Question

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?

Answer 1

Answer:

0.0520

Step-by-step explanation:

Given the data above of 15 past presidents, let (L) denote left handedness and (D) denote democrat

Therefore, 40% = L = 0.4

47% = D = 0.47

13% chance that a left handed president is a democrat = 0.13

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

1. 0.4×0.47 = 0.1880

2. 0.4×0.13 = 0.0520

3. 0.47×0.13 = 0.0611

4. 0.4/0.47 = 0.8510

5. 0.4/0.13 = 3.0769

6. 0.47/0.13 = 3.6154

But, we are trying to find P(L and D) = P(L)×P(D|L)

= 0.40 × 013

= 0.0520