I am in a game show.In this game show there are 3 doors. Behind one of these doors lies a brand new and spectacular Ferrari, however, behind two of these doors lie nothing.

These doors are represented by the letters A, B and C respectively.

The host of this game show gives me the opportunity to win a Ferrari, and tells me that if I open the door which has this Ferrari sitting behind it, I will get to keep it. The problem is, I will only get to open one door. If it turns out to have nothing behind it, I will go home empty handed.

So, the game show host asks me which door I'd like to open. I select the door B.

Before the host opens the door B for me, he opens up the door C and shows me that nothing lies behind it.

This means that the Ferrari is either behind door A or door B.

Now, the host asks me if I'd like to open door A instead of door B. He gives me the chance to open door A. Should I open door A or should I open door B?

Would my chances of winning a Ferrari increase if I were to open door A instead of door B?

Explain your answer and justify why you are correct...

Answers

Answer 1
Answer: This is the well known Monty Hall problem.

You have more chances of winning by switching doors.

Let A,B,C denote the events "The ferrari is behind A" and so on
Let HA,HB,HC denote the events "The host opens A" and so on

By Baye's formula, P(B)_(HC)=(P(HC)_BP(B))/(P(HC))=\frac{\frac{1}2\cdot\frac{1}3}{\frac{1}2}=\frac{1}3 whereas P(A)_(HC)=(P(HC)_AP(A))/(P(HC))=\frac{1\cdot\frac{1}3}{\frac{1}2}=\boxed{\frac{2}3} !!



Answer 2
Answer: if you switch after the host shows you the door without the ferrari, your chance of getting the car is greater.

we have 3 hypothetical situations.

if the car is in door A, but and then you pick door A, the host will open either door B or C and you will pick the other remaining door, resulting in a loss.

if the car is in door B, and you pick door A, the host will show you door C, and if you switch to door B, you will win.

if the car is in door C, and you pick door A, the host will show you door B, and if you switch to door C, you will win.

There is a 2/3 chance of winning a ferrari if you switch after the host shows you the door without.

:D

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......................................

Brainliest if correct hurryyy!!

Answers

Answer:

8.14

Step-by-step explanation:

Ok.So to do this, you have to know the Pytharomean Theroem.If you do not know how,then that is totally ok.I will tell.So the theroem has a formula.

Down below is a image showing the variables.

Now the hypotenuse is c.And b and a are the legs.

So we can put evaluses in the formula(formula:a^2 + b^2 = c):

(6.0)^2 + (5.5)^2 = c^2

36 + 30.25=c^2

66.25=c^2

8.14 = c

So 8.14 is the anwser!

Hoped it helped!

Solve and graph the inequality. 6.7 > - 0.2x 4.5

Answers

-11 > x is the final answer

What is the value of x in the equation 2x+7=1

Answers

2x + 7 = 1

Subtract 7 on both sides

2x = 1 - 7

2x = -6

Divide both sides by 2

x = -6/2

x = -3

Your answer is \boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\bf{x=-3}}}}}}}}}}}}}}}}}}}

2x+7=1 =>2x=1-7 => 2x=-6 => x= -6\2 => x=-3
Have a nice day!! ^^

Solve the following system using the substitution method.9x + 8y = 15
-5x + 12y = -107

(x, y) = ( , )

Answers

x = (15 - 8y)/9

-5[(15 - 8y)/9] + 12y = -107

(-75/9) + (40/9) + 12y = -107

y = -8.59

x = [15 - 8(-8.59)]/9

x = 9.3
(x,y) = (9.3, -8.59)

What numbers are divisble by 23 but are between 2 and 10?

Answers

That is impossible.

For a number to be divisible by 23, it has to be greater than 23.

A number cannot be greater than 23, if you have restricted it to be within the numbers 2 and 10.

The inequality x2 + 12x + 35 ≥ 0 has two critical points and three possible intervals for solutions. Choose each set of possible test points for the three intervals.–8, –6, –4

–10, –6, 0

–6, 0, 6

–6, 0, 10

Answers

The answers are...

-8, -6, -4
-10, -6, 0

Answer:

–8, –6, –4

-10, –6, 0

Step-by-step explanation:

a & b are the correct answers