Question

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

• A. $\frac{3}{4}$
• B. $\frac{1}{2}$
• C. $\frac{1}{3}$
• D. $\frac{2}{3}$

Answer 1

Answer: (A)

$A=$ coin show head
$B=$ dice show $5$ or $6$
$P\left(A\right)=\frac{1}{2},P\left(B\right)=\frac{2}{6}=\frac{1}{3}$
Hence, the required Probability $=P\left(A\right)+P\left(A^C\right)P\left(B^C\right)P\left(A\right)+{\left(P\left(A^C\right)P\left(B^C\right)\right)}^{2}P\left(A\right)+\dots$
$=\frac{1}{2}+\frac{1}{2}\times \frac{2}{3}\times \frac{1}{2}+{\left(\frac{1}{2}\times \frac{2}{3}\right)}^2\times \frac{1}{2}+\dots$
$=\frac{\frac{1}{2}}{1-\frac{1}{3}}=\frac{\frac{1}{2}}{\frac{2}{3}}=\frac{3}{4}$