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

3. parallelogram

step-by-step explanation:

we turn each of our sides into direction vectors

ab=b-a=(3 - 7, 1 - 0) = (-4, 1)

bc=c-b=(1,6)

cd=d-c=(4,-1)

da=a-d=(-1,-6)

we see how the opposite sides have negated direction vectors; this is characteristic of a parallelogram.   we have congruent sides ab=cd and bc=da with unequal lengths because 1^2+6^2 is different from (-4)^2+1^2.   so it's not a square or a rhombus.  

we check the dot product of direction vectors of adjacent sides.   we have

ab dot bc = -4(1) + 1(6) = 2

that's not zero so the sides are not perpendicular.   we've ruled out rectangle.

the direction vectors indicate two pairs of congruent parallel sides, so we choose parallelogram over trapezoid.

idk lol

step-by-step explanation: