LIS 4273 Module #4 Assignment

Hello,

Based on this table, what is the probability of:

	B	B1
A	10	20
A1	20	40

A1: Event A

Total occurrences for A = 10 (B) + 20 (B1) = 30

Total occurrences (all events) = 10 + 20 + 20 + 40 = 90

P(A)= 30/90 ​= 1/3 = 0.3333 = 33%

$A2:\; Event\; B$

$Total\; occurrences\; for\; B\; =\; 10\; (A)\; +\; 20\; (A1)\; =\; 30$

$P(B)=\; 30/90=0.3333\; =\; 33\%$

$A3:\; Event\; A\; or\; B$

$A\; happens\; in\; the\; first\; row:\; 10\; (B)\; +\; 20\; (B1)\; =\; 30$

$B\; happens\; in\; the\; first\; column:\; 10\; (A)\; +\; 20\; (A1)\; =\; 30$

$Subtract\; the\; overlap\; of\; A\; and\; B:\; 10$

P(A∪B) = 50/90 ​= 0.5556 = 56%

$A4:$P(A∪B) = P(A) + P(B) − P(A∩B) (changed this equation to make it work)

P(A)= 0.3333

P(B)= 0.3333

P(A∩B) = 10/90 ​= 0.1111

$P(A\cup B)=0.3333+0\; .3333-0.1111=0.5555\; =\; 56\%$

$B1:$Is this answer True or False?

$True$

$B2:$Please explain why?

The weatherman said it will rain, but the real chance of rain is still only 11%. This is because rain falls in this area very infrequently—only five days a year—so even on rare occasions when it does rain, the overall likelihood of rain is still low based on past frequency. 

C: Last assignment from our textbook, pp. 55 Exercise # 2.3.

> # Calculate the probability of 0 complications in 10 patients

> result <- dbinom(0, size = 10, prob = 0.2)
> print(result)
[1] 0.1073742