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

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

Hence answer option is

B: 0.40 * 0.13 = 0.0520

  the correct choice is marked

step-by-step explanation:

the first two graphs show multiple turning points. the last graphs matches the description in every detail.