Probabality.. (1 Viewer)

[Psychosomatical]

There are 30 tickets. I buy 5 tickets. There are two prizes. What is the probability of me winning at LEAST ONE prize?

Thanks.

 

[CountVonChocula]

gambling is an addiction that ruins lives

 

[Psychosomatical]

gambling is an addiction that ruins lives

Feck, i meant to post this in advanced. Constructive, thanks,

 

[annabackwards]

p(wins at least once) = 1 - p(no chance of winning)
= 1- (5/30)(5/29)
= 1 - 5/174
= 169/174

 

[Psychosomatical]

p(wins at least once) = 1 - p(no chance of winning)
= 1- (5/30)(5/29)
= 1 - 5/174
= 169/174

This is gonna blow your mind but i think you're wrong.

 

[Psychosomatical]

p(wins at least once) = 1 - p(no chance of winning)
= 1- (5/30)(5/29)
= 1 - 5/174
= 169/174

Isn't teh chance of NOT winning (25/30)(25/29)?

Therefore it's 1 - (25/30)(25/29)?

Please prove me wrong and make sure i'm not crazy.

 

[mira.mourad]

There are 30 tickets. I buy 5 tickets. There are two prizes. What is the probability of me winning at LEAST ONE prize?

Thanks.

it doesnt state which prize will be won or lost therefore the probability is win, lose or lose win

P(W,W)= 5/30 x 4/29
P(W,L)= 5/30 x 25/29
P(L,W)= 25/30 x 4/29
P(L,L)= 25/30 x 25/29

therefore probability of winning at least ONE is

P(W,L) + P(L,W)

= (5/30 x 25/29) + (25/30 x 4/29)
= 15/58 of winning at least one prize

PS: i think this is the right answer but u should double check

 

[Psychosomatical]

it doesnt state which prize will be won or lost therefore the probability is win, lose or lose win

P(W,W)= 5/30 x 4/29
P(W,L)= 5/30 x 25/29
P(L,W)= 25/30 x 4/29
P(L,L)= 25/30 x 25/29

therefore probability of winning at least ONE is

P(W,L) + P(L,W)

= (5/30 x 25/29) + (25/30 x 4/29)
= 15/58 of winning at least one prize

PS: i think this is the right answer but u should double check

2 is at least 1....

 

[Top Secret]

it doesnt state which prize will be won or lost therefore the probability is win, lose or lose win

P(W,W)= 5/30 x 4/29
P(W,L)= 5/30 x 25/29
P(L,W)= 25/30 x 4/29
P(L,L)= 25/30 x 25/29

therefore probability of winning at least ONE is

P(W,L) + P(L,W)

= (5/30 x 25/29) + (25/30 x 4/29)
= 15/58 of winning at least one prize

PS: i think this is the right answer but u should double check

You are right except that you needed to add P(W,L), P(L,W) AND P(W,W).

 

[annabackwards]

Isn't teh chance of NOT winning (25/30)(25/29)?

Therefore it's 1 - (25/30)(25/29)?

Please prove me wrong and make sure i'm not crazy.

You are correct lol. I think i screwed it up in my brain XD

So it should be
p(wins at least once) = 1 - p(no chance of winning)
= 1 - (25/30)(24/29)
= 9/29

it doesnt state which prize will be won or lost therefore the probability is win, lose or lose win

P(W,W)= 5/30 x 4/29
P(W,L)= 5/30 x 25/29
P(L,W)= 25/30 x 4/29
P(L,L)= 25/30 x 25/29

therefore probability of winning at least ONE is

P(W,L) + P(L,W)

= (5/30 x 25/29) + (25/30 x 4/29)
= 15/58 of winning at least one prize

PS: i think this is the right answer but u should double check

You can do it this way too, but it's just longer.

By the way as Top Secret pointed out, it should be
P(W,L) + P(L,W) + P(W,W)
= (5/30 x 25/29) + (25/30 x 5/29) + (5/30 x 4/29)
= 9/29
which is of course the same as the other method

 

[mira.mourad]

You are right except that you needed to add P(W,L), P(L,W) AND P(W,W).

hence the "PS: u should double check" lol

 

[annabackwards]

hence the "PS: u should double check" lol

Aha, everyone makes mistakes.

I can't believe i made such a stupid mistake >.>

 

[random-1005]

p(wins at least once) = 1 - p(no chance of winning)
= 1- (5/30)(5/29)
= 1 - 5/174
= 169/174

should be 4

 

[annabackwards]

should be 4

Nope, it should actually be:
p(wins at least once) = 1 - p(no chance of winning)
= 1 - (25/30)(24/29)
= 9/29

 

[Studentleader]

Am I doing it wrong thinking:

P(Win atleast one) = 1 - P(Not win anything)
= 1 - 28C5/30C5
= 9/29

Well if the last person is right I am... damn I feel good doing year 12 general maths

 

[Top Secret]

Am I doing it wrong thinking:

P(Win atleast one) = 1 - P(Not win anything)
= 1 - 28C5/30C5
= 9/29

Well if the last person is right I am... damn I feel good doing year 12 general maths

Haha, you basically found the possibility of selecting non-winning tickets from the total. Anna selected any tickets, and found the probability of them being winners. Same answers anyway.

 

[harry4]

is it just me or does every one hate 3 unit/4 unit probability
im lucky to get one out of five on those questions

 

[annabackwards]

is it just me or does every one hate 3 unit/4 unit probability
im lucky to get one out of five on those questions

Pretty much everyone hates it because it's a stupid and useless topic XD

 

[Studentleader]

Pretty much everyone hates it because it's a stupid and useless topic XD

Hope its better than WA.

In applicable mathematics we have a chapter in our books called 'Counting techniques' - basically factorials, commutations and permutations. We did an inclass test on this which basically everyone flunked out on THEN we had a seminar with the chief marker who said this is not included in the syllabus - mind you it ended up being in the trials AND the actual exam.

I think probability combined with some statistics (linear programming and time series) ended up taking me down from 84 to 77 :mad1: