急!!概率 (10分)

In 1-100,
24 odd and prime number (OP)
26 odd and non-prime number (OC)

Now we have to calculate the probability of "First drawing 3 odd number, then drawing 3 prime number"

Since which odd numbers are drawn will affect the number of prime remaining, we have to study the question case by case

Case (I): All 3 odd number drawn prime number (3 OP)
Case (II): 2 of those 3 odd number drawn are prime (2 OP + 1 OC)
Case (III): 1 of those 3 odd number drawn are prime (1 OP + 2 OC)
Case (IV): Non of those 3 odd number drawn are prime (3 OC)


=============================
Case (I): All 3 odd number drawn prime number (3 OP)
=============================

P(drawing 3 OP) = 24/100 * 23/99 * 22/98 ... (1)

Since there are 22 prime number remaining...

P(drawing 3 prime number after drawing 3 OP) = 22/97 * 21/96 * 20/95 ... (2)


Therefore, P(drawing 3 odd and 3 prime by Case (I)) = (1) * (2) = 0.000130739



=============================
Case (II): 2 of those 3 odd number drawn are prime (2 OP + 1 OC)
=============================

[OP OP OC]: Prob = 24/100 * 23/99 * 26/98
[OP OC OP]: Prob = 24/100 * 26/99 * 23/98
[OC OP OP]: Prob = 26/100 * 24/99 * 23/98

These 3 prob are equal.
Therefore, P(drawing 2 OP + 1 OC) = 3 * (26*24*23) / (100*99*98) ... (3)

Since there are 23 prime number remaining after that...

P(drawing 3 prime number after drawing 2 OP + 1 OC) = 23/97 * 22/96 * 21/95 ... (4)


Therefore, P(drawing 3 odd and 3 prime by Case (II)) = (3) * (4) = 0.000533059



=============================
Case (III): 1 of those 3 odd number drawn are prime (1 OP + 2 OC)
=============================

[OP OC OC]: Prob = 24/100 * 26/99 * 25/98
[OC OP OC]: Prob = 26/100 * 24/99 * 25/98
[OC OC OP]: Prob = 26/100 * 25/99 * 24/98

These 3 prob are equal.
Therefore, P(drawing 1 OP + 2 OC) = 3 * (26*25*24) / (100*99*98) ... (5)

Since there are 24 prime number remaining after that...

P(drawing 3 prime number after drawing 1 OP + 2 OC) = 24/97 * 23/96 * 22/95 ... (6)


Therefore, P(drawing 3 odd and 3 prime by Case (III)) = (5) * (6) = 0.000662186



=============================
Case (IV): 0 of those 3 odd number drawn are prime (3 OC)
=============================

P(drawing 3 OC) = 26/100 * 25/99 * 24/98 ... (7)

Since there are 25 prime number remaining after that...

P(drawing 3 prime number after drawing 3 OC) = 25/97 * 24/96 * 23/95 ... (8)


Therefore, P(drawing 3 odd and 3 prime by Case (IV)) = (7) * (8) = 0.000250828


=============================
Req'd answer
= SUM of prob of 4 cases
= 0.000130739 + 0.000533059 + 0.000662186 + 0.000250828
= 0.001576812
=============================

1-100的數值含質數=23
1-100的單數值含質數=22

A1.先抽3個單數值不含質數
=(50-22)/100*(49-22)/99*(48-22)/98
A2.先抽3個單數值含1個質數
=22/100*(49-21)/99*(48-21)/98
=(50-22)/100*22/99*(48-21)/98
=(50-22)/100*(49-22)/99*22/98
A3.先抽3個單數值含2個質數
=22/100*21/99*(50-20)/98
=22/100*(50-21)/99*21/98
=(50-22)/100*22/99*21/98
A4.先抽3個單數值含3個質數
=22/100*21/99*20/98

B1=23/97 x 22/96 x 21/95 (因繼續抽剩下97個數值,而質數沒有減少)
B2= 22/97 x 21/96 x 20/95(因繼續抽剩下97個數值,而質數減少了1個)
B3=21/97 x 20/96 x 19/95(因繼續抽剩下97個數值,而質數減少了2個)
B4= 20/97 x 19/96 x 18/95(因繼續抽剩下97個數值,而質數減少了3個)

剩下數列中繼續抽中3個質數值的概率
=(B1+B2+B3+B4)-(A1+A2+A3+A4)
答案不知道錯對,但較合理

2007-12-03 14:03:19 補充：
剩下數列中繼續抽中3個質數值的概率=(B1+B2+B3+B4)-(A1+A2+A3+A4)或不需-(A1+A2+A3+A4)答案=B1+B2+B3+B4(因繼續抽剩下97個數值的數列)

單數值不含質數50/100 x 49/99 x 48/98 x 23/97 x 22/96 x 21/95
單數值1個含質數50/100 x 49/99 x 48/98 x 22/97 x 21/96 x 20/95
單數值2個含質數50/100 x 49/99 x 48/98 x 21/97 x 20/96 x 19/95
單數值3個含質數50/100 x 49/99 x 48/98 x 20/97 x 19/96 x 18/95
以上相加

2,3,5,7,11,13,17,23,29,31,37,41,43,47,53,61,67,71,73,83,89,91,97
total 23

請您想想喎, 不知道對不對, 剛巧我也在學緊.謝謝.

2007-12-07 04:52:40 補充：
是您對的,因我未有周詳想,對不起啊. 我想您心中有答案,只是找人對對?