ã…—peachesã…— 2,148 Posted May 6, 2014 Share Posted May 6, 2014 this is so embarrassing because this problem is so easy but ihavent taken stats since high school and i cant remember for the life of me what this question is even asking for and how to solve it this is embarrassing but,, please help me.. il send wons.. Link to comment Share on other sites More sharing options...
ã…—peachesã…— 2,148 Posted May 6, 2014 Author Share Posted May 6, 2014 i solved A if that helps :( Link to comment Share on other sites More sharing options...
abra 40,910 Posted May 6, 2014 Share Posted May 6, 2014 i solved A if that helps :( damn it, I knew that one :( I'll take a look at the other ones But I don't get what they're asking in b) Link to comment Share on other sites More sharing options...
ã…—peachesã…— 2,148 Posted May 6, 2014 Author Share Posted May 6, 2014 damn it, I knew that one :( I'll take a look at the other ones But I don't get what they're asking in b) exactly :( im stuck on B, i literally dont understand what they are asking me to do Link to comment Share on other sites More sharing options...
abra 40,910 Posted May 6, 2014 Share Posted May 6, 2014 exactly :( im stuck on B, i literally dont understand what they are asking me to do I think C is something like: 6/16 [chance of winning half of the bets if winning is as probable as losing] * 18/38 [chance to win any bet] Link to comment Share on other sites More sharing options...
abra 40,910 Posted May 6, 2014 Share Posted May 6, 2014 And D would be 5/16*18/38 But I haven't had maths in years Link to comment Share on other sites More sharing options...
Slowpoke X 1,319 Posted May 6, 2014 Share Posted May 6, 2014 damn it, I knew that one :( I'll take a look at the other ones But I don't get what they're asking in b) if you already solved A, shouldn't you use the probability you calculated in A to solve B? Link to comment Share on other sites More sharing options...
abra 40,910 Posted May 6, 2014 Share Posted May 6, 2014 if you already solved A, shouldn't you use the probability you calculated in A to solve B? But I'd have to know what they're asking Link to comment Share on other sites More sharing options...
Slowpoke X 1,319 Posted May 6, 2014 Share Posted May 6, 2014 But I'd have to know what they're asking http://en.wikipedia.org/wiki/Statistical_probability were you using that formula to solve A, or either C, or D? Link to comment Share on other sites More sharing options...
Chungie 1,215 Posted May 6, 2014 Share Posted May 6, 2014 A) P = 18/38 B) Betting on read, chances of winning = 18/38, and losing chance is 20/38, therefore distributions: chances of winning 4 times= (18/38)^4 chances of winning 3 times = (18/38)^3*(20/38) * C4,1 2 times= (18//38)^2 * (20/38)^2 * C4,2 1 time= (18/38)*(20/38) * C4,1 0 time= (20/38)^4 C) (18/38)^2 * (20/38)^2 * C4,2 D) chances of winning 1 time+ chances of winning 0 time (the value of above calculations). Link to comment Share on other sites More sharing options...
abra 40,910 Posted May 6, 2014 Share Posted May 6, 2014 ^ http://en.wikipedia.org/wiki/Statistical_probability were you using that formula to solve A, or either C, or D? That doesn't help me solve B as it's too vague. I used it for A but not sure if I got it right for C or D. Link to comment Share on other sites More sharing options...
ã…—peachesã…— 2,148 Posted May 6, 2014 Author Share Posted May 6, 2014 I SOLVED THEM !whooW1 rofl just saw everyes responses. thanks guys! A) P = 18/38 B) Betting on read, chances of winning = 18/38, and losing chance is 20/38, therefore distributions: chances of winning 4 times= (18/38)^4 chances of winning 3 times = (18/38)^3*(20/38) * C^4,1 2 times= (18//38)^2 * (20/38)^2 * C^4,2 1 time= (18/38)*(20/38) * C^4,1 0 time= (20/38)^4 C) (18/38)^2 * (20/38)^2 * C^4,2 D) chances of winning 1 time+ chances of winning 0 time (the value of above calculations). yas ur right!! Link to comment Share on other sites More sharing options...
