Spider and Fly (Posted on 2003-11-25)

	Submitted by Ravi Raja
Rating: 3.6250 (8 votes)
Solution:	(Hide)
Using the binomial distribution, the probability that n flies out of 5 have been eaten is combin(5,n)*(1/2)^5, where combin(5,2)=5!/(n!*(5-n)!). The probability that 0 flies were eaten is combin(5,0)*(1/2)^5=1/32. The probability that 1 fly was eaten is combin(5,1)*(1/2)^5=5/32. The probability that 2 flies were eaten is combin(5,2)*(1/2)^5=10/32. So the probability that the spider is still hungry is 1/32 + 5/32 + 10/32 = 16/32 = 1/2. The probability the spider is full is 1-1/2=1/2. Thus the probability of a successful attempt to pass is (1/2)*1 + (1/2)*0.5 = 0.75 .

	Subject	Author	Date
	Puzzle Solution	K Sengupta	2009-02-26 15:34:03
	Answer	K Sengupta	2009-02-15 15:36:15
	re(3): simpler computation	SilverKnight	2003-12-04 16:55:15
	re(2): simpler computation	Bob Genisot	2003-12-04 16:31:08
	re: Easier than it sounds...	Lee	2003-12-03 23:57:12
	Easier than it sounds...	Magick321	2003-12-03 22:53:38
	Solution?	Prab	2003-11-28 00:30:00
	spider and fly	Ed Lunn	2003-11-26 18:19:49
	none	Lee	2003-11-26 13:53:19
	Solution	Clinton Heath	2003-11-26 12:24:57
	re: Solution to a strange question	krdmt5_000	2003-11-25 16:23:48
	Solution to a strange question	krdmt5_000	2003-11-25 16:18:30
	re(3): simpler computation	Robin Gatter	2003-11-25 12:13:45
	re(2): simpler computation	Charlie	2003-11-25 10:36:38
	re: simpler computation	SilverKnight	2003-11-25 10:10:59
	simpler computation	Charlie	2003-11-25 09:58:41
	re: just a little correction in my text	SilverKnight	2003-11-25 09:46:00
	well, I was wrong - corrected solution	Saso	2003-11-25 09:45:18
	just a little correction in my text	Saso	2003-11-25 09:23:40
	solution	SilverKnight	2003-11-25 09:19:08
	solution	Saso	2003-11-25 09:18:32