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Two points: First: try to find the probability for two heads occured and then to find the probability of being consecutive of any two heads occured. Then use the product rule to find the final answer: 9p^2*(1-p)^8

To turtle: This is a forum and everybody want to get help from others. I wish that we will be friends. Not like this! Thanx!

The solution: (a/b+b/c+c/d+d/a)/4>=4th sqrt[(a/b)(b/c)(c/d)(d/a)]=4 Same as: (a1+a2+...+an)/n>=nth sqrt(a1*a2*...*an)

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To Affinity: [1+(a+b)+ab][1-(a+b)]=1-(a+b)^2+ab-ab(a+b)=1-a^2-ab-b^2-a^2b-ab^2<1 Therefore: (a+1)(b+1)<1/[1-(a+b)] Could u offer more? Thank!

Thanx!

A biased coin, the probability of the head for a toss is p, where p not equal to 0.5 and 0<p<1, is tossed ten times. Q1) Find the probability for exact two heads Q2)Find the probability for exactly consecutive two heads only Do u want more, call 0413 197 399

Maths Ext2 I need more inequality question!:o

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