Question

$A$ has 3 tickets of a lottery containing 3 prizes and 9 blanks. $B$ has two tickets of another lottery containing 2 prizes and 6 blanks. The ratio of their chances of success is

• A. $\frac{32}{55}:\frac{15}{28}$
• B. $\frac{32}{55}:\frac{13}{28}$
• C. $\frac{34}{55}:\frac{13}{28}$
• D. $\frac{34}{55}:\frac{15}{28}$

Answer 1

Answer: (C)

$\frac{34}{55}:\frac{13}{28}$

Since $A$ has $3$ shares in a lottery,
his chance of success means that he gets at least $1$ prize or
$2$ prize or $3$ prize and his chance of failure means that he gets no prize.
It is certain that either he succeeds or fails.
If $p$ denotes his chance of success and $q$ the chance of his failure
then, $p+q=1$ or $p=1-q$
We know $q\times n=$ total number of ways
${=}^{12}{C}_1=\frac{12\times 11\times 10}{1\times 2\times 3}=220$
Since out of $12$ tickets in the lottery,
he can draw any $3$ tickets by virtue of his having $3$ shares in the lottery and
$m=$ favourable number of ways ${=}^9{C}_3=\frac{9\times 8\times 7}{1\times 2\times 3}=84$
Since he will fail to draw a prize if all the tickets draw by him are blanks.
$\therefore q=\frac{m}{n}=\frac{84}{220}=\frac{21}{55}$
$\therefore p=$ $A^{\prime }s$ chance of success $=1-\frac{21}{55}=\frac{34}{55}$
Similarly, $B^{\prime }s$ chance of success
$p^{\prime }=1-q^{\prime }=1-\frac{{}^6{C}_2}{{}^8{C}_2}=1-\frac{6\times 5}{8\times 7}$
$=1-\frac{15}{28}=\frac{13}{28}$
$\therefore$ $A^{\prime }s$ chance of success$=B^{\prime }s$ chance of success.
$=P:P^{\prime }=\frac{34}{35}:\frac{13}{28}$