Question

Suppose that the handedness of the last fifteen U.S. presidents is as follows: 40% were left-handed (L) 47% were Democrats (D) If a president is left-handed, there is a 13% chance that the president is a Democrat. Based on this information on the last fifteen U.S. presidents, is "being left-handed" independent of "being a Democrat"? Yes, since 0.40 * 0.47 is not equal to 0.13. Yes, since 0.47 is not equal to 0.13. No, since 0.47 is not equal to 0.13. No, since 0.40 is not equal to 0.13.

Answer 1

No, since 0.47 is not equal to 0.13.

Explanation:

We are given with the handedness of the last fifteen U.S. presidents as;

Probability of left-handed, P(L) = 0.40

Probability of Democrats, P(D) = 0.47

Probability that the president is a Democrat given a president is left- handed, P(D/L) = 0.13

Based on this information on the last fifteen U.S. presidents, "being left-handed" is not independent of "being a Democrat" because ;

For this both event L and D to be independent P(D/L) must be equal to P(D) as only then we can say that both events are independent as it is given that the probability of president being democratic is 0.47 and this probability changes to 0.13 when we know that president is left-handed then the president of being democratic changes from 0.47 to 0.13.

Answer 2

Correct option is: No, since 0.47 is not equal to 0.13.

Explanation:

Independent events are those events that are not affected by the occurrence of other events.

If events A and B are independent then, P (A ∩ B) = P (A) × P (B).

Then the conditional probability of B when A has already occurred is:

`P (B|A)=(P(A\cap B))/(P(A))=(P(A)* P(B))/(P(A))=P(B)`

It is provided that:

P (L) = 0.40

P (D) = 0.47

P (D|L) = 0.13.

If events D and L are independent then: P (D|L) = P (D).

But P (D|L) = 0.13 ≠ P (D) = 0.47.

Thus, the event of "being left-handed" is not independent of the event of being a Democrat.

The correct option is: No, since 0.47 is not equal to 0.13.