Question

A man throws a die and tosses a coin alternately, starting with the coin. Find the probability that he gets head before $5$ or $6$ on the die.

• A. $\frac{3}{4}$
• B. $\frac{3}{7}$
• C. $\frac{1}{2}$
• D. None of these
  

Answer 1

The correct option is A $\frac{3}{4}$

$P$ (Head)$=P\left(H\right)=\frac{1}{2}=P$(Tail )$=$$P\left(T\right)$

$P\left(A\right)=$ $P$($5$ or $6$ on a die) $=P\left(5\right)+P\left(6\right)=\frac{1}{6}+\frac{1}{6}=\frac{1}{3}$
So $P\left(A^{\prime }\right)=$ $P$ (not getting $5$ or $6$ on die) $=$ $P^{\prime }$($5$ or $6$ on die )$=$ $\frac{2}{3}$ $H$ or ($T$ and ($5$ or $6$)') or ($T$ and ($5$ or $6$)') and ($T$ and ($5$ and $6$)'$H$) or ...
These events $H$, ($T$ and ($5$ or $6$ )') etc are mutually exclusive events.
By addition theorem.
$P$(success) $=$ $P(H$ or $\left(T\right(5or6{)}^{\prime })$ or $\left(T\right(5$ or $6{)}^{\prime }\left)T\right(5$ or $6{)}^{\prime })$ or...
$=$ $P\left(H\right)+p\left(T\right).p\left(A^{\prime }\right)P\left(H\right)+P\left(T\right)P\left(A^{\prime }\right)P\left(T\right)P\left(A^{\prime }\right)P\left(H\right)+....$
Since $T,(5$ or $6)$' are independent events, apply multiplication rule.
$=\frac{1}{2}+\frac{1}{2}.\frac{2}{3}.\frac{1}{2}+\frac{1}{2}.\frac{2}{3}.\frac{1}{2}.\frac{2}{3}.\frac{1}{2}$ $=\frac{1}{2}[1+\frac{1}{3}+{\left(\frac{1}{3}\right)}^2+{\left(\frac{1}{3}\right)}^3+...]$
$=\frac{1}{2}\left[\frac{1}{1-\frac{1}{3}}\right]=\frac{3}{4}$