sssa)?

But it's combinations so order wouldn't matter?! (sa) would be the same as (as) Am I missing something?

Not sure what you mean by function. I've gone through it manually. In terms of a systematic way, I could do a four branched tree diagram and they eliminate the duplicates but this is tedious and time consuming. Are you suggesting there is an alternate systematic way to do post a or are you...

OK, thought it over, think I've got it. For each of the 14 possible combinations I will establish the number of ways that combination can be achieved. For example (cssa) can be achieved 6*3*2*4 ways, (cccc) can be achieved 6*5*4*3 ways, (sssa) can be achieved 3*2*1*4 ways etc. Do this for all...

Gotcha. I'll have a sit down and think of it. Thanks for your help

Hang on a second.. There are a total of 13 items, of which we are to pick 4. We have established there are 14 ways we can do this. From the sample space of 14 I'm identifying those with 0, 1, 2 and 3 strawberries. I count 5 of the 14 combinations have 0 strawberries. I count 4/14 with 1...

From part a I've established there are 14 combinations of picking the 4 items: {a,a,a,a} {a,a,a,s} {a,a,a,c} {a,a,s,s} {a,a,s,c} {a,a,c,c} {a,s,s,s} {a,s,s,c} {a,s,c,c} {a,c,c,c} {s,s,s,c} {s,s,c,c} {s,c,c,c} {c,c,c,c} Assuming parts b and c are still talking about selecting 4 items, 4 of the...

Ah, if it's selecting 4 then b and c will just be subsets of part a. Answer to b would be 4 and c would be 10. For d that makes the distribution very basic, pr(s=0)= 5/14, pr(s=1)=4/14, etc. Is the more direct way for b (assuming not restricted by 4 selections) given by (number of apples...

Hi all, Say you have 4 apples, 3 strawberries and 6 carrots... a) If 4 items are to be selected, how many combinations are there? b) How many combinations will have no carrots? c) How many combinations will have at least 1 apple? d) Create the probability distribution table for strawberries...

uart, please refer to post 9. I am asking: - whether you think this is a suitable test/exam question. - why the triangle on the right has been included. Is part of the question for students to recognize that the triangle on the right is over-defined? Please do not think I am not reading your...