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