A-coin-that-comes-up-head-with-probability-p-and-tail-with-probability-1-p-independently-of-each-flip-is-flipped-five-times-The-probability-of-two-heads-and-three-tails-is-equal-to-1-7-of-the-proba

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Question Number 111276 by Aina Samuel Temidayo last updated on 03/Sep/20

$$\mathrm{A}\mathrm{coin}\mathrm{that}\mathrm{comes}\mathrm{up}\mathrm{head}\mathrm{with}$$$$\mathrm{probability}\mathrm{p}\mathrm{and}\mathrm{tail}\mathrm{with}\mathrm{probability}$$$$1-\mathrm{p}\mathrm{independently}\mathrm{of}\mathrm{each}\mathrm{flip}\mathrm{is}\mathrm{flipped}\mathrm{five}\mathrm{times}.$$$$\mathrm{The}\mathrm{probability}\mathrm{of}\mathrm{two}\mathrm{heads}\mathrm{and}\mathrm{three}\mathrm{tails}\mathrm{is}\mathrm{equal}\mathrm{to}\frac{1}{7}\mathrm{of}\mathrm{the}\mathrm{probability}\mathrm{of}\mathrm{three}$$$$\mathrm{heads}\mathrm{and}\mathrm{two}\mathrm{tails}.\mathrm{Let}\mathrm{p}=\frac{\mathrm{x}}{\mathrm{y}},\mathrm{where}$$$$\gcd (\mathrm{x},\mathrm{y})=1.\mathrm{Find}\mathrm{x}+\mathrm{y}.$$

Answered by john santu last updated on 03/Sep/20

$$p(2H,3T)=C_2^5{p}^2{(1-p)}^3$$$$p(3H,2T)=C_3^5{p}^3{(1-p)}^2$$$$\iff C_2^5{p}^2{(1-p)}^3=\frac{1}{7}{p}^3{(1-p)}^2$$$$\iff 10.p^2{(1-p)}^3=\frac{10}{7}{p}^3{(1-p)}^2$$$$\iff 1-p=\frac{1}{7}p;7-7p=p$$$$\Rightarrow p=\frac{7}{8}=\frac{x}{y},wheregcd(x,y)=1$$$$Hencex+y=15.$$

Commented by Aina Samuel Temidayo last updated on 03/Sep/20

$$\mathrm{Thanks}.$$

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