Question

A man alternatively tosses a coin and throws a dice beginning with the coin. The probability that he gets head in the coin before he gets a $5$ or $6$ in the dice is?

• A. $3/4$
• B. $1/2$
• C. $1/3$
• D. None

Answer 1

The correct option is D None

$x=\frac{x_1+x_2+x_3.....x_9}{9}x=\frac{x_1+x_2+x_3+.....+x_9}{9}....\left(1\right)Afteradding1morepesameanlet=ykgx+\frac{6x}{100}=\frac{x_1+x_2+x_3+.....+x_9+y_{1}kg}{10}....\left(2\right)\Rightarrow \frac{100x+6x}{100}=\frac{x_1+x_2+x_3+.....+x_9+y_1}{10}\Rightarrow 10.6x=\frac{x_1+x_2+x_3+.....+x_9+y_1}{1}\Rightarrow 10.x=9x+y_1\Rightarrow y_1=1.6x\Rightarrow x=\frac{y_1}{1.6}=\frac{5y_1}{8}\Rightarrow x=\frac{x_1+x_2+x_3+.....+x_9}{11}+\frac{y_1}{11}+\frac{y_2}{11}....\left(3\right)Fromequation\left(1\right)x=\frac{9x}{11}+\frac{y_1}{11}+\frac{y_2}{11}\Rightarrow \frac{5y_1}{8}=\frac{9\times 5y_1}{11\times 8}+\frac{y_1}{11}+\frac{y_2}{11}\Rightarrow \frac{5y_1}{8}-\frac{45y_1}{11\times 8}-\frac{y_1}{11}=\frac{y_2}{11}\Rightarrow \frac{55y_1-45y_1-8y_1}{11\times 8}=\frac{y_2}{11}\Rightarrow \frac{2y_1}{8}=\frac{y_2}{1}{y}_1=4y_2$