Question

Select Independent or Not independent for each description of events.

P(A)=0.3, P(B)=0.5, P(A|B)=0.15

P(A)=0.2, P(B)=0.4, P(A|B)=0.2

P(A)=0.6, P(B)=0.2, P(A|B)=0.3

Answer 1

Answer:

When two events X and Y are independent,

Then, P(X ∩ Y) = P(X) × P(Y)

We know that,

$P(\frac{X}{Y})=\frac{P(X\cap Y)}{P(Y)}\text{ and }P(\frac{Y}{X})=\frac{P(X\cap Y)}{P(X)}$

Thus, by substituting the value of P(X ∩ Y),

$P(\frac{X}{Y})=P(X)\text{ and }P(\frac{Y}{X})=P(Y)$

Which is the required condition for independent events.

In first option,

P(A)=0.3, P(B)=0.5, P(A|B)=0.15

$\implies P(\frac{A}{B})\neq P(A)$

⇒ A and B are not independent events

In second option,

P(A)=0.2, P(B)=0.4, P(A|B)=0.2

$P(\frac{A}{B})=P(A)$

⇒ A and B are independent events.

In third option,

P(A)=0.6, P(B)=0.2, P(A|B)=0.3

$\implies P(\frac{A}{B})\neq P(A)$

⇒ A and B are not independent events

Answer 2

First Part: Not independent

Second Part: Independent

Third Part: Not independent

Two events A and B are independent if:

$P(A|B)=P(A|B') \\Or,P(B|A)=P(B|A')$

Given,

$P(A)=0.3, P(B)=0.5, P(A|B)=0.15$

$P(A)=0.2, P(B)=0.4, P(A|B)=0.2$

$P(A)=0.6, P(B)=0.2, P(A|B)=0.3$

$P(A\cap B)=P(A|B)P(B)$

Then by substituting we get,

$P(A|B)=P(A)$ and $P(B|A)=P(B)$

Computation of First probability:

So, if the probability is independent it will follow the above conditions.

$P(A|B)=0.15\\P(A)=0.3\\0.3\neq 0.15\\ P(A|B)\neq P(A)$

So, it is not independent

Computation of second probability:

So, if the probability is independent it will follow the above conditions.$P(A|B)=0.2\\P(A)=0.2\\0.2=0.2\\ P(A|B)= P(A)$

So, it is independent

Computation of second probability:

So, if the probability is independent it will follow the above conditions.

$P(A|B)=0.3\\P(A)=0.6\\0.3\neq 0.6\\ P(A|B)\neq P(A)$

So, it is not independent.

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