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January 30, 2015

January 30, 2015

Posted by **John** on Tuesday, February 19, 2013 at 2:44am.

- Maths -
**n**, Wednesday, February 20, 2013 at 9:16am9

- Maths -
**hehe**, Friday, February 22, 2013 at 9:13pmfail.

- Maths -
**Gwen**, Sunday, February 24, 2013 at 5:23amFirst, we will find the probability of at least one grandchild making consecutive tweets by using the inclusion-exclusion principle, and then subtract the result from 1.

From the inclusion-exclusion principle,

P(at least one grandchild makes consecutive tweets)

= (5 choose 1)P(a specified grandchild makes consecutive tweets) - (5 choose 2)P(two specified grandchildren makes consecutive tweets) + (5 choose 3)P(three specified grandchildren makes consecutive tweets) - (5 choose 4)P(four specified grandchildren makes consecutive tweets) + (5 choose 5)P(all five grandchildren makes consecutive tweets).

For 1 <= k <= 5, we need to find the probability that k specified grandchildren (and possibly others) make consecutive tweets.

There are 10! permutations of all 10 tweets.

Think of arranging k "blocks" of two tweets each, and (10 - 2k) single tweets. There are (10 - k)! ways of arranging these (10 - k) items, and 2 ways of arranging the two tweets within each of the k blocks. So (10 - k)!(2^k) of the 10! permutations result in k specified grandchildren (and possibly others) making consecutive tweets.

So the probability that k specified grandchildren (and possibly others) make consecutive tweets is (2^k)/[10*9*...*(10 - k + 1)].

Therefore, we have

P(at least one grandchild makes consecutive tweets)

= 5(2/10) - 10(4/(10*9)) + 10(8/(10*9*8)) - 5(16/(10*9*8*7)) + 1(32/(10*9*8*7*6))

= 1 - 4/9 + 1/9 - 1/63 + 1/945.

So the probability of no grandchild making consecutive tweets is

1 - (1 - 4/9 + 1/9 - 1/63 + 1/945) = 4/9 - 1/9 + 1/63 - 1/945 = 47/135.

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