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

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 probability of three heads and two tails. Let p=(x/y), where gcd(x,y) =1 . Find x+y.

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

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 ⇔ C_2 ^5 p^2 (1−p)^3 = (1/7)p^3 (1−p)^2 ⇔ 10.p^2 (1−p)^3 = ((10)/7)p^3 (1−p)^2 ⇔ 1−p = (1/7)p ; 7−7p = p ⇒ p = (7/8) = (x/y) , where gcd(x,y)=1 Hence x+y = 15.

$${p}\left(\mathrm{2}{H},\mathrm{3}{T}\right)\:={C}_{\mathrm{2}} ^{\mathrm{5}} \:{p}^{\mathrm{2}} \left(\mathrm{1}−{p}\right)^{\mathrm{3}} \\ $$$${p}\left(\mathrm{3}{H},\mathrm{2}{T}\right)\:=\:{C}_{\mathrm{3}} ^{\mathrm{5}} \:{p}^{\mathrm{3}} \left(\mathrm{1}−{p}\right)^{\mathrm{2}} \\ $$$$\Leftrightarrow\:{C}_{\mathrm{2}} ^{\mathrm{5}} \:{p}^{\mathrm{2}} \left(\mathrm{1}−{p}\right)^{\mathrm{3}} \:=\:\frac{\mathrm{1}}{\mathrm{7}}{p}^{\mathrm{3}} \left(\mathrm{1}−{p}\right)^{\mathrm{2}} \\ $$$$\Leftrightarrow\:\mathrm{10}.{p}^{\mathrm{2}} \left(\mathrm{1}−{p}\right)^{\mathrm{3}} \:=\:\frac{\mathrm{10}}{\mathrm{7}}{p}^{\mathrm{3}} \:\left(\mathrm{1}−{p}\right)^{\mathrm{2}} \\ $$$$\Leftrightarrow\:\mathrm{1}−{p}\:=\:\frac{\mathrm{1}}{\mathrm{7}}{p}\:;\:\mathrm{7}−\mathrm{7}{p}\:=\:{p} \\ $$$$\Rightarrow\:{p}\:=\:\frac{\mathrm{7}}{\mathrm{8}}\:=\:\frac{{x}}{{y}}\:,\:{where}\:{gcd}\left({x},{y}\right)=\mathrm{1} \\ $$$${Hence}\:{x}+{y}\:=\:\mathrm{15}. \\ $$

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

Thanks.

$$\mathrm{Thanks}. \\ $$

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