Probability 11

Any answer for verification?

2009-07-08 13:48:50 補充：
Numbers are chosen with repetition allowed.
(1) We ignore the odd numbers as they are discarded.
(2) We ignore the tenth digit as it has nothing to do with the analysis.
Total ways to choose the numbers = 55 = 3,125
Number of ways to choose 3 numbers from 5 is 5C3 = 10
Consider the case where only 0, 2 and 4 are the chosen digits.
Probable combinations for of the unit digits are:
00024, 00224, 00244, 02224, 02244 and 02444
Probable arrangement of digits 00024 (or 02224, 02444
)=
Fix 4 (5 ways) x arrange the 2 (4 ways) = 20 ways
Probable arrangement of digits 00224 (or 00244, 02244)=
Fix 4 (5 ways) x arrange the two 2's among 4 positions (4C2) = 30 ways
Total possible ways for (0,2,4) = 20x3 + 30x3 = 150
Total possible ways for any three = 150 x 10 = 1,500
Probability = 1,500/3,125 = 12/25

Are there 2 questions?

Q1: Derive, in terms of probabilities of intersections of the events A,B,C,D,E the probability that exactly two of these events occur.

2009-07-08 13:46:45 補充：
Q2: Two-digit numbers are chosen from tables of random numbers until five even numbers have been obtained, odd numbers being disregarded. Find the probability that the end digits of the numbers so chosen include exactly three of the dogits 0,2,4,6,8.

2009-07-08 13:48:22 補充：
for Q1, what are the events A, B, C, D, E are?

so, for this question, you want to find out the probability of the intersections of the event A, B, C, D, & E

or to find out the probability that exactly two of these events occur.
i.e. either
AB, AC, AD, AE, BC, BD, BE, CD, CE, DE

??

2009-07-08 13:49:52 補充：
for Q2, would odd nos. chosen be returned back to the table?