Probability Qn (1 Viewer)

[lpodtouch]

The numerals 1,2,4,5,7,8 are marked one on each side of three counters so that the sum of the numerals on any particular counter is 9. The counters are drawn at random one-by-one from box, each counter being tossed after it is drawn then with the uppermost faces unchanged, placed side-by-side on a table (in the order which they are drawn) to form a three digit number.

i. Show that the probability the three-digit number formed is a multiple of 3 is 1/4.
ii. If the random trial described above is performed several times, find the probability that the second three-digit number multiple of 3 occurs on the 5th trial.

Thanks

 

[braintic]

(i) Just count outcomes: There are 2^3 = 8 possibilities for the digits (ignoring order): 124, 125, 174, 175, 824, 825, 874, 875

Two of those have digits that add up to a multiple of 3.

2/8 = 1/4

(ii) What is this "second three-digit number multiple of 3" ?

 

[rand_althor]

(ii) What is this "second three-digit number multiple of 3" ?

I think this is what it means: the random trial is repeated several times, so you end up with several three-digit numbers by the end. For the 5th trial to produce the second three-digit number which is a multiple of 3, there has to have been only one three-digit number which is a multiple of 3 in the 1st, 2nd, 3rd and 4th trials.

 

[simpleetal]

Are the counters returned to the box after each trial? If so, call a number which satisfies the given conditions as a 'good' number. So the required probability is (probability of exactly one 'good' number produced in the first four trials) * (probability of a 'good' number produced in the fifth trial)= 4C1 * (1/4)*(3/4)^3 x (1/4)