Choosing multiple probabilities

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Loganblahtimes2

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"(a) In how many ways can a sample of 3 pears be taken from a basket containing 12 pears? Part 2: If 2 pears are spoiled, in how many ways can the sample of 3 include (b) 1 spoiled pear (c) no spoiled pears (d) at most 1 spoiled pear."

I am having trouble on exclusively part D for this problem. I have figured out the answers to the other sections, those being:
A: 1,320
B: 90
C: 120
I found these by doing the following:
A)
12!/(12-3)!
B)
(2 nCr 1)*(10 nCr 2)
C)
(2 nCr 0)*(10 nCr 3)

I am confused on part D as I do not know how I could enter multiple variables in a TI-84 to find out the probability. Once again, the class this is from is a starting level college math class. Sorry for the lack of details, but I don't know where to start.
 
Loganblahtimes2 said:
"(a) In how many ways can a sample of 3 pears be taken from a basket containing 12 pears? Part 2: If 2 pears are spoiled, in how many ways can the sample of 3 include (b) 1 spoiled pear (c) no spoiled pears (d) at most 1 spoiled pear."

I am having trouble on exclusively part D for this problem. I have figured out the answers to the other sections, those being:
A: 1,320
B: 90
C: 120
I found these by doing the following:
A)
12!/(12-3)!
B)
(2 nCr 1)*(10 nCr 2)
C)
(2 nCr 0)*(10 nCr 3)

I am confused on part D as I do not know how I could enter multiple variables in a TI-84 to find out the probability. Once again, the class this is from is a starting level college math class. Sorry for the lack of details, but I don't know where to start.
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How many ways - 2 spoiled pears were chosen?
 
What does at most 1 spoiled pear mean in this problem. That is exactly how many pears might be spoiled if at most 1 pear is spoiled?
 
Subhotosh Khan said:
What is the probability that 2 spoiled pears were chosen.
Click to expand...

(2/12)*(1/11)=1/66. The probability for 2 spoiled pears to be chosen is 1/66.

"At most 1 spoiled pear" means 0 or 1 pears chosen from the 3 can be spoiled. If 1 pear is spoiled, 0 more will be spoiled...?
 
Loganblahtimes2 said:
(2/12)*(1/11)=1/66. The probability for 2 spoiled pears to be chosen is 1/66.

"At most 1 spoiled pear" means 0 or 1 pears chosen from the 3 can be spoiled. If 1 pear is spoiled, 0 more will be spoiled...?
Click to expand...
Yes, if 1 pear is spoiled, 0 more will be spoiled. But why did you write that? What are you thinking? Why did you question that statement.

Yes, at most one spoiled pear means 0 or 1 pear from the the chosen 3 can be spoiled.

So p(at most one spoiled pear) = p( exactly 0 pears are spoiled OR exactly 1 pear is spoiled). Can you run with this?
 
Jomo said:
Yes, if 1 pear is spoiled, 0 more will be spoiled. But why did you write that? What are you thinking? Why did you question that statement.

Yes, at most one spoiled pear means 0 or 1 pear from the the chosen 3 can be spoiled.

So p(at most one spoiled pear) = p( exactly 0 pears are spoiled OR exactly 1 pear is spoiled). Can you run with this?
Click to expand...

I wrote "If 1 pear is spoiled, 0 more will be spoiled...?" Because I was unsure if you meant "0 more will be spoiled" in the basket total or in the selected amount. I understand you mean the selected 3.

I understand 0 or 1 pears would be spoiled. However, how would this inserted into a probability equation?
 
Loganblahtimes2 said:
I understand 0 or 1 pears would be spoiled. However, how would this inserted into a probability equation?
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I gave you the probability which you need to solve. Here it is again. p( exactly 0 pears are spoiled OR exactly 1 pear is spoiled)=....
 
I understand the probability of 0 or 1 pear being spoiled. I'm unsure of how this would be put into an equation, apologies. I'm unsure of how p(0 or 1) would go into said equation. I believe it would be (2 nCr 1)*(10 nCr 2) at this point, being ([total number of spoiled pears] nCr [exactly 1 pear being spoiled])*([total number of good pears] nCr [the other 2 being good in the picking])
 
P( A or B) = P(A) + P(B) - P( A and B). What are A and B in your problem?

Based on your earlier posts you should be able to compute these probabilities. If not, then please tell us where you are getting stuck.
 
I believe B would be the number of spoiled pears and A would be the number of regular pears in the basket. B = 2 and A = 10. What I'm confused on is the format in how I enter this. Would I use the equation I used in part A, or in parts B and C?
 
You agreed that you want the probability of 0 spoiled pears or 1 spoiled pear. In that statement to the left it states what A and B should be. Please try again.
 
Match up the A and B:

p( exactly 0 pears are spoiled OR exactly 1 pear is spoiled)
p ( A OR B )
 
Loganblahtimes2 said:
Do you mean A and B being the answers to my previous equation? If so, they would be 1,320 and 90.
Click to expand...
Forget earlier questions. In fact forget this entire problem for a moment.

Can you match up A and B (they are even colored coded) below.

p( exactly 0 pears are spoiled OR exactly 1 pear is spoiled)
p ( A OR B )
 
Jomo said:
Forget earlier questions. In fact forget this entire problem for a moment.

Can you match up A and B (they are even colored coded) below.

p( exactly 0 pears are spoiled OR exactly 1 pear is spoiled)
p ( A OR B )
Click to expand...

Okay lol, A is 0 pears spoiled and B is 1 pear spoiled. P(0 or 1) = P(0) + P(1) - P(0 and 1, would A+B=1?).
 
Loganblahtimes2 said:
Okay lol, A is 0 pears spoiled and B is 1 pear spoiled. P(0 or 1) = P(0) + P(1) - P(0 and 1, would A+B=1?).
Click to expand...
Are you saying that A+B = 1 since 0+1=1?? So the probability of getting 12 heads on 40 tosses is the same as getting (3 heads and 9 heads) on 40 tosses since 3+9 = 12? Not quite.

Seriously, think about the difference between OR and AND. What does it mean to pick 0 spoiled pears AND 1 spoiled pear? Exactly how can that happen? Finish this sentence. You go to the 12 pears knowing that you will pick 3 and you ....
 
I was saying that A+B=1, yes.
I know the difference between AND and OR. Picking 0 spoiled pears means picking 3 regular pears, and picking 1 spoiled pear means 2 regular ones alongside it. I understand that I mixed up OR with AND previously. "You go to the 12 knowing that you will pick 3 and you pick 1 spoiled pear or 0 spoiled pears." Apologies for the confusion.
 
How can you, if possible, pick exactly 0 spoiled pears AND exactly 1 spoiled pears if you go over to the pears and pick three of them?
 
Okay, than an and. "You go to the 12 knowing that you will pick 3 and you pick 1 spoiled pear and 0 spoiled pears."
 
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