Suppose that the handedness of the last 15 U.S. presidents is as follows:

(i) 40% were left-handed (L)
(ii) 47% were democrats (D)
(iii) If a president is left-handed, there is a 13% chance that the president is a Democrat.

What is the probability that a randomly chosen U.S. president is left-handed and a democrat?

A: 0.40 * 0.47 = 0.1880
B: 0.40 * 0.13 = 0.0520
C: 0.47 * 0.13 = 0.0611
D: 0.40/0.47 = 0.8510
E: 0.40/0.13 = 3.0769
F: 0.47/0.13 = 3.6154

Respuesta :

Answer:

B: 0.40 * 0.13 = 0.0520

Step-by-step explanation:

Given that the handedness of the last 15 U.S. presidents is as follows:

(i) 40% were left-handed (L)

(ii) 47% were democrats (D)

(iii) If a president is left-handed, there is a 13% chance that the president is a Democrat

Let A = left handed

B = democrat

P(A) = 0.40 : P(B) =0.47: P(B/A) = 0.13

By definition of conditional probability we have

[tex]P(B/A) = 0.13 \\P(AB)/P(A) = 0.13\\P(AB) = 0.13 *0.40 \\=0.052[/tex]

Hence answer option is

B: 0.40 * 0.13 = 0.0520

The probability that a randomly chosen U.S. president is left-handed and a democrat is 0.052. Then the correct option is B.

What is probability?

Probability means possibility. It deals with the occurrence of a random event. The value of probability can only be from 0 to 1. Its basic meaning is something is likely to happen. It is the ratio of the favorable event to the total number of events.

Given

The last 15 U.S. presidents are as follows:

  1. 40% were left-handed (L)
  2. 47% were democrats (D)
  3. If a president is left-handed, there is a 13% chance that the president is a Democrat.

Let A = left-handed (L)

B = democrat (B)

P(A) = 0.40

P(B) = 0.43

P(B/A) = 0.13

By definition of the conditional probability we have

[tex]\rm P(B/A) = 0.13\\\\\dfrac{P(AB)}{P(A)} = 0.13\\\\P(AB)\ = 0.13*0.40\\\\P(AB) \ =0.052[/tex]

Thus, the correct option is B.

More about the probability link is given below.

https://brainly.com/question/795909

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