Question

In a fair at a game stall, slips with numbers 3,3,5,7,7,7,9,9,9, and 11 are placed in a lot. A person wins if the mean of the numbers is written on the slip drawn. If the draw is uniform and random, determine the probability of losing.

Answer 1

Hint: Find the mean of the numbers using the fact that mean is equal to the ratio of the sum of all observations to the number of all observations. Assume that E is the event of drawing out the mean of the numbers. Use the fact that the probability of an event E is given by $P\left( E \right)=\dfrac{n\left( E \right)}{n\left( S \right)}$ where n(E) is the number of cases favourable to E and n(S) is the total number of cases. Hence find the probability of drawing out the mean of the numbers.

Complete step-by-step solution:
We know that mean is equal to the ratio of the sum of all observations to the number of all observations.
Let $\mu $ be the mean of the numbers.
Hence, we have
$\mu =\dfrac{3+3+5+7+7+7+9+9+9+11}{10}=7$
Let E be the event that the number drawn is equal to the mean of the numbers (7). Hence, we have
$E=\left\{ 7,7,7 \right\}$.
Also, we have
$S=\left\{ 3,3,5,7,7,7,9,9,9,11 \right\}$
We know that if E is the event then the probability of the event is given by $P\left( E \right)=\dfrac{n\left( E \right)}{n\left( S \right)}$
Hence, we have
$P\left( E \right)=\dfrac{3}{10}=0.3$
Hence the probability of winning the game is 0.3
We know that $P\left( E \right)+P\left( E' \right)=1$
Hence, we have
$P\left( E' \right)=1-P\left( E \right)=1-0.3=0.7$
Hence the probability of losing the game is 0.7


Note: [1] Drawing out uniformly is necessary for the application of the formula $P\left( E \right)=\dfrac{n\left( E \right)}{n\left( S \right)}$. If the draw is not uniform, then there is a bias factor in the drawing. Such cases are solved using conditional probability
[2] The sum of probabilities of an event and its complement is 1. This formula is very useful when it is difficult to calculated P(E) but P(E’) can be easily calculated.