The University of Oklahoma has a renowned radar research program within the School of Electrical and Computer Engineering. Say that there are two radar prototypes under development, A1 and A2, and the effectiveness of each is determined by their ability to spatially detect a particular object. A1 has a probability of 0.5 of detecting the object, and A2 has probability 0.3 of detecting the object. a. Find the probability that the object is detected. b. Find the probability that the object is detected by exactly one of the radar prototypes. c. Given that the object was detected by exactly one of the prototypes, find the probability that it was A1 that detected it.

a) = 0.65

b) = 0.5

c) = 0.7

Step-by-step explanation:

a)

First, using either A1 detected the object or A2 detected it or both did, the probability that the object is detected can be determined  by any of it

so

P( probability that object is detected) = (p1 * q2) + (p2 * q1) + (p1 * p2)

so we substitute

P( probability that object is detected = (0.5 * 0.7) + (0.3 * 0.5) + (0.5 * 0.3)

= 0.65

b)

probability that object is detected by exactly 1 of the radar prototypes.

P( object is detected by exactly one of the radar prototypes ) = (p1 * q2) + (p2*q1)

= (0.5*0.7) + (0.3*0.5)

= 0.5

c)

probability that A1 that detected it.

using the Bayes theorem,  

P(detected by A1 / detected by exactly one of the prototypes) =  (p1 * q2) / (p1 * q2 + p2*q1) = (0.5 * 0.7) / (0.5 * 0.7) + (0.3 * 0.5)

= 0.35 / 0.5

= 0.7

0.16 or 16%

step-by-step explanation:

lets take a coin for example.

if you toss it twice, it has a 1/4 chance of landing on heads both times because 1/2 * 1/2 is 1/4.

with the cards, the chance of picking dog is 0.4. having two people pick one is like picking a card twice in a row. so 0.4 * 0.4 is 0.16, or 16%

cheater

step-by-step explanation: