Amachine produces photodetectors in pairs. tests show that the first photo detector is acceptable with probability 3/5. when the first photo detector is acceptable, the second photo detector is acceptable with probability 4/5. if the first photo detector is defective, the second photo detector is acceptable with probability 2/5.
(a) find the probability that exactly one photo detector of a pair is acceptable.

P(A or B) = 7/25

the probability that exactly one photo detector of a pair is acceptable is 7/25

Step-by-step explanation:

Given:

The probability that the first photo detector is acceptable P(A)= 3/5

The Probability that the first photo detector is defective P(A') = 1 - P(A) = 1 - 3/5 = 2/5

The probability that the second photo detector is acceptable given that the first one is defective P(B|A')= 2/5

The probability that the second photo detector is defective given that the first one is acceptable P(B'|A) = 1 - 4/5 = 1/5

The probability that exactly one photo detector of a pair is acceptable is given as:

P(A/B) = P(A)P(B'|A) + P(A')P(B|A')

P(A/B) = 3/5 × 1/5 + 2/5 × 2/5

P(A/B) = 7/25

there are lots of ways.   pick's theorem tells us if the vertices are lattice points the area a is

a = b/2 + i - 1

where b is the number of boundary lattice points and i is the number of interior lattice points.

in the figure i count b=16, i=50 so

a = 8 + 50 - 1 = 57

let's try the shoelace formula.   the area is half the sum of the cross products of the sides.  

a(-2,6)   -2(2)-6(-5) = 26

b(-5,2)   -5(-3) - 2(-2) = 19

c(-2,-3) -2(-3) - (-3)(4) = 18

d(4,-3)   4(3) - (-3)(5) = 27

e(5,3)   5(3) - 3(1) = 12

f(1,3)     1(6) - 3(-2) = 12

a(-2,6)

each of the cross products is twice the area of the triangle with the associated side and one vertex the origin. so the area of abo=26/2=13.

area = (1/2) (26 + 19 +   18 + 27 + 12 + 12) = 57

that checks!

no

step-by-step explanation:

answer