Question

checking for independence
you know that if two events, a and b, are independent events, then
p(a and b)=p(a)×p(b). the converse is also true. if
p(c and d)=p(c)×p(d), then events c and d are independent events.
it is sometimes difficult to know whether two events are independent just by
considering the situation. so, you can check to see if two events are
independent by multiplying their probabilities. if the product is equal to the
probability of the two events happening together, then the events are
independent.
consider the cards shown. are the events \choosing a number card\ and
\choosing a circle card\ independent events?
the two events, \choosing a letter card\ and \choosing a square card\ are
independent events.
a meteorologist says that there is an 80% chance of rain and a 50% chance of
damaging winds. she also says that there is a 20% chance of rain and
damaging winds. are the events
ain\ and \damaging winds\ independent
events?
p(rain)=
p(damaging winds)=
p(rain and damaging winds)=
the two events
ain\ and \damaging winds\ are not independent events

Explanation:

Step1: Determine the probabilities

The probability of rain \(P(\text{rain})\) is given as \(80\%=\frac{80}{100} = 0.8\).
The probability of damaging winds \(P(\text{damaging winds})\) is given as \(50\%=\frac{50}{100}=0.5\).
The probability of rain and damaging winds \(P(\text{rain and damaging winds})\) is given as \(20\%=\frac{20}{100} = 0.2\).

Step2: Calculate the product of the individual probabilities

Using the formula \(P(A)\times P(B)\) for independent - like situation (where \(A\) is rain and \(B\) is damaging winds), we have \(P(\text{rain})\times P(\text{damaging winds})=0.8\times0.5\).

$$0.8\times0.5=(8\times5)\times10^{- 2}=40\times10^{-2}=0.4$$

Step3: Compare the product and the joint - probability

We compare \(P(\text{rain})\times P(\text{damaging winds})\) and \(P(\text{rain and damaging winds})\).
Since \(P(\text{rain})\times P(\text{damaging winds}) = 0.4\) and \(P(\text{rain and damaging winds})=0.2\), and \(0.4
eq0.2\)

Answer:

\(P(\text{rain}) = 0.8\); \(P(\text{damaging winds})=0.5\); \(P(\text{rain and damaging winds}) = 0.2\)