According to a survey, 52% of the residents of a city A oppose a downtown casino. Of these 52%, about 6 out of 10 strongly oppose the casino.
(a) Find the probability that a random selected who opposes the casino and strongly oppose the casino.
(b) Find the probability that a random selected resident who opposes the casino does not strongly oppose the casino.
(c) Would it be unusual for a random selected resident to oppose the casino and strongly oppose the casino?
Solution
Given that 52% of the residents of a city A oppose a downtown casino.
So
$$ \begin{aligned} P(\text{Who oppose Casino}) &= 52/100 \\ &= 0.52. \end{aligned} $$
The probability of the residents who strongly oppose casino is
$$ \begin{aligned}P(\text{ Who oppose and strongly Oppose Casino}) &= (52/100)*(6/10)\\ &= 0.312. \end{aligned} $$
(a) The probability that a random selected who opposes the casino and strongly oppose the casino is
$$ \begin{aligned} P(\text{ Who oppose and strongly Oppose Casino}) &= (52/100)*(6/10)\\ &= 0.312. \end{aligned} $$
(b) The probability that a random selected resident who opposes the casino does not oppose the casino is
$$ \begin{aligned} P(\text{Who oppose Casino}) - P(\text{ Who strongly Oppose Casino}) &=0.52-0.312\\ &= 0.2. \end{aligned} $$
(c) It would bot be unusual for a random selected resident to oppose the casino and strongly oppose the casino, because the probability 0.20 is not too small.
Further Reading
- Statistics
- Descriptive Statistics
- Probability Theory
- Probability Distribution
- Hypothesis Testing
- Confidence interval
- Sample size determination
- Non-parametric Tests
- Correlation Regression
- Statistics Calculators
