probability

thesocialnetwork · **Joined:** Wed, 15 Feb 2012 07:19:33 UTC **Posts:** 31

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!

outermeasure · **Posted:** Sat, 28 Apr 2012 13:58:47 UTC

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).

fam · **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 $\frac {1}{10}$

S.O.S. Mathematics CyberBoard

probability

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