Problem 11
Question
\(A\) has 3 shares in a lottery containing 3 prizes and 9 blanks. \(B\) has two shares in a lottery containing 2 prizes and 6 blanks; Find theratio of \(A\) 's chance of success to \(B\) 's chance of success. (a) \(927: 715\) (b) \(972: 751\) (c) \(925: 715\) (d) \(715: 972\) olution (c) Since \(A\) has 3 shares in a lottery, his chance of success means that he gets at least 1 prize, that is, he gets either 1 prize or 2 prizes or 3 prizes 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 now find \(q \times n=\) total number of ways $$ ={ }^{12} C_{3}=\frac{12 \times 11 \times 10}{1 \times 2 \times 3}=220 $$ 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 drawn by him are blanks. $$ \therefore q=\frac{m}{n}=\frac{84}{220}=\frac{21}{55} $$ \(\therefore p=A\) 's chance of success \(=1-\frac{21}{55}=\frac{34}{55}\)Similarly B's chance of success $$ \begin{aligned} p^{\prime} &=1-q^{\prime}=1-\frac{{ }^{6} C_{2}}{{ }^{8} C_{2}}=1-\frac{6 \times 5}{8 \times 7} \\ &=1-\frac{15}{28}=\frac{13}{28} \end{aligned} $$ \(\therefore A\) 's chance of success: B's chance of success $$ =p: p^{\prime}=\frac{34}{35}: \frac{13}{28}=\frac{952}{1540}: \frac{715}{1540}=952: 715 $$
Step-by-Step Solution
VerifiedKey Concepts
Combinatorics
In this specific lottery context, combinations are used to determine the number of ways tickets can be chosen. The notation \({^n}C_r\) represents the number of combinations where \( n \) is the total number of items to choose from, and \( r \) is the number of items to choose. For instance, to calculate \\( {^{12}C_3} \,\) in the problem, we find the number of ways 3 tickets can be drawn from the 12 available tickets. This is depicted by the formula:
- \( \binom{n}{r} = \frac{n!}{r!(n-r)!} \)
Ratio and Proportion
In this exercise, we are calculating the ratio of the chances of success for two lottery players, A and B. To find this ratio, we need to first calculate their individual chances of success using the concept of probability:
- The probability of an event happening is calculated as the ratio of the number of favorable outcomes to the total number of possible outcomes.
Lottery Problems
For example, consider player A who participates with multiple lottery shares. Their task is to calculate the probability of drawing at least one prize from an array of prize and blank tickets. This involves:
- Understanding the total number of possible outcomes (all combinations of tickets),
- Calculating the number of favorable outcomes (drawing a prize),
- Finding the ratio of favorable outcomes to total possible outcomes.