What If You Paint the Balls?
Date: 03/25/2002 at 05:31:17 From: John Midgley Subject: Probability - what if you *paint* the balls? Hi, You have 800 blue balls in a bag. You take out 35 of them, paint them red, and replace them. Do this 23 times and you *might* paint all of them - what are the odds ? As the number of repetitions increases, how can you model the improving odds of painting all the balls? Alternatively, is there an *intelligent* question I should be asking about this scenario? I'm stuck because: a) I should have been paying attention when they tried to teach me; b) The damned things change colour. Incidentally, if Dr Math had been around before those 240 months elapsed, I *might* have passed 'A' level maths (I'm in England).  About the same probability as in the sentence above. Regards, John Midgley
Date: 03/25/2002 at 08:31:50 From: Doctor Anthony Subject: Re: Probability - what if you *paint* the balls? We can use a difference equation to represent what is happening. Every painting operation covers 35/800 of the total. Let u(n) be the number that are red after n painting turns. 800-u(n) will be the number that are still blue. Then u(n+1) = u(n) + (35/800)[800-u(n)] u(n+1) = u(n)[1 - 35/800] + 35 u(n+1) = (153/160).u(n) + 35 u(n+1) - (153/160).u(n) = 35 So we must solve the difference equation to find u(n) in terms of n. Let u(n) = v(n) + w(n) where v(n) is the complementary function and w(n) the particular solution. The complementary function is x - 153/160 = 0 giving x = 153/160 so C.F. is v(n) = A.(153/160)^n For the particular solution we assume w(n) = a.n + b and substituting into the difference equation a.(n+1) + b - (153/160)[a.n + b] = 35 n(a - 153/160) + a + b - (153/160).b = 35 n(a - 153/160) + a + (7/160)b = 35 This is an identity, and so terms in n and the constant can be equated on both sides of the identity. Coefficient of n = 0 so a = 153/160 and equating the constant term on both sides gives 153/160 + (7/160)b = 35 (7/160)b = 5447/160 b = 5447/7 adding v(n) and w(n) we get u(n) = A.(153/160)^n + (153/160)n + 5447/7 and u(1) = 35 so 35 = (A+1).(153/160) + 5447/7 (153/160)(A+1) = -5202/7 A+1 = -5440/7 A = -5447/7 u(n) = (5447/7)[1 - (153/160)^n] + (153/160)n and if n = 23 this gives u(n) = 500.0418 + 21.99375 = 522.035 So after 23 paintings there will be 522 of the 800 balls painted red. - Doctor Anthony, The Math Forum http://mathforum.org/dr.math/
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