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 Post subject: probability
PostPosted: Sat, 28 Apr 2012 10:21:40 UTC 
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At a games stall, an electronic device displays a three-digit number from 000 to 999. When a button is pressed, each number has an equal chance of being displayed. Find the probabilityh that the three digits displayed

i) Have a sum of at least 24,

ii) are identical, given that the sum of the three numbers is at least 24.

answer is 1/50 and 1/10 for part i) and part ii) but i don't know how to get there!


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 Post subject: Re: probability
PostPosted: Sat, 28 Apr 2012 13:58:47 UTC 
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thesocialnetwork wrote:
At a games stall, an electronic device displays a three-digit number from 000 to 999. When a button is pressed, each number has an equal chance of being displayed. Find the probabilityh that the three digits displayed

i) Have a sum of at least 24,

ii) are identical, given that the sum of the three numbers is at least 24.

answer is 1/50 and 1/10 for part i) and part ii) but i don't know how to get there!


Just count for (i), and use the definition of conditional probability for (ii).

_________________
\begin{aligned}
Spin(1)&=O(1)=\mathbb{Z}/2&\quad&\text{and}\\
Spin(2)&=U(1)=SO(2)&&\text{are obvious}\\
Spin(3)&=Sp(1)=SU(2)&&\text{by }q\mapsto(\mathop{\mathrm{Im}}\mathbb{H}\ni p\mapsto qp\bar{q})\\
Spin(4)&=Sp(1)\times Sp(1)&&\text{by }(q_1,q_2)\mapsto(\mathbb{H}\ni p\mapsto q_1p\bar{q_2})\\
Spin(5)&=Sp(2)&&\text{by }\mathbb{HP}^1\cong S^4_{round}\hookrightarrow\mathbb{R}^5\\
Spin(6)&=SU(4)&&\text{by the irrep }\Lambda_+\mathbb{C}^4
\end{aligned}


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 Post subject: Re: probability
PostPosted: Sat, 28 Apr 2012 19:43:56 UTC 
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Joined: Tue, 16 Aug 2011 23:38:56 UTC
Posts: 124
thesocialnetwork wrote:
At a games stall, an electronic device displays a three-digit number from 000 to 999. When a button is pressed, each number has an equal chance of being displayed. Find the probabilityh that the three digits displayed

i) Have a sum of at least 24,

ii) are identical, given that the sum of the three numbers is at least 24.


i) P(sum of at least 24) = Number of Favorable Cases / Number of Possible Cases

Favorable cases: 999; 998; 997; 996; 988; 987; 888

If you do the permutation you will get 20 cases, then:

P(sum of at least 24) = Number of Favorable Cases / Number of Possible Cases =\frac {20}{1000}=\frac{1}{50}

ii) Favorable cases: 999; 888 just 2, as the possible cases are 20 then the result will be \frac {1}{10}


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