Atelevision game show has 6 doors, of which the contestant must pick 2. behind 2 of the doors ate expensive cars, and behind the other 4 doors are consolation prizes. the contestant get to keep the items behind the 2 doors she select. determine the probability that the contestant wins both cars.

In this question, there are 4 consolation door and 2 car doors of total 6 doors. That means for the first door the chance to get consolation is 4/6 and car is 2/6. In second door, only 5 remains so the chance to get consolation is either 4/5 (if 1st door car) or 3/5(if 1st door consolation)

Win 1 car condition:  

1. 1st door car, 2nd door consolation = 2/6 * 4/5 =8/30

2. 1st door consolation, 2nd door car= 4/6 * 2/5 = 8/30

8/30+8/30= 16/30

No car condition

1. 1st door consolation, 2nd door consolation= 4/6 * 3/5= 12/30

At least one car condition

1. 1st door car, 2nd door car = 2/6 * 1/5 =2/30

2. 1st door car, 2nd door consolation = 2/6 * 4/5 =8/30

3. 1st door consolation, 2nd door car= 4/6 * 2/5 = 8/30

2/30 + 8/30 + 8/30= 18/30

160

step-by-step explanation:

idk

i hope i

[2, 2.5)

(2.5, 3]

step-by-step explanation:

i think its these two not sure [2, 2.5)

(2.5, 3]