Probability Math Problems

[ViciousShadow21]

Any math gurus out there that could help me with these probability problems i'd appreciate it. Im terrible at probability haha. and i want to check my answers. Thanks

1. A package of 25 zinnia seeds contain 8 seeds for red flowers, 12 seeds for white flowers, and 5 seeds for yellow flowers. Three seeds are randomly selected and planted. Find the probability of each of the following:

- All three seeds will produce red flowers.

-The first seed will produce a red flower, the second seed will produce a white flower, and the third seed will produce a red flower.

-None of the seeds will produce red flowers.

2. Drug A is given to 5 patients. Drug B is given to 4 patients. Drug C is given to 6 patients. If four of these patients are selected at random, find the probability that 2 were give drug A and 2 were given drug C.

3. At the royal Dynasty Chinese Restaurant, dinner for 8 consists of three items from column A, four items from column B, and three items from column C. If columns A B and C have five, seven, and six items, respectively, how many different dinner combinations are possible.

 

[iStudentUK]

1.

(a) Three red seeds.

The seeds are not replaced each time as they are planted. The first seed must be red (8/25) the second as well (7/24) and the third (6/23).

Times all three together = 14/575 (2.43%)

(b) Red -> White -> Red

(8/25) * (12/24) * (7/23) = 28/575 (4.87%)

(c) No red.

Four ways to do this. All white, all yellow, 1 white 2 yellow, 2 yellow 1 white. Order not important.

All white = 11/115

All yellow = 1/230

1W 2Y. This is (12*5*4)/(25*24*23) then *3 as it can be done three ways (YWW, WYW, WWY) so 6/115

2W 1Y. (5*12*11)/(25*24*23)*3 = 33/230

Total = 34/115 (29.57%)


2.

I'm assuming each patient gets only one drug, so there are 15 in total. Need 2 drug A and 2 C. Again, this is a non-replacement question as each patient is counted once.

So lets just take AACC. This is (5/15)*(4/14)*(6/13)*(5/12) = 600/32760 = 5/273

But this can be done in (4!)/(2!*2!) ways, or 24/4 = 6.

So 30/273 = 10/91 (10.99%)


3.

A has 5 items, pick 4. B has 7, pick 4. C has 6, pick 3. I'll have to assume they have to pick different menu items, not the same one twice.

So there are 5 ways to pick 4 items from A, 35 to pick 4 from B and 20 to pick 3 from C.

So for each of the 5 ways you can pick A, there are 35 ways to pick B. For each of the 35 ways to pick from menu B there are 20 ways to pick C. So in my head I'm picturing a tree diagram. There are 5 branches, then 35 on each of those, then 20 on each of those. Which leaves 35*20*5, or 3500 combinations. (EDIT- Misread the OP, it should be 7000, see below).



I hope this helps. It's been a while since I did probability, and Q3 was a bit weird so not very sure. Hope I agree with you?

 

[Purestang]

As i was taught, i believe that iStudentUK is correct. Its been about 3 years since i took a basic stats course.

 

[samiwas]

I got the same answer for 1C, but went a completely different route. What DOES bloom doesn't matter. The only point is that they are not red. So it would just be 17*16*15 / 13,800 = 29.57%. 17 is the number of total seeds minus the 8 red seeds. 13,800 being the product of 25*24*23.

The way iStudentUK gave seemed overly complicated.

 

[iStudentUK]

<snip>

That does seem simpler! Don't know why I went the long way around!

 

[samiwas]

I should have waited until I had looked at 3 before responding...but here's what I got for 3, so let me know if I thought this out wrong:

Column A = 3 choices from 5 selections, 5*4*3 = 60 possibilities
Column B = 4 choices from 7 selections, 7*6*5*4 = 840 possibilities
Column C = 3 choices from 6 selections, 6*5*4 = 120 possibilities

Total = 60*840*120 = 6,048,000 total possibilities

 

[swiftaw]

Any math gurus out there that could help me with these probability problems i'd appreciate it. Im terrible at probability haha. and i want to check my answers. Thanks

1. A package of 25 zinnia seeds contain 8 seeds for red flowers, 12 seeds for white flowers, and 5 seeds for yellow flowers. Three seeds are randomly selected and planted. Find the probability of each of the following:

- All three seeds will produce red flowers.

-The first seed will produce a red flower, the second seed will produce a white flower, and the third seed will produce a red flower.

-None of the seeds will produce red flowers.

2. Drug A is given to 5 patients. Drug B is given to 4 patients. Drug C is given to 6 patients. If four of these patients are selected at random, find the probability that 2 were give drug A and 2 were given drug C.

3. At the royal Dynasty Chinese Restaurant, dinner for 8 consists of three items from column A, four items from column B, and three items from column C. If columns A B and C have five, seven, and six items, respectively, how many different dinner combinations are possible.

1. P( 3 red flowers) = 8C3 / 25C3 = [8!/(5!*3!)]/[25!/(22!*3!)] = 56/2300 = 14/575 = .02435

P(first red, second white, third red) = 8/25 * 12/24 * 7/23 = 28/575 = .0487

P(2 red, 1 white in any order) = [8!/(6!*2!)]*[12!/[(11!*1!)]/[25!/(22!*3!)]=84/575

P(no red) = [17!/(14!*3!)]/[25!/(22!*3!)] = 34/115

2. P(2 drug A, 2 drug C) = [5!/(2!*3!)]*[6!/(2!*4!)]/[15!/(4!*11!)] = 10/91

3 5C3 * 7C4 * 6C3 = 10*35*20 = 7000

 

[iStudentUK]

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I noticed I misread Q3, I thought it was pick 4 from 5, but it's 3 so there is an error there.

I think it depends on if order matters. I don't think it should. So if the 5 items in menu A are called a,b,c,d,e then a,b,c=a,c,b etc. In which case there are 10 ways to pick 3 items from 5.

So my answer should be 7000 as I misread the question.

 

[samiwas]

Wirelessly posted (Mozilla/5.0 (iPhone; U; CPU iPhone OS 4_3 like Mac OS X; en-gb) AppleWebKit/533.17.9 (KHTML, like Gecko) Version/5.0.2 Mobile/8F190 Safari/6533.18.5)

I noticed I misread Q3, I thought it was pick 4 from 5, but it's 3 so there is an error there.

I think it depends on if order matters. I don't think it should. So if the 5 items in menu A are called a,b,c,d,e then a,b,c=a,c,b etc. In which case there are 10 ways to pick 3 items from 5.

So my answer should be 7000 as I misread the question.

Ah, yes....I didn't take the order of selection into account. Kind of a big error.

So, forgive me since it's been almost 20 years since I've been in a math class. How do you mathematically figure 10 possibilities, with 3 out of 5 choices? I can write out the possibilities and figure it. But I don't know how 5C3 or 6C3 that swiftaw writes above means anything.

 

[ViciousShadow21]

Thanks all for your replies. I matched up with with everyone except for the last one. Thanks again

 

[iStudentUK]

How do you mathematically figure 10 possibilities, with 3 out of 5 choices? I can write out the possibilities and figure it. But I don't know how 5C3 or 6C3 that swiftaw writes above means anything.

The general equation for "how many ways are there to pick r objects from a total of n objects, where the order doesn't matter" is given by-

n!/(r!(n-r!))

So for "5 pick 3" it is 5!/(3!2!) = 10

However, as this comes up a lot in maths the equation is often written as nCr (the n is superscript and the r subscript). Scientific calculators usually have it somewhere (not to be confused with nPr). So 5C3 = 10. I would pronounce 5C3 as "5 pick 3".

 

[siurpeeman]

How do you mathematically figure 10 possibilities, with 3 out of 5 choices? I can write out the possibilities and figure it. But I don't know how 5C3 or 6C3 that swiftaw writes above means anything.

Let's say we have five people: A, B, C, D, and E. They're going to a concert but there are only three tickets. We need to figure how many ways three people can go. We have five choices to pick our first person, four for the next and three for the third; that's 5*4*3, or 60.

But this number is too big, because it includes all the different ways we could have picked the same people. Let's say A, B, and C go to concert, so the 60 above includes all the iterations of A, B, and C: ABC, ACB, BCA, BAC, CAB, CBA. It doesn't matter who gets chosen first; ABC is just as good as CBA. So, the 60 needs to be divided by the ways we can arrange each three-person group, namely 3*2*1, or 6. 60 divided by 6 is 10. 10 ways of choosing 3 from 5, given that order does not matter.

I'm reading this response back to myself, and while the original intent was to explain the 10 without involving formulas for combinations, I'm not sure if I made it any easier to understand.

The general equation for "how many ways are there to pick r objects from a total of n objects, where the order doesn't matter" is given by-

n!/(r!(n-r!))

So for "5 pick 3" it is 5!/(3!2!) = 10

However, as this comes up a lot in maths the equation is often written as nCr (the n is superscript and the r subscript). Scientific calculators usually have it somewhere (not to be confused with nPr). So 5C3 = 10. I would pronounce 5C3 as "5 pick 3".

In my experience, when people use the word "pick" they often do so to refer to permutations, nPr, and not combinations, nCr. If someone is to tell me 5C3, they'll either say "5-C-3" or "combinations of five things taken three at a time." The latter may be long-winded, but it eliminates ambiguity.

 

[iStudentUK]

In my experience, when people use the word "pick" they often do so to refer to permutations, nPr, and not combinations, nCr. If someone is to tell me 5C3, they'll either say "5-C-3" or "combinations of five things taken three at a time." The latter may be long-winded, but it eliminates ambiguity.

Fair enough. Although, if I were actually in this restaurant picking menu items I'd probably say it as "Come on guys, I'm starving! Pick your 3 items already!"