Camne Nak kire probability phone number ni???..Uwaaa

LimpBizkutMerri

The last three digit of a XYZ company have been erased and all the we know is that the number was 06-2325XXX. Assuming that all possibilities are equally likely, find the probability of the following events.

a.        The missing digits are 7, 4, 0 (in this order)
b.        The set of missing digits is {0, 4, 7}.
c.        The missing digits are all equal to each other
d.        Two of the missing digits ate equal to each other, but the third is different from the other two.
e.        All three digits are different to each other


waaa... pening aku pikir camne nak jawap soalan niii.. yg aku tau range numbe die xxx tu dari 000 -999

yg c tu aku dpt laa.. 10/1000 = 0.01
yg e tu.. 990/1000 = 0.99

yg lain aku blurrrr.. tatau camneee:kant::kant:

sape sape leh tolon???

eemerics

meh sini aku tulun jawab:

a. probability =1 , sbb mana ada choice yg lain, dah fixed 740

b. {0,4,7} meaning dalam XXX nih ada no 0, 4 & 7
    kalu buat simple calculation 3*3*3 = total ada 27 possibilities.
    so ans = 1/27

c. yeaa betul 10/1000

d. utk ni plak ada total 270 possibilities (cth; 112, 334, 676, etc..)
    camner aku dpt 270 nih, ko bukak balik buku add math.
    so ans = 270/1000

e. sorii bukan 990/1000 yee
    aku tau camner ko dpt 990, sbb 1000 - 10( darik soklan c)
    salah sbb ko x consider combination cam 988, 878, 343, etc
    please read the question carefully "ALL three digits are different to each other"
    total 720 possibilities (cth; 123, 456, 789, etc..)
    camner aku dpt 720 nih, ko bukak lagik buku add math. jgn malas
    so ans = 720/1000

ok ka? kalu ader silap tolong betolkan ek..:hatdown:

[ Last edited by  eemerics at 20-9-2006 12:40 AM ]

LimpBizkutMerri

oo ye ker??? ok ok... mane aa aku nak cari buku add math woooo.. dah 5 taun tinggal kan sekolahhh...hahaha

a) 1/1000

b) depends, kalau boleh guna nombor tu sekali je....6/1000. kalau boleh guna byk kali 27/1000

c) 1/100

d) 27/100

e) 72/100