Consider the following 7 door version of the monty hall problem. there are 7 doors, behind one of which there is a car (which you want), and behind the rest of which there are goats (which you don? t want). initially, all possibilities are equally likely for where the car is. you choose a door. monty hall then opens 3 goat doors, and offers you the option of switching to any of the remaining 3 doors. assume that monty hall knows which door has the car, will always open 3 goat doors and offer the option of switching, and that monty chooses with equal probabilities from all his choices of which goat doors to open. should you switch? what is your probability of success if you switch to one of the remaining 3 doors?

2/7

Step-by-step explanation:

Like it says it is the same as monty hall with 3 doors.

You have 7 doors, the probability of the car being in each door is 1/7. After you pick one the probability of the car being in the other doors is 6/7, then he opens 3 doors showing only goats but the probability of it being in one of all the remaining doors is still 6/7 because you made a 1/7 probability choice when you picked your door. Meaning the remaining 3 doors have a probability of 6/7, then 2/7 each

honestly bro aint no one got time to answer these dumb question just do it p***y

step-by-step explanation:

just do 5.2 + 2 ( 7.2) and 1/3 x 3 (.9) and youv'e got your equation. 0.9( 9/10) + 7.2 ^2 = 16.4 hope i could !

step-by-step explanation: