Question

If $\text{P(A)}=\frac{1}{4}$, $\text{P(B)}=\frac{7}{4}$, and $\text{P(A and B)}=\frac{5}{14}$, then what types of events are A and B?

• A. Independent
• B. Dependent

Answer 1

The correct option is B Dependent

Here,
$\text{P(A)}=\frac{1}{4}$
$\text{P(B)}=\frac{7}{4}$, and
$\text{P(A and B)}=\frac{5}{14}$

For events $\text{A}$ and $\text{B}$ to be independent,
$\text{P(A and B) = P(A)}\times \text{P(B)}$

Now, check for the given events:
$\text{P(A)}\times \text{P(B)}=\frac{1}{4}\times \frac{7}{4}=\frac{1\times 7}{4\times 4}=\frac{7}{16}$

But it is given $\text{P(A and B)}=\frac{5}{14}$

Hence, $\text{P(A and B)}\ne \text{P(A)}\times \text{P(B)}$

$\therefore$ $\text{A}$ and $\text{B}$ are not independent events; they are $\text{dependent events}$.

$\underset{–}{\text{Dependent Events}}$
Compound events in which the outcome of one depends upon the outcome of the other

$\underset{–}{\text{Note:}}$
For any dependent events $\text{E1}$ and $\text{E1}$,
$\text{P(E1 and E2)}=\text{P(E1)}\times \text{P(E2 knowing that E1 already occured)}$