Probability Questions (1 Viewer)

[Skeptyks]

I am really bad at probability...
So please help me with these two questions I have attempted to solve them/cannot do them outright.

15) In a raffle in which 200 tickets are sold, there are 3 prizes- first prize of $100, second prize of $50, third prize of $30.
I) A girl has one ticket in the raffle. What is the probability that she wins:
b) at least $50
I understand the principle that to get at least one prize is to subtract the possibility of getting no prizes e.g. 1- (199/200 x 198/199 x 197/198), but for at least $50, I don't quite understand.

16) In a bag of balloons, the ratio of red balloons to other colours is 1:5. If 3 balloons are selected from the bag at random, what is the probability that:
a) at least one balloon is red
b) exactly one balloon is red
For (a), I got the answer by saying P(no red) = (5/6)^3 so therefore P(at least one red) = 1 - (5/6)^3 = 91/216. Now that's the correct answer but I just don't understand the thought behind it so if anyone could give a clear explanation, that would be great.
For (b), I have no idea how to approach this question.

I know these are really basic questions and I have done permutations/combinations in Year 11 (Term 3), but I felt that my probability was extremely weak and I needed to re-do some of these questions to get in the right thought process when attempting these questions.
Thanks greatly in advance.

 

[pwoh]

Q15 - at least $50. It is the same as the method you used for "at least one prize", except now the third prize doesn't matter because it's less than $50. So 1 - (199/200 x 198/199).

Q16
a) Think about all the possibilities related to red balloons. You can have no red balloons, one red balloon, two red balloons or three red balloons. The probability of all of these add up to 1. We want "at least one red balloon", so this is all of the possibilities except for "no red balloons". So 1-P(no red) = 1-(5/6)^3

b) The chance of getting one red balloon is 1/6. The chance of getting a non-red balloon is 5/6. We want one red balloon and two non-red balloons. There are three ways this can occur: the red balloon can be chosen 1st, 2nd or 3rd. Therefore (1/6) * (5/6) * (5/6) * 3.

(This is assuming that the balloons are put back into the bag once they have been chosen so that the ratio remains the same. I don't think there is enough information to calculate the probability if the balloons were not replaced, since we are not given the total number of balloons - but someone feel free to correct me if this assumption is wrong.)

 

[Nooblet94]

This should really be in the maths forum, but I'll answer them anyway.

For the first one, to win at least $50 she needs to win first or second prize.
probability of first prize=199/200
probability of second prize=198/199
You then multiply the two together and subtract that from 1 to get the answer. You don't use the third draw because it doesn't affect the outcome of the first two.

For 16-a), no red and at least one red are complements (either one or the other must to happen), so when you add their probabilities, the answer is 1. You can then rearrange algebraically to say that the probability of an event is equal to 1 minus the probability of the complement.

16-b) uses binomial probability as you've got two complementary events, red or not red. By the sounds of it you haven't done it in class, so here's the general equation:

EDIT: for some reason it's not letting me paste the html code in here, so I'll have to link you to it: http://latex.codecogs.com/gif.latex?P(X=r)=~^nC_rp^rq^{n-r}\\&space;\textrm{Where:}\\&space;r=\textrm{the&space;number&space;of&space;succesful&space;trials}\\&space;n=\textrm{the&space;totalnumber&space;of&space;trials}\\&space;p=\textrm{the&space;probability&space;of&space;success&space;in&space;a&space;single&space;trial}\\&space;q=\textrm{the&space;probability&space;of&space;failure&space;in&space;a&space;single&space;trial}

So you let r=1, n=3, p=1/6, q=5/6 and get the answer.

 

[Skeptyks]

Q15 - at least $50. It is the same as the method you used for "at least one prize", except now the third prize doesn't matter because it's less than $50. So 1 - (199/200 x 198/199).

Q16
a) Think about all the possibilities related to red balloons. You can have no red balloons, one red balloon, two red balloons or three red balloons. The probability of all of these add up to 1. We want "at least one red balloon", so this is all of the possibilities except for "no red balloons". So 1-P(no red) = 1-(5/6)^3

b) The chance of getting one red balloon is 1/6. The chance of getting a non-red balloon is 5/6. We want one red balloon and two non-red balloons. There are three ways this can occur: the red balloon can be chosen 1st, 2nd or 3rd. Therefore (1/6) * (5/6) * (5/6) * 3.

(This is assuming that the balloons are put back into the bag once they have been chosen so that the ratio remains the same. I don't think there is enough information to calculate the probability if the balloons were not replaced, since we are not given the total number of balloons - but someone feel free to correct me if this assumption is wrong.)

Thankyou, the descriptions were really clear. I can't believe I didn't understand how to do them before.

This should really be in the maths forum, but I'll answer them anyway.

For the first one, to win at least $50 she needs to win first or second prize.
probability of first prize=199/200
probability of second prize=198/199
You then multiply the two together and subtract that from 1 to get the answer. You don't use the third draw because it doesn't affect the outcome of the first two.

For 16-a), no red and at least one red are complements (either one or the other must to happen), so when you add their probabilities, the answer is 1. You can then rearrange algebraically to say that the probability of an event is equal to 1 minus the probability of the complement.

16-b) uses binomial probability as you've got two complementary events, red or not red. By the sounds of it you haven't done it in class, so here's the general equation:

EDIT: for some reason it's not letting me paste the html code in here, so I'll have to link you to it: http://latex.codecogs.com/gif.latex?P(X=r)=~^nC_rp^rq^{n-r}\\&space;\textrm{Where:}\\&space;r=\textrm{the&space;number&space;of&space;succesful&space;trials}\\&space;n=\textrm{the&space;totalnumber&space;of&space;trials}\\&space;p=\textrm{the&space;probability&space;of&space;success&space;in&space;a&space;single&space;trial}\\&space;q=\textrm{the&space;probability&space;of&space;failure&space;in&space;a&space;single&space;trial}

So you let r=1, n=3, p=1/6, q=5/6 and get the answer.

Thanks to you to. I didn't notice that I could use the general equation of binomial probability (we have infact done it last year, but I havn't revised/touched it in so long I completely forgot about it). I will put it in the math's forum next time. I just completely forgot about it because I always just look around this particular section.