Chungie 1,215 Posted May 6, 2014 Share Posted May 6, 2014 ^ That doesn't help me solve B as it's too vague. I used it for A but not sure if I got it right for C or D. For the B, the questions asks for distributions. Therefore the formula , P(winning)^n*C(n,r)*P(losing)^n-1 for each chances. I SOLVED THEM !whooW1 rofl just saw everyes responses. thanks guys! yas ur right!! lol good luck Link to comment Share on other sites More sharing options...
abra 40,910 Posted May 6, 2014 Share Posted May 6, 2014 For the B, the questions asks for distributions. Therefore the formula , P(winning)^n*C(n,r)*P(losing)^n-1 for each chances. oh ok how far off was I on C and D? Link to comment Share on other sites More sharing options...
Chungie 1,215 Posted May 6, 2014 Share Posted May 6, 2014 oh ok how far off was I on C and D? Quite off. The C can also be solved by using distribution formula. So chances of winning 2 times is P(winning)^2*P(losing)^2*C(4,2). For D, it says fewer than "2" of the 4 plays, therefore it's just addition of the 1 and 0 winning chances. Link to comment Share on other sites More sharing options...
abra 40,910 Posted May 6, 2014 Share Posted May 6, 2014 Quite off. The C can also be solved by using distribution formula. So chances of winning 2 times is P(winning)^2*P(losing)^2*C(4,2). For D, it says fewer than "2" of the 4 plays, therefore it's just addition of the 1 and 0 winning chances. We were only taught really basic statistical probability Why the C(4,2)? Link to comment Share on other sites More sharing options...
Chungie 1,215 Posted May 6, 2014 Share Posted May 6, 2014 We were only taught really basic statistical probability Why the C(4,2)? Awww these are all taught in schools here. Binomial coefficients, where C stands for combination. C(n,r), n is a superscript on the left side whereas r is a subscript on the right side. n stands for number of trials, r is number of success. Link to comment Share on other sites More sharing options...
abra 40,910 Posted May 6, 2014 Share Posted May 6, 2014 Awww these are all taught in schools here. Binomial coefficients, where C stands for combination. C(n,r), n is a superscript on the left side whereas r is a subscript on the right side. n stands for number of trials, r is number of success. oh I need to look it up some time... What would the chances be for winning 1, 2 and 3 times if you simplify it? Link to comment Share on other sites More sharing options...
Chungie 1,215 Posted May 6, 2014 Share Posted May 6, 2014 oh I need to look it up some time... What would the chances be for winning 1, 2 and 3 times if you simplify it? Like what do you mean? Winning three times? Link to comment Share on other sites More sharing options...
abra 40,910 Posted May 6, 2014 Share Posted May 6, 2014 Like what do you mean? Winning three times? No, as in, could you simplify this: chances of winning 3 times = (18/38)^3*(20/38) * C4,1 2 times= (18//38)^2 * (20/38)^2 * C4,2 1 time= (18/38)*(20/38) * C4,1 Because as things stand I don't get it. Link to comment Share on other sites More sharing options...
Chungie 1,215 Posted May 6, 2014 Share Posted May 6, 2014 No, as in, could you simplify this: chances of winning 3 times = (18/38)^3*(20/38) * C4,1 2 times= (18//38)^2 * (20/38)^2 * C4,2 1 time= (18/38)*(20/38) * C4,1 Because as things stand I don't get it. Ahh, C=n!/r!(n-r)!, using this formula, C can be solved. 4!=24, 3!=6, 2!=2, 1!=1. So, C(4,1)=4!/1!(4-1)! = 4!/3! = 4. The others are simply solved by using multiplications. Link to comment Share on other sites More sharing options...
abra 40,910 Posted May 6, 2014 Share Posted May 6, 2014 Ahh, C=n!/r!(n-r)!, using this formula, C can be solved. 4!=24, 3!=6, 2!=2, 1!=1. So, C(4,1)=4!/1!(4-1)! = 4!/3! = 4. The others are simply solved by using multiplications. thanks Link to comment Share on other sites More sharing options...
Chungie 1,215 Posted May 6, 2014 Share Posted May 6, 2014 thanks np Link to comment Share on other sites More sharing options...
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